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Balbharti Maharashtra State Board 12th Physics Textbook Solutions Chapter 4 Thermodynamics Textbook Exercise Questions and Answers.

Maharashtra State Board 12th Physics Solutions Chapter 4 Thermodynamics

1. Choose the correct option.

i) A gas in a closed container is heated with 10J of energy, causing the lid of the container to rise 2m with 3N of force. What is the total change in the internal energy of the system?
(A) 10J
(B) 4J
(C) -10J
(D) – 4J
Answer:
(B) 4J

ii) Which of the following is an example of the first law of thermodynamics?
(A) The specific heat of an object explains how easily it changes temperatures.
(B) While melting, an ice cube remains at the same temperature.
(C) When a refrigerator is unplugged, everything inside of it returns to room temperature after some time.
(D) After falling down the hill, a ball’s kinetic energy plus heat energy equals the initial potential energy.
Answer:
(B) While melting, an ice cube remains at the same temperature. [Here, ∆u = 0, W = Q]
(C) When a refrigerator is unplugged, everything inside of it returns to room temperature after some time.
(D) After falling down the hill, a ball’s kinetic energy plus heat energy equals the initial potential energy.

iii) Efficiency of a Carnot engine is large when
(A) T_H is large
(B) T_C is low
(C) T_H – T_C is large
(D) T_H – T_C is small
Answer:
(A) T_H is large
(B) T_C is low
(C) T_H – T_C is large
[η = \(\frac{T_{\mathrm{H}}-T_{\mathrm{c}}}{T_{\mathrm{H}}}\) = 1 – \(\frac{T_{\mathrm{c}}}{T_{\mathrm{H}}}\)]

iv) The second law of thermodynamics deals with transfer of:
(A) work done
(B) energy
(C) momentum
(D) mass
Answer:
(B) energy

v) During refrigeration cycle, heat is rejected by the refrigerant in the :
(A) condenser
(B) cold chamber
(C) evaporator
(D) hot chamber
Answer:
closed tube[See the textbook]

2. Answer in brief.

i) A gas contained in a cylinder surrounded by a thick layer of insulating material is quickly compressed.
(a) Has there been a transfer of heat?
(b) Has work been done?
Answer:
(a) There is no transfer of heat.
(b) The work is done on the gas.

ii) Give an example of some familiar process in which no heat is added to or removed form a system, but the temperature of the system changes.
Answer:
Hot water in a container cools after sometime. Its temperature goes on decreasing with time and after sometime it attains room temperature.
[Note : Here, we do not provide heat to the water or remove heat from the water. The water cools on exchange of heat with the surroundings. Recall the portion covered in chapter 3.]

iii) Give an example of some familiar process in which heat is added to an object, without changing its temperature.
Answer:

1. Melting of ice
2. Boiling of water.

iv) What sets the limits on efficiency of a heat engine?
Answer:
The temperature of the cold reservoir sets the limit on the efficiency of a heat engine.
[Notes : (1) η = 1 – \(\frac{T_{\mathrm{C}}}{T_{\mathrm{H}}}\)
This formula shows that for maximum efficiency, T_C should be as low as possible and T_H should be as high as possible.
(2) For a Carnot engine, efficiency
η = 1 – \(\frac{T_{\mathrm{C}}}{T_{\mathrm{H}}}\).η → 1 T_C → 0.]

v) Why should a Carnot cycle have two isothermal two adiabatic processes?
Answer:
With two isothermal and two adiabatic processes, all reversible, the efficiency of the Carnot engine depends only on the temperatures of the hot and cold reservoirs.
[Note : This is not so in the Otto cycle and Diesel cycle.]

3. Answer the following questions.

i) A mixture of hydrogen and oxygen is enclosed in a rigid insulting cylinder. It is ignited by a spark. The temperature and the pressure both increase considerably. Assume that the energy supplied by the spark is negligible, what conclusions may be drawn by application of the first law of thermodynamics?
Answer:
The internal energy of a system is the sum of potential energy and kinetic energy of all the constituents of the system. In the example stated above, conversion of potential energy into kinetic energy is responsible for a considerable rise in pressure and temperature of the mixture of hydrogen and oxygen ignited by the spark.

ii) A resistor held in running water carries electric current. Treat the resistor as the system
(a) Does heat flow into the resistor?
(b) Is there a flow of heat into the water?
(c) Is any work done?
(d) Assuming the state of resistance to remain unchanged, apply the first law of thermodynamics to this process.
Answer:
(a) Heat is generated into the resistor due to the passage of electric current. In the usual notation, heat generated = I2Rt.
(b) Yes. Water receives heat from the resistor.

Here, I = current through the resistor, R = resistance of the resistor, t = time for which the current is passed through the resistor, M = mass of the water, S = specific heat of water, T = rise in the temperature of water, P = pressure against which the work is done by the water, ∆u= increase in the volume of the water.

iii) A mixture of fuel and oxygen is burned in a constant-volume chamber surrounded by a water bath. It was noticed that the temperature of water is increased during the process. Treating the mixture of fuel and oxygen as the system,
(a) Has heat been transferred ?
(b) Has work been done?
(c) What is the sign of ∆u ?
Answer:
(a) Heat has been transferred from the chamber to the water bath.
(b) No work is done by the system (the mixture of fuel and oxygen) as there is no change in its volume.
(c) There is increase in the temperature of water. Therefore, ∆u is positive for water.
For the system (the mixture of fuel and oxygen), ∆u is negative.

iv) Draw a p-V diagram and explain the concept of positive and negative work. Give one example each.
Answer:
Consider some quantity of an ideal gas enclosed in a cylinder fitted with a movable, massless and frictionless piston.

Suppose the gas is allowed to expand by moving the piston outward extremely slowly. There is decrease in pressure of the gas as the volume of the gas increases. Below figure shows the corresponding P-V diagram.

In this case, the work done by the gas on its surroundings,
W = \(\int_{V_{\mathrm{i}}}^{V_{\mathrm{f}}} P d V\) (= area under the curve) is positive as the volume of the gas has increased from V_i to V_f.
Let us now suppose that starting from the same

initial condition, the piston is moved inward extremely slowly so that the gas is compressed. There is increase in pressure of the gas as the volume of the gas decreases. Figure shows the corresponding P-V diagram.
In this case, the work done by the gas on its surroundings, W = \(\int_{V_{\mathrm{i}}}^{V_{\mathrm{f}}} P d V\) (= area under the curve) is negative as the volume of the gas has decreased from V_i to V_f.

v) A solar cooker and a pressure cooker both are used to cook food. Treating them as thermodynamic systems, discuss the similarities and differences between them.
Answer:
Similarities :

1. Heat is added to the system.
2. There is increase in the internal energy of the system.
3. Work is done by the system on its environment.

Differences : In a solar cooker, heat is supplied in the form of solar radiation. The rate of supply of heat is relatively low.

In a pressure cooker, usually LPG is used (burned) to provide heat. The rate of supply of heat v is relatively high.

As a result, it takes very long time for cooking when a solar cooker is used. With a pressure cooker, it does not take very long time for cooking.
[Note : A solar cooker can be used only when enough solar radiation is available.]

Question 4.
A gas contained in a cylinder fitted with a frictionless piston expands against a constant external pressure of 1 atm from a volume of 5 litres to a volume of 10 litres. In doing so it absorbs 400 J of thermal energy from its surroundings. Determine the change in internal energy of system. [Ans: -106.5 J]
Answer:
Data : P = 1 atm = 1.013 × 10⁵ Pa, V₁ = 5 litres = 5 × 10^-3 m³ V₂ = 10 litres = 10 × 10^-3 m³, Q = 400J.
The work done by the system (gas in this case) on its surroundings,
W = P(V₂ – V₁)
= (1.013 × 10⁵ Pa) (10 × 10^-3 m³ – 5 × 10^-3 m³)
= 1.013 (5 × 10²)J = 5.065 × 10²J
The change in the internal energy of the system, ∆u = Q – W = 400J – 506.5J = -106.5J
The minus sign shows that there is a decrease in the internal energy of the system.

Question 5.
A system releases 130 kJ of heat while 109 kJ of work is done on the system. Calculate the change in internal energy.
[Ans: ∆U = 21 kJ]
Answer:
Data : Q = -130kj, W= – 109kJ
∆u = Q – W = – 130kJ – ( – 109kJ)
= (-130 + 104) kJ = – 26 kj.
This is the change (decrease) in the internal energy.

Question 6.
Efficiency of a Carnot cycle is 75%. If temperature of the hot reservoir is 727ºC, calculate the temperature of the cold reservoir. [Ans: 23ºC]
Data : η = 75% = 0.75, T_H = (273 + 727) K = 1000 K
η = 1 – \(\frac{T_{\mathrm{C}}}{T_{\mathrm{H}}}\) ∴ \(\frac{T_{\mathrm{C}}}{T_{\mathrm{H}}}\) = 1 – η
∴ T_C = T_C(1 – η) = 1000 K (1 – 0.75)
= 250K = (250 – 273)°C
= -23 °C
This is the temperature of the cold reservoir.

Question 7.
A Carnot refrigerator operates between 250K and 300K. Calculate its coefficient of performance. [Ans: 5]
Answer:
Data : T_C = 250 K, T_H = 300 K
K = \(\frac{T_{\mathrm{C}}}{T_{\mathrm{H}}-T_{\mathrm{C}}}\) = \(\frac{250 \mathrm{~K}}{300 \mathrm{~K}-250 \mathrm{~K}}\) = \(\frac{250}{50}\) = 5
This is the coefficient of performance of the refrigerator.

Question 8.
An ideal gas is taken through an isothermal process. If it does 2000 J of work on its environment, how much heat is added to it? [Ans: 2000J]
Answer:
Data : W = 2000 J, isothermal process
In this case, the change in the internal energy of the gas, ∆u, is zero as the gas is taken through an isothermal process.
Hence, the heat added to it,
Q = ∆ u + W = 0 + W = 200J

Question 9.
An ideal monatomic gas is adiabatically compressed so that its final temperature is twice its initial temperature. What is the ratio of the final pressure to its initial pressure? [Ans: 5.656]
Answer:
Data : T_f = 2T_i, monatomic gas ∴ γ = 5/3


This is the ratio of the final pressure (P_f) to the initial pressure (P_i).

Question 10.
A hypothetical thermodynamic cycle is shown in the figure. Calculate the work done in 25 cycles.

[Ans: 7.855 × 10⁴ J]
Answer:


25 cycles
The work done in one cycle, \(\oint\)PdV
= πab = (3.142) (2 × 10^-3 m³) (5 × 10⁵ Pa)
= 3.142 × 10³J
Hence, the work done in 25 cycles
= (25) (3.142 × 10³ J) = 7.855 × 10⁴J

Question 11.
The figure shows the V-T diagram for one cycle of a hypothetical heat engine which uses the ideal gas. Draw the p-V diagram and P-V diagram of the system. [Ans: (a)]

[Ans: (b)]
Answer:
(a) P-V diagram (Schematic)

ab: isobaric process,
bc : isothermal process,
cd : isobaric process,
da : isothermal process
\(\frac{P_{\mathrm{a}} V_{\mathrm{a}}}{T_{\mathrm{a}}}\) = \(\frac{P_{\mathrm{b}} V_{\mathrm{b}}}{T_{\mathrm{b}}}\) = \(\frac{P_{\mathrm{c}} V_{\mathrm{c}}}{T_{\mathrm{c}}}\) = \(\frac{P_{\mathrm{d}} V_{\mathrm{d}}}{T_{\mathrm{d}}}\) = nR

(b) P—T diagram (Schematic)

Question 12.
A system is taken to its final state from initial state in hypothetical paths as shown figure. Calculate the work done in each case.

[Ans: AB = 2.4 × 10⁶ J, CD = -8 × 10⁵ J, BC and DA zero, because constant volume change]
Answer:

Data: P_A = P_B = 6 × 10⁵ Pa, P_C = P_D = 2 × 10⁵ Pa V_A = V_D = 2 L, V_B = V_C6 L, 1 L = 10^-3m³
(i) The work done along the path A → B (isobaric process), W_AB = P_A (V_B – V_A) = (6 × 10⁵ Pa)(6 – 2)(10^-3 m³) = 2.4 × 10³ J
(ii) W_BC = zero as the process is isochoric (V = constant).
(iii) The work done along the path C → D (isobaric process), W_CD = P_C (V_D – V_C)
= (2 × 10⁵ Pa) (2 – 6) (10^-3m³) = -8 × 10²J
(iv) W_DA = zero as V = constant.

12th Physics Digest Chapter 4 Thermodynamics Intext Questions and Answers

Can you tell? (Textbook Page No. 76)

Question 1.
Why is it that different objects kept on a table at room temperature do not exchange heat with the table ?
Answer:
The objects do exchange heat with the table but there is no net transfer of energy (heat) as the objects and the table are at the same temperature.

Can you tell? (Textbook Page No. 77)

Question 1.
Why is it necessary to make a physical contact between a thermocouple and the object for measuring its temperature ?
Answer:
For heat transfer to develop thermoemf.

Can you tell? (Textbook Page No. 81)

Question 1.
Can you explain the thermodynamics involved in cooking food using a pressure cooker ?
Answer:
Basically, heat is supplied by the burning fuel causing increase in the internal energy of the food (system), and the system does some work on its surroundings. In the absence of any data about the components of food and their thermal and chemical properties, we cannot evaluate changes in internal energy and work done.

Use your brain power (Textbook Page No. 85)

Question 1.
Verify that the area under the P-V curve has dimensions of work.
Answer:
Area under the P-V curve is \(\int_{V_{\mathrm{i}}}^{V_{\mathrm{f}}} P d V\), where P is the pressure and V is the volume.

Use your brain power (Textbook Page No. 91)

Question 1.
Show that the isothermal work may also be expressed as W = nRT ln\(\left(\frac{\boldsymbol{P}_{\mathrm{i}}}{\boldsymbol{P}_{\mathrm{f}}}\right)\),
Answer:
In the usual notation,
W = nRT In \(\frac{V_{\mathrm{f}}}{V_{\mathrm{i}}}\) and P_i V_i = P_f V_f = nRT in an isothermal process
∴ \(\frac{V_{\mathrm{f}}}{V_{\mathrm{i}}}\) = \(\frac{P_{\mathrm{i}}}{P_{\mathrm{f}}}\) and W = nRT ln \(\frac{P_{\mathrm{i}}}{P_{\mathrm{f}}}\)

Use your brain power (Textbook Page No. 94)

Question 1.
Why is the P-V curve for an adiabatic process steeper than that for an isothermal process ?
Answer:
Adiabatic process : PV^γ = constant
∴ V^γdP + γPV^γ-1 dV = 0
∴ \(\frac{d P}{d V}\) = – \(\frac{\gamma P}{V}\)
Isothermal process : PV = constant
∴ pdV + VdP = 0 ∴ \(\frac{d P}{d V}\) = \(\frac{P}{-V}\)
Now, γ > 1

∴ \(\frac{d P}{d V}\) is the slope of the P – V curve.
∴The P – V curve for an adiabatic process is steeper than that for an isothermal process.

Question 2.
Explain formation of clouds at high altitude.
Answer:
As the temperature of the earth increases due to absorption of solar radiation, water from rivers, lakes, oceans, etc. evaporates and rises to high altitude. Water vapour forms clouds as water molecules come together under appropriate conditions. Clouds are condensed water vapour and are of various type, names as cumulus clouds, nimbostratus clouds, stratus clouds and high-flying cirrus clouds.

Can you tell? (Textbook Page No. 95)

Question 1.
How would you interpret Eq. 4.21 (Q = W) for a cyclic process ?
Answer:
It means ∆ u = 0 for a cyclic process as the system returns to its initial state.

Question 2.
An engine works at 5000 rpm, and it performs 1000 J of work in one cycle. If the engine runs for 10 min, how much total work is done by the engine ?
Answer:
The total work done by the engine = (1000 J/cycle) (5000 cycles/min) (10 min) = 5 × 10⁷ J.

Do you know? (Textbook Page No. 101)

Question 1.
Capacity of an air conditioner is expressed in tonne. Do you know why?
Answer:
Before refrigerator and AC were invented, cooling was done by using blocks of ice. When cooling machines were invented, their capacity was expressed in terms of the equivalent amount of ice melted in a day (24 hours). The same term is used even today.
(Note : 1 tonne = 1000 kg = 2204.6 pounds, 1 ton (British) = 2240 pounds = 1016.046909 kg, 1 ton (US) = 2000 pounds = 907.184 kg.]

Use your brain power (Textbook Page No. 105)

Question 1.
Suggest a practical way to increase the efficiency of a heat engine.
Answer:
The efficiency of a heat engine can be increased by choosing the hot reservoir at very high temperature and cold reservoir at very low temperature.

Balbharti Maharashtra State Board 12th Physics Textbook Solutions Chapter 14 Dual Nature of Radiation and Matter Textbook Exercise Questions and Answers.

Maharashtra State Board 12th Physics Solutions Chapter 14 Dual Nature of Radiation and Matter

1. Choose the correct answer.

i) A photocell is used to automatically switch on the street lights in the evening when the sunlight is low in intensity. Thus it has to work with visible light. The material of the cathode of the photocell is
(A) zinc
(B) aluminum
(C) nickel
(D) potassium
Answer:
(D) potassium

ii) Polychromatic (containing many different frequencies) radiation is used in an experiment on the photoelectric effect. The stopping potential
(A) will depend on the average wavelength
(B) will depend on the longest wavelength
(C) will depend on the shortest wavelength
(D) does not depend on the wavelength
Answer:
(C) will depend on the shortest wavelength

iii) An electron, a proton, an α-particle and a hydrogen atom are moving with the same kinetic energy. The associated de Broglie wavelength will be longest for
(A) electron
(B) proton
(C) α-particle
(D) hydrogen atom
Answer:
(A) electron

iv) If N_Red and N_Blue are the number of photons emitted by the respective sources of equal power and equal dimensions in unit time, then
(A) N_Red < N_Blue
(B) N_Red = N_Blue
(C) N_Red > N_Blue
(D) N_Red ≈ N_Blue
Answer:
(C) N_Red > N_Blue

v) The equation E = pc is valid
(A) for all sub-atomic particles
(B) is valid for an electron but not for a photon
(C) is valid for a photon but not for an electron
(D) is valid for both an electron and a photon
Answer:
(C) is valid for a photon but not for an electron

2. Answer in brief.

i) What is photoelectric effect?
Answer:
The phenomenon of emission of electrons from a metal surface when electromagnetic radiation of appropriate frequency is incident on it is known as photoelectric effect.

ii) Can microwaves be used in the experiment on photoelectric effect?
Answer:
No

iii) Is it always possible to see photoelectric effect with red light?
Answer:
No

iv) Using the values of work function given in Table 14.1, tell which metal will require the highest frequency of incident radiation to generate photocurrent.

Answer:
Gold.
[ Note : W₀ = hv₀, where h is Planck’s constant. The larger the work function (W₀), the higher is the threshold frequency (v₀). ]

v) What do you understand by the term wave-particle duality? Where does it apply?
Answer:
Depending upon experimental conditions or structure of matter, electromagnetic radiation and material particles exhibit wave nature or particle nature. This is known as wave-particle duality.

It applies to all phenomena. The wave nature and particle nature are liked by the de Broglie relation λ = h/p, where λ is the wavelength of matter waves, also called de Broglie waves / Schrodinger waves, p is the magnitude of the momentum of a particle or quantum of radiation and h is the universal constant called Planck’s constant.

[Note : It is the smallness of h (= 6.63 × 10^-34 J∙s) that is very significant in wave-particle duality.]

Question 3.
Explain the inverse linear dependence of stopping potential on the incident wavelength in a photoelectric effect experiment.
Answer:
We have V₀e = \(\frac{h c}{\lambda}\) – Φ, where V₀ is the stopping potential, e is the magnitude of the charge on the electron, h is Planck’s constant, c is the speed of light in free space, λ is the wavelength of the electromagnetic radiation incident on a metal surface and Φ is the work function for the metal, h, c and e are constants. Φ is constant for a particular metal.
Hence, it follows that as \(\frac{1}{\lambda}\) increases, V₀ increases.
The plot of V₀ verses \(\frac{1}{\lambda}\) is linear. This is because the energy associated with a quantum of radiation (photon) is directly proportional to the frequency of radiation and hence inversely proportional to the wavelength of radiation.

Question 4.
It is observed in an experiment on photoelectric effect that an increase in the intensity of the incident radiation does not change the maximum kinetic energy of the electrons. Where does the extra energy of the incident radiation go? Is it lost? State your answer with explanatory reasoning.
Answer:
When electromagnetic radiation with frequency greater than the threshold frequency is incident on a metal surface, there is emission of electrons. It is observed that not every incident photon is effective in liberating an electron. In fact, the number of electrons emitted per second is far less than the number of photons incident per second. The photons that are not effective in liberation of electrons are reflected (or scattered) or absorbed resulting in rise in the temperature of the metal surface. The maximum kinetic energy of a photoelectron depends on the frequency of the incident radiation and the threshold frequency for the metal. It has nothing to do with the intensity of the incident radiation. The increase in intensity results in increase in the number of electrons emitted per second.

Question 5.
Explain what do you understand by the de Broglie wavelength of an electron. Will an electron at rest have an associated de Broglie wavelength? Justify your answer.
Answer:
Under certain conditions an electron exhibits wave nature. Waves associated with a moving electron are called matter waves or de Broglie waves or- Schrodinger waves. The de Broglie wavelength of these matter waves is given by X = h/p, where h is Planck’s constant and p is the magnitude of the momentum of the electron.

If an electron is at rest, its momentum would be zero, and hence the corresponding de Broglie wavelength would be infinite indicating absence of a matter wave. However, according to quantum mechanics/wave mechanics, this is not possible.

Question 6.
State the importance of Davisson and Germer experiment.
Answer:
The Davisson and Germer experiment directly indicated the wave nature of material particles and quantitatively verified the de Broglie hypothesis for the existence of matter waves.

[Note : The aim of the experiment was not to verify wave like properties of electrons. The realisation came only later, an example of serendipity.]

[Note : Like X-rays, electrons exhibit wave nature under suitable conditions. When the wavelength of matter waves associated with moving electrons is comparable to the inter-atomic spacing in a crystal, electrons show diffraction effects. In 1927, Sir George Thomson (1892 – 1975), British physicist, with his student Alex Reid, observed electron diffraction with a metal foil. It is found that neutrons, atoms, molecules, Œ-particles, etc. show wave nature under suitable conditions.]

Question 7.
What will be the energy of each photon in monochromatic light of frequency 5 × 10¹⁴ Hz?
Answer:
Data: y = 5 × 10¹⁴ Hz, h = 6.63 × 10^-34 Js,
1eV=1.6 × 10^-19 J
The energy of each photon,
E = hv = (6.63 × 10^-34 J.s)(5 × 10¹⁴ Hz)
= 3.315 × 10^-19 J
= \(\frac{3.315 \times 10^{-19} \mathrm{~J}}{1.6 \times 10^{-19} \mathrm{~J} / \mathrm{eV}}\) = 2.072 eV

Question 8.
Observations from an experiment on photoelectric effect for the stopping potential by varying the incident frequency were plotted. The slope of the linear curve was found to be approximately 4.1 × 10^-15 V s. Given that the charge of an electron is 1.6 × 10^-19 C, find the value of the Planck’s constant h.
Answer:
Data : Slope=4.1 × 10^-15 V∙s, e = 1.6 ×10^-19 C
V₀e = hv – hv₀
∴ V₀ =\(\left(\frac{h}{e}\right) v-\left(\frac{h v_{0}}{e}\right)\)
∴ Slope = \(\frac{h}{e}\) ∴ Planck’s constant,
h = (slope) (e)=(4.1 × 10^-15 V∙s)(1.6 × 10^-19 C)
= 6.56 × 10 34J. (as 1 V = \(\frac{1 \mathrm{~J}}{1 \mathrm{C}}\))

Question 9.
The threshold wavelength of tungsten is 2.76 × 10^-5 cm. (a) Explain why no photoelectrons are emitted when the wavelength is more than 2.76 × 10^-5 cm.(b) What will be the maximum kinetic energy of electrons ejected in each of the following cases
(i) if ultraviolet radiation of wavelength λ = 1.80 × 10^-5 cm and
(ii) radiation of frequency 4 × 10¹⁵ Hz is made incident on the tungsten surface.
Answer:
Data: λ₀ = 2.76 × 10^-5 cm = 2.76 × 10^-7 m,
λ =1.80 × 10^-5 cm = 1.80 × 10^-7 m,
v = 4 × 10¹⁵ Hz, h = 6.63 × 10^-34 J∙s,c = 3 × 10⁸ m/s
(a) For λ > λ₀, v < v₀ (threshold frequency).
∴ hv < hv₀. Hence, no photoelectrons are emitted.

(b) Maximum kinetic energy of electrons ejected
= hc\(\left(\frac{1}{\lambda}-\frac{1}{\lambda_{0}}\right)\)
=(6.63 × 10^-34)(3 × 10⁸)\(\left(\frac{10^{7}}{1.8}-\frac{10^{7}}{2.76}\right)\)J
= (6.63 × 10^-19)(0.5555 – 0.3623)
= (6.63)(0.1932 × 10^-19)J = 1.281 × 10^-19 J
= \(\frac{1.281 \times 10^{-19} \mathrm{~J}}{1.6 \times 10^{-19} \mathrm{~J} / \mathrm{eV}}\) = 0.8006 eV

(c) Maximum kinetic energy of electrons ejected
= hv – \(\frac{h c}{\lambda_{0}}\)
=(6.63 × 10^-34(4 × 10¹⁵) – \(\frac{\left(6.63 \times 10^{-34}\right)\left(3 \times 10^{8}\right)}{2.76 \times 10^{-7}}\)
= 26.52 × 10^-19 – 7.207 × 10^-19
= 19.313 × 10^-19 J
= \(\frac{19.313 \times 10^{-19} \mathrm{~J}}{1.6 \times 10^{-19} \mathrm{~J} / \mathrm{eV}}\) = 12.07eV

Question 10.
Photocurrent recorded in the micro ammeter in an experimental set-up of photoelectric effect vanishes when the retarding potential is more than 0.8 V if the wavelength of incident radiation is 4950 Å. If the source of incident radiation is changed, the stopping potential turns out to be 1.2 V. Find the work function of the cathode material and the wavelength of the second source.
Answer:
Data: V₀ = 0.8 V, λ = 4950 Å = 4.950 × 10^-7 m,
V₀‘ = 1.2V, h = 6.63 × 10^-34 J∙s, c = 3 × 10⁸ m/s.
(i) V₀e = hv – Φ = \(\frac{h c}{\lambda}\) – Φ
∴ The work function of the cathode material,

Question 11.
Radiation of wavelength 4500 Å is incident on a metal having work function 2.0 eV. Due to the presence of a magnetic field B, the most energetic photoelectrons emitted in a direction perpendicular to the field move along a circular path of radius 20 cm. What is the value of the magnetic field B?
Answer:
Data: λ = 4500Å = 4.5 × 10^-7 m,
Φ = 2.0eV = 2 × 1.6 × 10^-19 J = 3.2 × 10^-19 J,
h = 6.63 × 10^-34 J∙s, c = 3 × 10⁸ m/s,
r = 20 cm = 0.2 m, e= 1.6 × 10^-19 C,
m = 9.1 × 10^-31kg

This is the value of the magnetic field.

Question 12.
Given the following data for incident wavelength and the stopping potential obtained from an experiment on photoelectric effect, estimate the value of Planck’s constant and the work function of the cathode material. What is the threshold frequency and corresponding wavelength? What is the most likely metal used for emitter?

Answer:
Data: λ = 2536Å = 2.536 × 10^-7 m,
λ’ = 3650Å = 3.650 ×10^-7 m, V₀ = 1.95V, V₀‘ = 0.5V,
c = 3 × 10⁸ mIs, e = 1.6 × 10^-19 C

(i) V₀e = \(\frac{h c}{\lambda}\) – Φ and V₀‘e =\(\frac{h c}{\lambda^{\prime}}\) – Φ
∴ (V₀ – V₀‘)e = hc\(\left(\frac{1}{\lambda}-\frac{1}{\lambda^{\prime}}\right)\)
∴ (1.95 – 0.5(1.6 × 10^-19)
= h (3 × 10⁸\(\left(\frac{10^{7}}{2.536}-\frac{10^{7}}{3.650}\right)\)
∴ 2.32 × 10^-19 = h(3 × 10¹⁵)(0.3943 – 0.2740)
∴ h = \(\frac{2.32 \times 10^{-34}}{0.3609}\) = 6.428 × 10^-34 J∙s
This is the value of Planck’s constant.

(ii) Φ = \(\frac{h c}{\lambda}\) – V₀e

This is the work function of the cathode material.

(iii) Φ = hv₀
∴ The threshold frequency, v₀ = \(\frac{\phi}{h}\)
= \(\frac{4.484 \times 10^{-19} \mathrm{~J}}{6.428 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}}\) = 6.976 × 10¹⁴ Hz

(iv) v₀ = \(\frac{c}{\lambda_{\mathrm{o}}}\) ∴ The threshold frequency, λ₀ = \(\frac{c}{v_{\mathrm{o}}}\)
= \(\frac{3 \times 10^{8}}{6.976 \times 10^{14}}\) = 4.300 × 10^-7 m = 4300 Å

(v) The most likely metal used for emitter : calcium

Question 13.
Calculate the wavelength associated with an electron, its momentum and speed
(a) when it is accelerated through a potential of 54 V
Answer:
Data : V = 54 V, m = 9.1 × 10^-31 kg, e
e = 1.6 × 10^-19 C, h = 6.63 × 10^-34 J.s, KE = 150 eV
(a) We assume that the electron is initially at rest.
∴ Ve = \(\frac{1}{2}\)mv²
∴ v = \(\sqrt{\frac{2 V e}{m}}=\sqrt{\frac{2(54)\left(1.6 \times 10^{-19}\right)}{9.1 \times 10^{-31}}}\)
= \(\sqrt{19 \times 10^{12}}\) = 4.359 × 10⁶ m/5
This is the speed of the electron.
p = mv = (9.1× 10^-31)(4.359 × 10⁶)
= 3.967 × 10^-24 kg∙m/s
This is the momentum of the electron. The wavelength associated with the electron,
λ = \(\frac{h}{p}=\frac{6.63 \times 10^{-34}}{3.967 \times 10^{-24}}\) = 1.671 × 10^-10 m
= 1.671 Å = 0.1671 nm

(b) when it is moving with kinetic energy of 150 eV.
Answer:
As KE ∝ \(\sqrt{V}\), we get
\(\frac{v^{\prime}}{v}=\sqrt{\frac{150}{54}}\) = 1.666
∴ v’ = 1.666v = (1.666)(4.356 × 10⁶)
= 7.262 × 10⁶ m/s
This is the speed of the electron.
p’ = mv’’=(9.1 × 10^-31)(7.262 × 10⁶)
= 6.608 × 10^-24 kg∙m/s
This is the momentum of the electron. The
wavelength associated with the electron,
λ = \(\frac{h}{p^{\prime}}=\frac{6.63 \times 10^{-34}}{6.608 \times 10^{-24}} \) = 1.003 × 10^-10 m
= 1.003 Å = 0.1003 nm

Question 14.
The de Broglie wavelengths associated with an electron and a proton are same. What will be the ratio of
(i) their momenta
(ii) their kinetic energies?
Answer:
Data : λ (electron) = λ (proton)
m (proton) = 1836 m (electron)
(i) λ = \(\frac{h}{p}\) As λ (electron) = λ (proton),
\(\frac{p(\text { electron })}{p \text { (proton) }}\) = 1, where p denotes the magnitude of momentum.

(ii) Assuming v «c,
KE = \(\frac{1}{2}\)mv² = \(\frac{1}{2} \frac{m^{2} v^{2}}{m}=\frac{p^{2}}{2 m}\)
∴ \(\frac{\mathrm{KE} \text { (electron) }}{\mathrm{KE} \text { (proton) }}=\frac{m \text { (proton) }}{m \text { (electron) }}\) = 1836 as p is the same for the electron and the proton.

Question 15.
Two particles have the same de Broglie wavelength and one is moving four times as fast as the other. If the slower particle
is an α-particle, what are the possibilities for the other particle?
Answer:
Data : λ₁ = λ₂, v₁ = 4v₂
λ = \(\frac{h}{p}=\frac{h}{m v}\) ∴ λ₁ = \(\frac{h}{m_{1} v_{1}}\), λ₂ = \(\frac{h}{m_{2} v_{2}}\)
∴ m₁ = m₂ \(\frac{v_{2}}{v_{1}}\) = m₂\(\left(\frac{1}{4}\right)=\frac{m_{2}}{4}\)
As particle 2 is the a-particle, particle 1 (having the mass \(\frac{1}{4}\) times that of the a-particle) may be a proton or neutron.

Question 16.
What is the speed of a proton having de Broglie wavelength of 0.08 Å?
Answer:
Data : λ = 0.08 Å = 8 × 10^-12m, h = 6.63 × 10^-34 J∙s, m = 1.672 × 10^-27 kg
λ = \(\frac{h}{m v}\) ∴ v = \(\frac{h}{\lambda m}=\frac{6.63 \times 10^{-34}}{\left(8 \times 10^{-12}\right)\left(1.672 \times 10^{-27}\right)}\)
∴ v = 4.957 × 10⁴ m/s
This is the speed of the proton.

Question 17.
In nuclear reactors, neutrons travel with energies of 5 × 10^-21 J. Find their speed and wavelength.
Answer:
Data : KE = 5 × 10^-21 J, m = 1.675 × 10^-27 kg, h = 6.63 × 10^-34 J∙s
KE = \(\frac{1}{2}\) mv² = 5 × 10^-21 J
∴ v = \(\sqrt{\frac{2 \mathrm{KE}}{m}}=\sqrt{\frac{(2)\left(5 \times 10^{-21}\right)}{1.675 \times 10^{-27}}}\)
= 2.443 × 10³ m/s
This is the speed of the neutrons. The de Broglie wavelength associated with the neutron,
λ = \(\frac{h}{m v}=\frac{6.63 \times 10^{-34}}{\left(1.675 \times 10^{-27}\right)\left(2.443 \times 10^{3}\right)}\)
= 1.620 × 10^-10 m = 1.620 Å

Question 18.
Find the ratio of the de Broglie wavelengths of an electron and a proton when both are moving with the (a) same speed, (b) same energy and (c) same momentum? State which of the two will have the longer wavelength in each case?
Answer:
Data: m_p = 1836 m_e

12th Physics Digest Chapter 14 Dual Nature of Radiation and Matter Intext Questions and Answers

Remember This (Textbook Page No. 316)

Question 1.
Is solar cell a photocell?
Answer:
Yes

Remember This (Textbook Page No. 317)

Question 1.
Can you estimate the de Broglie wavelength of the Earth?
Answer:
Taking the mass of the Earth as (about) 6 × 10²⁴ kg, and the linear speed of the earth around the Sun as (about) 3 × 10⁴ m/s, we have, the de Brogue wave length of the Earth as
λ = \(\frac{h}{p}=\frac{h}{M v}=\frac{6.63 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}}{\left(6 \times 10^{24} \mathrm{~kg}\right)\left(3 \times 10^{4} \mathrm{~m} / \mathrm{s}\right)}\)
= 3.683 × 10^-63 m (extremely small)

Question 2.
The expression p = E/c defines the momentum of a photon. Can this expression be used for momentum of an electron or proton?
Answer:
No

Remember This (Textbook Page No. 319)

Diffraction results described above can be produced in the laboratory using an electron diffraction tube as shown in figure. It has a filament which on heating produces electrons. This filament acts as a cathode. Electrons are accelerated to quite high speeds by creating large potential difference between the cathode and a positive electrode. On its way, the beam of electrons comes across a thin sheet of

graphite. The electrons are diffracted by the atomic layers in the graphite and form diffraction rings on the phosphor screen. By changing the voltage between the cathode and anode, the energy, and therefore the speed, of the electrons can be changed. This will change the wavelength of the electrons and a change will be seen in the diffraction pattern. By increasing the voltage, the radius of the diffraction rings will decrease. Try to explain why?
Answer:
When the accelerating voltage is increased, the kinetic energy and hence the momentum of the electron increases. This decreases the de Brogue wavelength of the electron. Hence, the radius of the diffraction ring decreases.

Remember This (Textbook Page No. 320)

Question 1.
On what scale or under which circumstances are the wave nature of matter apparent?
Answer:
When the de Brogue wavelength of a particle such as an electron, atom, or molecule is comparable to the interatomic spacing in a crystal, the wave nature of matter is revealed in diffraction/interference.

Balbharti Maharashtra State Board 12th Physics Textbook Solutions Chapter 2 Mechanical Properties of Fluids Textbook Exercise Questions and Answers.

Maharashtra State Board 12th Physics Solutions Chapter 2 Mechanical Properties of Fluids

1) Multiple Choice Questions

i) A hydraulic lift is designed to lift heavy objects of maximum mass 2000 kg. The area of cross-section of piston carrying
the load is 2.25 × 10^-2 m². What is the maximum pressure the piston would have to bear?
(A) 0.8711 × 10⁶ N/m²
(B) 0.5862 × 10⁷ N/m²
(C) 0.4869 × 10⁵ N/m²
(D) 0.3271 × 10⁴ N/m²
Answer:
(A) 0.8711 × 10⁶ N/m²

ii) Two capillary tubes of radii 0.3 cm and 0.6 cm are dipped in the same liquid. The ratio of heights through which the liquid will rise in the tubes is
(A) 1:2
(B) 2:1
(C) 1:4
(D) 4:1
Answer:
(B) 2:1

iii) The energy stored in a soap bubble of diameter 6 cm and T = 0.04 N/m is nearly
(A) 0.9 × 10^-3 J
(B) 0.4 × 10^-3 J
(C) 0.7 × 10^-3 J
(D) 0.5 × 10^-3 J
Answer:
(A) 0.9 × 10^-3 J

iv) Two hail stones with radii in the ratio of 1:4 fall from a great height through the atmosphere. Then the ratio of their terminal velocities is
(A) 1:2
(B) 1:12
(C) 1:16
(D) 1:8
Answer:
(C) 1:16

v) In Bernoulli’s theorem, which of the following is conserved?
(A) linear momentum
(B) angular momentum
(C) mass
(D) energy
Answer:
(D) energy

2) Answer in brief.

i) Why is the surface tension of paints and lubricating oils kept low?
Answer:
For better wettability (surface coverage), the surface tension and angle of contact of paints and lubricating oils must below.

ii) How much amount of work is done in forming a soap bubble of radius r?
Answer:
Let T be the surface tension of a soap solution. The initial surface area of soap bubble = 0
The final surface area of soap bubble = 2 × 4πr²
∴ The increase in surface area = 2 × 4πr^2-
The work done in blowing the soap bubble is W = surface tension × increase in surface area = T × 2 × 4πr² = 8πr²T

iii) What is the basis of the Bernoulli’s principle?
Answer:
Conservation of energy.

iv) Why is a low density liquid used as a manometric liquid in a physics laboratory?
Answer:
An open tube manometer measures the gauge pressure, p — p₀ = hpg, where p₀ is the pressure being measured, p₀ is the atmospheric pressure, h is the difference in height between the manometric liquid of density p in the two arms. For a given pressure p, the product hp is constant. That is, p should be small for h to be large. Therefore, for noticeably large h, laboratory manometer uses a low density liquid.

v) What is an incompressible fluid?
Answer:
An incompressible fluid is one which does not undergo change in volume for a large range of pressures. Thus, its density has a constant value throughout the fluid. In most cases, all liquids are incompressible.

Question 3.
Why two or more mercury drops form a single drop when brought in contact with each other?
Answer:
A spherical shape has the minimum surface area- to-volume ratio of all geometric forms. When two drops of a liquid are brought in contact, the cohesive forces between their molecules coalesces the drops into a single larger drop. This is because, the volume of the liquid remaining the same, the surface area of the resulting single drop is less than the combined surface area of the smaller drops. The resulting decrease in surface energy is released into the environment as heat.

Proof : Let n droplets each of radius r coalesce to form a single drop of radius R. As the volume of the liquid remains constant, volume of the drop = volume of n droplets
∴ \(\frac{4}{3}\)πR³ = n × \(\frac{4}{3}\)πr³
∴ R³ = nr³ ∴ R = \(\sqrt[3]{n}\)r
Surface area of n droplets = n × πR²
Surface area of the drop = 4πR² = n^2/3 × πR²
∴ The change in the surface area = surface area of drop – surface area of n droplets
= πR²(n^2/3 – n)
Since the bracketed term is negative, there is a decrease in surface area and a decrease in surface energy.

Question 4.
Why does velocity increase when water flowing in broader pipe enters a narrow pipe?
Answer:
When a tube narrows, the same volume occupies a greater length, as schematically shown in below figure. A₁ is the cross section of the broader pipe and that of narrower pipe is A₂. By the equation of continuity, V₂ = (A₁/A₂)V₁

Since A₁/A₂ > v₂ > v₁. For the same volume to pass points 1 and 2 in a given time, the speed must be greater at point 2.
The process is exactly reversible. If the fluid flows in the opposite direction, its speed decreases when the tube widens.

Question 5.
Why does the speed of a liquid increase and its pressure decrease when a liquid passes through constriction in a horizontal pipe?
Answer:

Consider a horizontal constricted tube.
Let A₁ and A₂ be the cross-sectional areas at points 1 and 2, respectively. Let v₁ and v₂ be the corresponding flow speeds, ρ is the density of the fluid in the pipeline. By the equation of continuity,
v₁A₁ = v₂A₂ …… (1)
∴ \(\frac{v_{2}}{v_{1}}\) = \(\frac{A_{1}}{A_{2}}\) > 1 (∵ A₁ > A₂)
Therefore, the speed of the liquid increases as it passes through the constriction. Since the meter is assumed to be horizontal, from Bernoulli’s equation we get,

Again, since A₁ > A₂, the bracketed term is positive so that p₁ > p₂. Thus, as the fluid passes through the constriction or throat, the higher speed results in lower pressure at the throat.

Question 6.
Derive an expression of excess pressure inside a liquid drop.
Answer:
Consider a small spherical liquid drop with a radius R. It has a convex surface, so that the pressure p on the concave side (inside the liquid) is greater than the pressure p0 on the convex side (outside the liquid). The surface area of the drop is
A = 4πR² … (1)
Imagine an increase in radius by an infinitesimal amount dR from the equilibrium value R. Then, the differential increase in surface area would be dA = 8πR ∙ dR …(2)
The increase in surface energy would be equal to the work required to increase the surface area :
dW = T∙dA = 8πTRdR …..(3)

We assume that dR is so small that the pressure inside remains the same, equal to p. All parts of the surface of the drop experience an outward force per unit area equal to ρ — ρ₀. Therefore, the work done by this outward pressure-developed force against the surface tension force during the increase in radius dR is
dW = (excess pressure × surface area) ∙ dR
= (ρ – ρ₀) × 4πnR² ∙ dR …..(4)
From Eqs. (3) and (4),
(ρ — ρ₀) × 4πR² ∙ dR = 8πTRdR
∴ ρ – ρ₀ = \(\frac{2 T}{R}\) …… (5)
which is called Laplace’s law for a spherical membrane (or Young-Laplace equation in spherical form).
[Notes : (1) The above method is called the principle of virtual work. (2) Equation (5) also applies to a gas bubble within a liquid, and the excess pressure in this case is also called the gauge pressure. An air or gas bubble within a liquid is technically called a cavity because it has only one gas-liquid interface. A bubble, on the other hand, such as a soap bubble, has two gas-liquid interfaces.]

Question 7.
Obtain an expression for conservation of mass starting from the equation of continuity.
Answer:

Consider a fluid in steady or streamline flow, that is its density is constant. The velocity of the fluid within a flow tube, while everywhere parallel to the tube, may change its magnitude. Suppose the velocity is \(\vec{v}_{1}\), at point P and \(\vec{v}_{2}\) at point. Q. If A₁ and A₂ are the cross-sectional areas of the tube at these two points, the volume flux across A₁, \(\frac{d}{d t}\)(V₂) = A₁v₁ and that across A₂, \(\frac{d}{d t}\)(V₂) = A₂v₂
By the equation of continuity of flow for a fluid, A₁v₁ = A₂V₂
i.e., \(\frac{d}{d t}\)(V₁) = \(\frac{d}{d t}\)(V₂)
If ρ₁ and ρ₁ are the densities of the fluid at P and Q, respectively, the mass flux across A₁, \(\frac{d}{d t}\)(m₁) = \(\frac{d}{d t}\)(ρ₁ v₁) = A₁ρ₁v₁
and that across A₂, \(\frac{d}{d t}\)(m₂) = \(\frac{d}{d t}\)(ρ₂V₂) = A₂ρ₂v₂
Since no fluid can enter or leave through the boundary of the tube, the conservation of mass requires the mass fluxes to be equal, i.e.,
\(\frac{d}{d t}\)(m₁) = \(\frac{d}{d t}\)(m₂)
i.e., A₁ρ₁v₁ = A₂ρ₂v₂
i. e., Apv = constant
which is the required expression.

Question 8.
Explain the capillary action.
Answer:
(1) When a capillary tube is partially immersed in a wetting liquid, there is capillary rise and the liquid meniscus inside the tube is concave, as shown in below figure.

Consider four points A, B, C, D, of which point A is just above the concave meniscus inside the capillary and point B is just below it. Points C and D are just above and below the free liquid surface outside.

Let P_A, P_B, P_C and P_D be the pressures at points A, B, C and D, respectively.
Now, P_A = P_C = atmospheric pressure
The pressure is the same on both sides of the free surface of a liquid, so that

The pressure on the concave side of a meniscus is always greater than that on the convex side, so that
P_A > P_B
∴ P_D > P_B (∵ P_A = P_D)

The excess pressure outside presses the liquid up the capillary until the pressures at B and D (at the same horizontal level) equalize, i.e., P_B becomes equal to P_D. Thus, there is a capillary rise.

(2) For a non-wetting liquid, there is capillary depression and the liquid meniscus in the capillary tube is convex, as shown in above figure.

Consider again four points A, B, C and D when the meniscus in the capillary tube is at the same level as the free surface of the liquid. Points A and B are just above and below the convex meniscus. Points C and D are just above and below the free liquid surface outside.

The pressure at B (P_B) is greater than that at A (P_A). The pressure at A is the atmospheric pressure H and at D, P_D \(\simeq\) H = P_A. Hence, the hydrostatic pressure at the same levels at B and D are not equal, P_B > P_D. Hence, the liquid flows from B to D and the level of the liquid in the capillary falls. This continues till the pressure at B’ is the same as that D’, that is till the pressures at the same level are equal.

Question 9.
Derive an expression for capillary rise for a liquid having a concave meniscus.
Answer:
Consider a capillary tube of radius r partially immersed into a wetting liquid of density p. Let the capillary rise be h and θ be the angle of contact at the edge of contact of the concave meniscus and glass. If R is the radius of curvature of the meniscus then from the figure, r = R cos θ.

Surface tension T is the tangential force per unit length acting along the contact line. It is directed into the liquid making an angle θ with the capillary wall. We ignore the small volume of the liquid in the meniscus. The gauge pressure within the liquid at a depth h, i.e., at the level of the free liquid surface open to the atmosphere, is
ρ – ρ_o = ρgh …. (1)
By Laplace’s law for a spherical membrane, this gauge pressure is
ρ – ρ_o = \(\frac{2 T}{R}\) ….. (2)
∴ hρg = \(\frac{2 T}{R}\) = \(\frac{2 T \cos \theta}{r}\)
∴ h = \(\frac{2 T \cos \theta}{r \rho g}\) …. (3)
Thus, narrower the capillary tube, the greater is the capillary rise.
From Eq. (3),
T = \(\frac{h \rho r g}{2 T \cos \theta}\) … (4)
Equations (3) and (4) are also valid for capillary depression h of a non-wetting liquid. In this case, the meniscus is convex and θ is obtuse. Then, cos θ is negative but so is h, indicating a fall or depression of the liquid in the capillary. T is positive in both cases.
[Note : The capillary rise h is called Jurin height, after James Jurin who studied the effect in 1718. For capillary rise, Eq. (3) is also called the ascent formula.]

Question 10.
Find the pressure 200 m below the surface of the ocean if pressure on the free surface of liquid is one atmosphere. (Density of sea water = 1060 kg/m³) [Ans. 21.789 × 10⁵ N/m²]
Answer:
Data : h = 200 m, p = 1060 kg/m³,
p₀ = 1.013 × 10⁵ Pa, g = 9.8 m/s²
Absolute pressure,
p = p₀ + hρg
= (1.013 × 10³) + (200)(1060)(9.8)
= (1.013 × 10⁵) + (20.776 × 10⁵)
= 21.789 × 10⁵ = 2.1789 MPa

Question 11.
In a hydraulic lift, the input piston had surface area 30 cm² and the output piston has surface area of 1500 cm². If a force of 25 N is applied to the input piston, calculate weight on output piston. [Ans. 1250 N]
Answer:
Data : A₁ = 30 cm² = 3 × 10^-3 m²,
A₂ = 1500 cm² = 0.15 m², F₁ = 25 N
By Pascal’s law,
\(\frac{F_{1}}{A_{1}}\) = \(\frac{F_{2}}{A_{2}}\)
∴ The force on the output piston,
F₂ = F₁\(\frac{A_{2}}{A_{1}}\) = (25)\(\frac{0.15}{3 \times 10^{-3}}\) = 25 × 50 = 1250 N

Question 12.
Calculate the viscous force acting on a rain drop of diameter 1 mm, falling with a uniform velocity 2 m/s through air. The coefficient of viscosity of air is 1.8 × 10^-5 Ns/m².
[Ans. 3.393 × 10^-7 N]
Answer:
Data : d = 1 mm, v₀ = 2 m / s,
η = 1.8 × 10^-5 N.s/m²
r = \(\frac{d}{2}\) = 0.5 mm = 5 × 10^-4 m
By Stokes’ law, the viscous force on the raindrop is f = 6πηrv₀
= 6 × 3.142 (1.8 × 10^-5 N.s/m² × 5 × 10^-4 m)(2 m/s)
= 3.394 × 10^-7 N

Question 13.
A horizontal force of 1 N is required to move a metal plate of area 10^-2 m² with a velocity of 2 × 10^-2 m/s, when it rests on a layer of oil 1.5 × 10^-3 m thick. Find the coefficient of viscosity of oil. [Ans. 7.5 Ns/m²]
Answer:
Data : F = 1 N, A = 10^-2m², v₀ = 2 × 10^-2 y = 1.5 × 10^-3 m
Velocity gradient, \(\frac{d v}{d y}\) = \(\frac{2 \times 10^{-2}}{1.5 \times 10^{-3}}\) = \(\frac{40}{3}\)s^-1

Question 14.
With what terminal velocity will an air bubble 0.4 mm in diameter rise in a liquid of viscosity 0.1 Ns/m² and specific gravity 0.9? Density of air is 1.29 kg/m³. [Ans. – 0.782 × 10^-3 m/s, The negative sign indicates that the bubble rises up]
Answer:
Data : d = 0.4 mm, η = 0.1 Pa.s, ρ_L = 0.9 × 10³ kg/m³ = 900 kg/m³, ρ_air = 1.29 kg/m³, g = 9.8 m/s².
Since the density of air is less than that of oil, the air bubble will rise up through the liquid. Hence, the viscous force is downward. At terminal velocity, this downward viscous force is equal in magnitude to the net upward force.
Viscous force = buoyant force – gravitational force

Question 15.
The speed of water is 2m/s through a pipe of internal diameter 10 cm. What should be the internal diameter of nozzle of the pipe if the speed of water at nozzle is 4 m/s?
[Ans. 7.07 × 10^-2m]
Answer:
Data : d₁ = 10 cm = 0.1 m, v₁ = 2 m/s, v₂ = 4 m/s
By the equation of continuity, the ratio of the speed is

Question 16.
With what velocity does water flow out of an orifice in a tank with gauge pressure 4 × 10⁵ N/m² before the flow starts? Density of water = 1000 kg/m³. [Ans. 28.28 m/s]
Answer:
Data : ρ — ρ₀ = 4 × 10⁵ Pa, ρ = 10³ kg/m³
If the orifice is at a depth h from the water surface in a tank, the gauge pressure there is
ρ – ρ₀ = hρg … (1)
By Toricelli’s law of efflux, the velocity of efflux,
v = \(\sqrt{2 g h}\) …(2)
Substituting for h from Eq. (1),

Question 17.
The pressure of water inside the closed pipe is 3 × 10⁵ N/m². This pressure reduces to 2 × 10⁵ N/m² on opening the value of the pipe. Calculate the speed of water flowing through the pipe. (Density of water = 1000 kg/m³). [Ans. 14.14 m/s]
Answer:
Data : p₁ = 3 × 10⁵ Pa, v₁ = 0, p₂ = 2 × 10⁵ Pa, ρ = 10³ kg/m³
Assuming the potential head to be zero, i.e., the pipe to be horizontal, the Bernoulli equation is

Question 18.
Calculate the rise of water inside a clean glass capillary tube of radius 0.1 mm, when immersed in water of surface tension 7 × 10^-2 N/m. The angle of contact between water and glass is zero, density of water = 1000 kg/m³, g = 9.8 m/s².
[Ans. 0.1429 m]
Answer:
Data : r = 0.1 mm = 1 × 10^-4m, θ = 0°,
T = 7 × 10^-2 N/m, r = 10³ kg/m³, g = 9.8 m/s²
cos θ = cos 0° = 1

= 0.143 m

Question 19.
An air bubble of radius 0.2 mm is situated just below the water surface. Calculate the gauge pressure. Surface tension of water = 7.2 × 10^-2 N/m.
[Ans. 720 N/m²]
Answer:
Data : R = 2 × 10^-4m, T = 7.2 × 10^-2N/m, p = 10³ kg/m³
The gauge pressure inside the bubble = \(\frac{2 T}{R}\)
= \(\frac{2\left(7.2 \times 10^{-2}\right)}{2 \times 10^{-4}}\) = 7.2 × 10² = 720 Pa

Question 20.
Twenty seven droplets of water, each of radius 0.1 mm coalesce into a single drop. Find the change in surface energy. Surface tension of water is 0.072 N/m. [Ans. 1.628 × 10^-7 J = 1.628 erg]
Answer:
Data : r = 1 mm = 1 × 10^-3 m, T = 0.472 J/m²
Let R be the radius of the single drop formed due to the coalescence of 8 droplets of mercury.
Volume of 8 droplets = volume of the single drop as the volume of the liquid remains constant.
∴ 8 × \(\frac{4}{3}\)πr³ = \(\frac{4}{3}\)πR³
∴ 8r³ = R³
∴ 2r = R
Surface area of 8 droplets = 8 × 4πr²
Surface area of single drop = 4πR²
∴ Decrease in surface area = 8 × 4πr² – 4πR²
= 4π(8r² – R²)
= 4π[8r² – (2r)²]
= 4π × 4r²
∴ The energy released = surface tension × decrease in surface area = T × 4π × 4r²
= 0.472 × 4 × 3.142 × 4 × (1 × 10^-3)²
= 2.373 × 10^-5 J
The decrease in surface energy = 0.072 × 4 × 3.142 × 18 × (1 × 10^-4)²
= 1.628 × 10^-7 J

Question 21.
A drop of mercury of radius 0.2 cm is broken into 8 identical droplets. Find the work done if the surface tension of mercury is 435.5 dyne/cm. [Ans. 2.189 × 10^-5J]
Answer:
Let R be the radius of the drop and r be the radius of each droplet.
Data : R = 0.2 cm, n = 8, T = 435.5 dyn/cm
As the volume of the liquid remains constant, volume of n droplets = volume of the drop
∴ n × \(\frac{4}{3}\)πr³ = \(\frac{4}{3}\)πR³

Surface area of the drop = 4πR²
Surface area of n droplets = n × 4πR²
∴ The increase in the surface area = surface area of n droplets-surface area of drop
= 4π(nr² – R²) = 4π(8 × \(\frac{R^{2}}{4}\) – R²)
= 4π(2 — 1)R² = 4πR²
∴ The work done
= surface tension × increase in surface area
= T × 4πR² = 435.5 × 4 × 3.142 × (0.2)²
= 2.19 × 10² ergs = 2.19 × 10^-5 J

Question 22.
How much work is required to form a bubble of 2 cm radius from the soap solution having surface tension 0.07 N/m.
[Ans. 0.7038 × 10^-3 J]
Answer:
Data : r = 4 cm = 4 × 10^-2 m, T = 25 × 10^-3 N/m
Initial surface area of soap bubble = 0
Final surface area of soap bubble = 2 × 4πr²
∴ Increase in surface area = 2 × 4πr²
The work done
= surface tension × increase in surface area
= T × 2 × 4πr²
= 25 × 10^-3 × 2 × 4 × 3.142 × (4 × 10^-2)²
= 1.005 × 10^-3 J
The work done = 0.07 × 8 × 3.142 × (2 × 10^-2)²
= 7.038 × 10^-4 J

Question 23.
A rectangular wire frame of size 2 cm × 2 cm, is dipped in a soap solution and taken out. A soap film is formed, if the size of the film is changed to 3 cm × 3 cm, calculate the work done in the process. The surface tension of soap film is 3 × 10^-2 N/m. [Ans. 3 × 10^-5 J]
Answer:
Data : A₁ = 2 × 2 cm² = 4 × 10^-4 m²,
A₂ = 3 × 3 cm² =9 × 10^-4 m², T = 3 × 10^-2 N/m
As the film has two surfaces, the work done is W = 2T(A₂ – A₁)
= 2(3 × 10^-2)(9 × 10^-4 × 10^-4)
= 3.0 × 10^-5 J = 30 µJ

12th Physics Digest Chapter 2 Mechanical Properties of Fluids Intext Questions and Answers

Can you tell? (Textbook Page No. 27)

Question 1.
Why does a knife have a sharp edge or a needle has a sharp tip ?
Answer:
For a given force, the pressure over which the force is exerted depends inversely on the area of contact; smaller the area, greater the pressure. For instance, a force applied to an area of 1 mm² applies a pressure that is 100 times as great as the same force applied to an area of 1 cm². The edge of a knife or the tip of a needle has a small area of contact. That is why a sharp needle is able to puncture the skin when a small force is exerted, but applying the same force with a finger does not.

Use your brain power

Question 1.
A student of mass 50 kg is standing on both feet. Estimate the pressure exerted by the student on the Earth. Assume reasonable value to any quantity you need; justify your assumption. You may use g = 10 m/s², By what factor will it change if the student lies on back ?
Answer:
Assume area of each foot = area of a 6 cm × 25 cm rectangle.
∴ Area of both feet = 0.03 m²
∴ The pressure due to the student’s weight
= \(\frac{m g}{A}\) = \(\frac{50 \times 10}{0.03}\) = 16.7 kPa

According to the most widely used Du Bois formula for body surface area (BSA), the student’s BSA = 1.5 m², so that the area of his back is less than half his BSA, i.e., < 0.75 m². When the student lies on his back, his area of contact is much smaller than this. So, estimating the area of contact to be 0.3 m², i.e., 10 times more than the area of his feet, the pressure will be less by a factor of 10 or more, [Du Bois formula : BSA = 0.2025 × W^0.425 × H^0.725, where W is weight in kilogram and H is height in metre.]

Can you tell? (Textbook Page No. 30)

Question 1.
The figures show three containers filled with the same oil. How will the pressures at the reference compare ?

Answer:
Filled to the same level, the pressure is the same at the bottom of each vessel.

Use Your Brain Power (Textbook Page 35)

Question 1.
Prove that equivalent SI unit of surface tension is J/m².
Answer:
The SI unit of surface tension =

Try This (Textbook Page No. 36)

Question 1.
Take a ring of about 5 cm in diameter. Tie a thread slightly loose at two diametrically opposite points on the ring. Dip the ring into a soap solution and take it out. Break the film on any one side of the thread. Discuss what happens.
Answer:
On taking the ring out, there is a soap film stretched over the ring, in which the thread moves about quite freely. Now, if the film is punctured with a pin on one side-side A in below figure-then immediately the thread is pulled taut by the film on the other side as far as it can go. The thread is now part of a perfect circle, because the surface tension on the side F of the film acts everywhere perpendicular to the thread, and minimizes the surface area of the film to as small as possible.

Can You Tell ? (Textbook Page No. 38)

Question 1.
How does a waterproofing agent work ?
Answer:
Wettability of a surface, and thus its propensity for penetration of water, depends upon the affinity between the water and the surface. A liquid wets a surface when its contact angle with the surface is acute. A waterproofing coating has angle of contact obtuse and thus makes the surface hydrophobic.

Brain Teaser (Textbook Page No. 41)

Question 1.
Can you suggest any method to measure the surface tension of a soap solution? Will this method have any commercial application?
Answer:
There are more than 40 methods for determining equilibrium surface tension at the liquid-fluid and solid-fluid boundaries. Measuring the capillary rise (see Unit 2.4.7) is the laboratory method to determine surface tension.

Among the various techniques, equilibrium surface tension is most frequently measured with force tensiometers or optical (or the drop profile analysis) tensiometers in customized measurement setups.
[See https: / / www.biolinscientific.com /measurements /surface-tension]

Question 2.
What happens to surface tension under different gravity (e.g., aboard the International Space Station or on the lunar surface)?
Answer:
Surface tension does not depend on gravity.
[Note : The behaviour of liquids on board an orbiting spacecraft is mainly driven by surface tension phenomena. These make predicting their behaviour more difficult than under normal gravity conditions (i.e., on the Earth’s surface). New challenges appear when handling liquids on board a spacecraft, which are not usually present in terrestrial environments. The reason is that under the weightlessness (or almost weightlessness) conditions in an orbiting spacecraft, the different inertial forces acting on the bulk of the liquid are almost zero, causing the surface tension forces to be the dominant ones.

In this ‘micro-gravity’ environment, the surface tension forms liquid drops into spheres to minimize surface area, causes liquid columns in a capillary rise up to its rim (without over flowing). Also, when a liquid drop impacts on a dry smooth surface on the Earth, a splash can be observed as the drop disintegrates into thousands of droplets. But no splash is observed as the drop hits dry smooth surface on the Moon. The difference is the atmosphere. As the Moon has no atmosphere, and therefore no gas surrounding a falling drop, the drop on the Moon does not splash.
(See http://mafija.fmf.uni-Ij.si/]

Can you tell? (Textbook Page No. 45)

Question 1.
What would happen if two streamlines intersect?
Answer:
The velocity of a fluid molecule is always tangential to the streamline. If two streamlines intersect, the velocity at that point will not be constant or unique.

Activity

Question 1.
Identify some examples of streamline flow and turbulent flow in everyday life. How would you explain them ? When would you prefer a streamline flow?
Answer:
Smoke rising from an incense stick inside a wind-less room, air flow around a car or aeroplane in motion are some examples of streamline flow, Fish, dolphins, and even massive whales are streamlined in shape to reduce drag. Migratory birds species that fly long distances often have particular features such as long necks, and flocks of birds fly in the shape of a spearhead as that forms a streamlined pattern.

Turbulence results in wasted energy. Cars and aeroplanes are painstakingly streamlined to reduce fluid friction, and thus the fuel consumption. (See ‘Disadvantages of turbulence’ in the following box.) Turbulence is commonly seen in washing machines and kitchen mixers. Turbulence in these devices is desirable because it causes mixing. (Also see ‘Advantages of turbulence’ in the following box.) Recent developments in high-speed videography and computational tools for modelling is rapidly advancing our understanding of the aerodynamics of bird and insect flights which fascinate both physicists and biologists.

Use your Brain power (Textbook Page No. 46)

Question 1.
The CGS unit of viscosity is the poise. Find the relation between the poise and the SI unit of viscosity.
Answer:
By Newton’s law of viscosity,
\(\frac{F}{A}\) = η\(\frac{d v}{d y}\)
where \(\frac{F}{A}\) is the viscous drag per unit area, \(\frac{d v}{d y}\) is the velocity gradient and η is the coefficient of viscosity of the fluid. Rewriting the above equation as

SI unit : the pascal second (abbreviated Pa.s), 1 Pa.s = 1 N.m^-2.s
CGS unit: dyne.cm^-2.s, called the poise [symbol P, named after Jean Louis Marie Poiseuille (1799 -1869), French physician].
[Note : Thè most commonly used submultiples are the millipascalsecond (mPa.s) and the centipoise (cP). 1 mPa.s = 1 cP.]

Use your Brain power (Textbook Page No. 49)

Question 1.
A water pipe with a diameter of 5.0 cm is connected to another pipe of diameter 2.5 cm. How would the speeds of the water flow compare ?
Answer:
Water is an incompressible fluid (almost). Then, by the equation of continuity, the ratio of the speeds, is

Do you know? (Textbook Page No. 50)

Question 1.
How does an aeroplane take off?
Answer:
A Venturi meter is a horizontal constricted tube that is used to measure the flow speed through a pipeline. The constricted part of the tube is called the throat. Although a Venturi meter can be used for a gas, they are most commonly used for liquids. As the fluid passes through the throat, the higher speed results in lower pressure at point 2 than at point 1. This pressure difference is measured from the difference in height h of the liquid levels in the U-tube manometer containing a liquid of density ρ_m. The following treatment is limited to an incompressible fluid.

Let A₁ and A₂ be the cross-sectional areas at points 1 and 2, respectively. Let v₁ and v₂ be the corresponding flow speeds. ρ is the density of the fluid in the pipeline. By the equation of continuity,
v₁A₁ = v₂A₂ … (1)
Since the meter is assumed to be horizontal, from Bernoufli’s equation we get,

The pressure difference is equal to ρ_mgh, where h is the differences in liquid levels in the manometer.
Then,

Equation (3) gives the flow speed of an incompressible fluid in the pipeline. The flow rates of practical interest are the mass and volume flow rates through the meter.
Volume flow rate =A₁v₁ and mass flow rate = density × volume flow
rate = ρA₁v₁
[Note When a Venturi meter is used in a liquid pipeline, the pressure difference is measured from the difference in height h of the levels of the same liquid in the two vertical tubes, as shown in the figure. Then, the pressure difference is equal to ρgh.


The flow meter is named after Giovanni Battista Venturi (1746—1822), Italian physicist.]

Question 2.
Why do racer cars and birds have typical shape ?
Answer:
The streamline shape of cars and birds reduce drag.

Question 3.
Have you experienced a sideways jerk while driving a two wheeler when a heavy vehicle overtakes you ?
Answer:
Suppose a truck passes a two-wheeler or car on a highway. Air passing between the vehicles flows in a narrower channel and must increase its speed according to Bernoulli’s principle causing the pressure between them to drop. Due to greater pressure on the outside, the two-wheeler or car veers towards the truck.

When two ships sail parallel side-by-side within a distance considerably less than their lengths, since ships are widest toward their middle, water moves faster in the narrow gap between them. As water velocity increases, the pressure in between the ships decreases due to the Bernoulli effect and draws the ships together. Several ships have collided and suffered damage in the early twentieth century. Ships performing At-sea refueling or cargo transfers performed by ships is very risky for the same reason.

Question 4.
Why does dust get deposited only on one side of the blades of a fan ?
Answer:
Blades of a ceiling/table fan have uniform thickness (unlike that of an aerofoil) but are angled (cambered) at 8° to 12° (optimally, 10°) from their plane. When they are set rotating, this camber causes the streamlines above/behind a fan blade to detach away from the surface of the blade creating a very low pressure on that side. The lower/front streamlines however follow the blade surface. Dust particles stick to a blade when it is at rest as well as when in motion both by intermolecular force of adhesion and due to static charges. However, they are not dislodged from the top/behind surface because of complete detachment of the streamlines.

The lower/front surface retains some of the dust because during motion, a thin layer of air remains stationary relative to the blade.

Question 5.
Why helmets have specific shape?
Answer:
Air drag plays a large role in slowing bike riders (especially, bicycle) down. Hence, a helmet is aerodynamically shaped so that it does not cause too much drag.

Use your Brain power (Textbook Page No. 52)

Question 1.
Does the Bernoulli’s equation change when the fluid is at rest ? How ?
Answer:
Bernoulli’s principle is for fluids in motion. Hence, it is pointless to apply it to a fluid at rest. Nevertheless, for a fluid is at rest, the Bernoulli equation gives the pressure difference due to a liquid column.

For a static fluid, v₁ = v₂ = 0. Bernoulli’s equation in that case is p₁ + ρgh₁ = ρ₂ + ρgh₂

Further, taking h₂ as the reference height of zero, i.e., by setting h₂ = 0, we get p₂ = p₁ + ρgh₁

This equation tells us that in static fluids, pressure increases with depth. As we go from point 1 to point 2 in the fluid, the depth increases by h₁ and consequently, p₂ is greater than p₁ by an amount ρgh₁.

In the case, p₁ = p₀, the atmospheric pressure at the top of the fluid, we get the familiar gauge pressure at a depth h₁ = ρgh₁. Thus, Bernoulli’s equation confirms the fact that the pressure change due to the weight of a fluid column of length h is ρgh.

Balbharti Maharashtra State Board 12th Physics Textbook Solutions Chapter 3 Kinetic Theory of Gases and Radiation Textbook Exercise Questions and Answers.

Maharashtra State Board 12th Physics Solutions Chapter 3 Kinetic Theory of Gases and Radiation

1. Choose the correct option.

i) In an ideal gas, the molecules possess
(A) only kinetic energy
(B) both kinetic energy and potential energy
(C) only potential energy
(D) neither kinetic energy nor potential energy
Answer:
(A) only kinetic energy

ii) The mean free path λ of molecules is given by
(A) \(\sqrt{\frac{2}{\pi n d^{2}}}\)
(B) \(\frac{1}{\pi n d^{2}}\)
(C) \(\frac{1}{\sqrt{2} \pi n d^{2}}\)
(D) \(\frac{1}{\sqrt{2 \pi n d^{1}}}\)
where n is the number of molecules per unit volume and d is the diameter of the molecules.
Answer:
(C) \(\frac{1}{\sqrt{2} \pi n d^{2}}\)

iii) If pressure of an ideal gas is decreased by 10% isothermally, then its volume will
(A) decrease by 9%
(B) increase by 9%
(C) decrease by 10%
(D) increase by 11.11%
Answer:
(D) increase by 11.11% [Use the formula P₁V₁ = P₂V₂. It gives \(\frac{V_{2}}{V_{1}}\) = \(\frac{1}{0.9}\) = 1.111 ∴ \(\frac{V_{2}-V_{1}}{V_{1}}\) = 0.1111, i.e., 11.11%

iv) If a = 0.72 and r = 0.24, then the value of t_r is
(A) 0.02
(B) 0.04
(C) 0.4
(D) 0.2
Answer:
(B) 0.04

v) The ratio of emissive power of a perfect blackbody at 1327°C and 527°C is
(A) 4 : 1
(B) 16 : 1
(C) 2 : 1
(D) 8 : 1
Answer:
(B) 16 : 1

2. Answer in brief.

i) What will happen to the mean square speed of the molecules of a gas if the temperature of the gas increases?
Answer:
If the temperature of a gas increases, the mean square speed of the molecules of the gas will increase in the same proportion.
[Note: \(\overline{v^{2}}\) = \(\frac{3 n R T}{N m}\) ∴ \(\overline{v^{2}}\) ∝ T for a fixed mass of gas.]

ii) On what factors do the degrees of freedom depend?
Answer:
The degrees of freedom depend upon
(i) the number of atoms forming a molecule
(ii) the structure of the molecule
(iii) the temperature of the gas.

iii) Write ideal gas equation for a mass of 7 g of nitrogen gas.
Answer:
In the usual notation, PV = nRT.

Therefore, the corresponding ideal gas equation is
PV = \(\frac{1}{4}\)RT.

iv) What is an ideal gas ? Does an ideal gas exist in practice ?.
Answer:
An ideal or perfect gas is a gas which obeys the gas laws (Boyle’s law, Charles’ law and Gay-Lussac’s law) at all pressures and temperatures. An ideal gas cannot be liquefied by application of pressure or lowering the temperature.

A molecule of an ideal gas is an ideal particle having only mass and velocity. Its structure and size are ignored. Also, intermolecular forces are zero except during collisions

v) Define athermanous substances and diathermanous substances.
Answer:

1. A substance which is largely opaque to thermal radiations, i.e., a substance which does not transmit heat radiations incident on it, is known as an athermanous substance.
2. A substance through which heat radiations can pass is known as a diathermanous substance.

Question 3.
When a gas is heated its temperature increases. Explain this phenomenon based on kinetic theory of gases.
Answer:
Molecules of a gas are in a state of continuous random motion. They possess kinetic energy. When a gas is heated, there is increase in the average kinetic energy per molecule of the gas. Hence, its temperature increases (the average kinetic energy per molecule being proportional to the absolute temperature of the gas).

Question 4.
Explain, on the basis of kinetic theory, how the pressure of gas changes if its volume is reduced at constant temperature.
Answer:
The average kinetic energy per molecule of a gas is constant at constant temperature. When the volume of a gas is reduced at constant temperature, the number of collisions of gas molecules per unit time with the walls of the container increases. This increases the momentum transferred per unit time per unit area, i.e., the force exerted by the gas on the walls. Hence, the pressure of the gas increases.

Question 5.
Mention the conditions under which a real gas obeys ideal gas equation.
Answer:
A real gas obeys ideal gas equation when temperature is vey high and pressure is very low.
[ Note : Under these conditions, the density of a gas is very low. Hence, the molecules, on an average, are far away from each other. The intermolecular forces are then not of much consequence. ]

Question 6.
State the law of equipartition of energy and hence calculate molar specific heat of mono-and di-atomic gases at constant volume and constant pressure.
Answer:
Law of equipartition of energy : For a gas in thermal equilibrium at absolute temperature T, the average energy for a molecule, associated with each quadratic term (each degree of freedom), is \(\frac{1}{2}\)k_BT,
where k_B is the Boltzmann constant. OR

The energy of the molecules of a gas, in thermal equilibrium at a thermodynamic temperature T and containing large number of molecules, is equally divided among their available degrees of freedom, with the energy per molecule for each degree of freedom equal to \(\frac{1}{2}\)k_BT, where k_B is the Boltzmann constant.
(a) Monatomic gas : For a monatomic gas, each atom has only three degrees of freedom as there can be only translational motion. Hence, the average energy per atom is \(\frac{3}{2}\)k_BT. The total internal energy per mole of the gas is E = \(\frac{3}{2}\)N_Ak_BT, where N_A is the Avogadro
number.
Therefore, the molar specific heat of the gas at constant volume is
C_V = \(\frac{d E}{d T}\) = \(\frac{3}{2}\)N_Ak_B = \(\frac{3}{2}\)R,
where R is the universal gas constant.
Now, by Mayer’s relation, C_p — C_v = R, where C_p is the specific heat of the gas at constant pressure.
∴ C_P = C_V + R = \(\frac{3}{2}\)R + R = \(\frac{5}{2}\)R

(b) Diatomic gas : Treating the molecules of a diatomic gas as rigid rotators, each molecule has three translational degrees of freedom and two rotational degrees of freedom. Hence, the average energy per molecule is
3(\(\frac{1}{2}\)k_BT) + 2(\(\frac{1}{2}\)k_BT) = \(\frac{5}{2}\)k_BT
The total internal energy per mole of the gas is
E = \(\frac{5}{2}\)N_Ak_BT.
∴ C_V = \(\frac{d E}{d T}\) = \(\frac{5}{2}\)N_Ak_B = \(\frac{5}{2}\)R and
C_P = C_V + R = \(\frac{5}{2}\)R + R = \(\frac{7}{2}\)R
A soft or non-rigid diatomic molecule has, in addition, one frequency of vibration which contributes two quadratic terms to the energy.
Hence, the energy per molecule of a soft diatomic molecule is

Therefore, the energy per mole of a soft diatomic molecule is
E = \(\frac{7}{2}\)k_BT × N_A = \(\frac{7}{2}\) RT
In this case, C_V = \(\frac{d E}{d T}\) = \(\frac{7}{2}\)R and
C_P = C_V + R = \(\frac{7}{2}\)R + R = \(\frac{9}{2}\)R
[Note : For a monatomic gas, adiabatic constant,
γ = \(\frac{C_{p}}{C_{V}}\) = \(\frac{5}{3}\). For a diatomic gas, γ =\(\frac{7}{5}\) or \(\frac{9}{7}\).]

Question 7.
What is a perfect blackbody ? How can it be realized in practice?
Answer:
A perfect blackbody or simply a blackbody is defined as a body which absorbs all the radiant energy incident on it.

Fery designed a spherical blackbody which consists of a hollow double-walled, metallic sphere provided with a tiny hole or aperture on one side, in below figure. The inside wall of the sphere is blackened with lampblack while the outside is silver-plated. The space between the two walls is evacuated to minimize heat loss by conduction and convection.

Any radiation entering the sphere through the aperture suffers multiple reflections where about 97% of it is absorbed at each incidence by the coating of lampblack. The radiation is almost completely absorbed after a number of internal reflections. A conical projection on the inside wall opposite the hole minimizes probability of incident radiation escaping out.

When the sphere is placed in a bath of suitable fused salts, so as to maintain it at the desired temperature, the hole serves as a source of black-body radiation. The intensity and the nature of the radiation depend only on the temperature of the walls.

A blackbody, by definition, has coefficient of absorption equal to 1. Hence, its coefficient of reflection and coefficient of transmission are both zero.

The radiation from a blackbody, called blackbody radiation, covers the entire range of the electromagnetic spectrum. Hence, a blackbody is called a full radiator.

Question 8.
State (i) Stefan-Boltmann law and
(ii) Wein’s displacement law.
Answer:
(i) The Stefan-Boltzmann law : The rate of emission of radiant energy per unit area or the power radiated per unit area of a perfect blackbody is directly proportional to the fourth power of its absolute temperature. OR
The quantity of radiant energy emitted by a perfect blackbody per unit time per unit surface area of the body is directly proportional to the fourth power of its absolute temperature.

(ii) Wien’s displacement law : The wavelength for which the emissive power of a blackbody is maximum, is inversely proportional to the absolute temperature of the blackbody.
OR
For a blackbody at an absolute temperature T, the product of T and the wavelength λ_m corresponding to the maximum radiation of energy is a constant.
λ_mT = b, a constant.
[Notes: (1) The law stated above was stated by Wilhelm Wien (1864-1928) German Physicist. (2) The value of the constant b in Wien’s displacement law is 2.898 × 10^-3 m.K.]

Question 9.
Explain spectral distribution of blackbody radiation.
Answer:
Blackbody radiation is the electromagnetic radiation emitted by a blackbody by virtue of its temperature. It extends over the whole range of wavelengths of electromagnetic waves. The distribution of energy over this entire range as a function of wavelength or frequency is known as the spectral distribution of blackbody radiation or blackbody radiation spectrum.

If R_λ is the emissive power of a blackbody in the wavelength range λ and λ + dλ, the energy it emits per unit area per unit time in this wavelength range depends on its absolute temperature T, the wavelength λ and the size of the interval dλ.

Question 10.
State and prove Kirchoff’s law of heat radiation.
Answer:
Kirchhoff’s law of heat radiation : At a given temperature, the ratio of the emissive power to the coefficient of absorption of a body is equal to the emissive power of a perfect blackbody at the same temperature for all wavelengths.
OR
For a body emitting and absorbing thermal radiation in thermal equilibrium, the emissivity is equal to its absorptivity.

Theoretical proof: Consider the following thought experiment: An ordinary body A and a perfect black body B are enclosed in an athermanous enclosure as shown in below figure.

According to Prevost’s theory of heat exchanges, there will be a continuous exchange of radiant energy between each body and its surroundings. Hence, the two bodies, after some time, will attain the same temperature as that of the enclosure.

Let a and e be the coefficients of absorption and emission respectively, of body A. Let R and R_b be the emissive powers of bodies A and B, respectively. ;

Suppose that Q is the quantity of radiant energy incident on each body per unit time per unit surface area of the body.

Body A will absorb the quantity aQ per unit time per unit surface area and radiate the quantity R per unit time per unit surface area. Since there is no change in its temperature, we must have,
aQ = R … (1)
As body B is a perfect blackbody, it will absorb the quantity Q per unit time per unit surface area and radiate the quantity R_b per unit time per unit surface area.

Since there is no change in its temperature, we must have,
Q = R_b ….. (2)
From Eqs. (1) and (2), we get,
a = \(\frac{R}{Q}\) = \(\frac{R}{R_{\mathrm{b}}}\) ….. (3)
From Eq. (3), we get, \(\frac{R}{a}\) = R_b OR

By definition of coefficient of emission,
\(\frac{R}{R_{\mathrm{b}}}\) …(4)
From Eqs. (3) and (4), we get, a = e.
Hence, the proof of Kirchhoff ‘s law of radiation.

Question 11.
Calculate the ratio of mean square speeds of molecules of a gas at 30 K and 120 K. [Ans: 1:4]
Answer:
Data : T₁ = 30 K, T₂ = 120 K

This is the required ratio.

Question 12.
Two vessels A and B are filled with same gas where volume, temperature and pressure in vessel A is twice the volume, temperature and pressure in vessel B. Calculate the ratio of number of molecules of gas in vessel A to that in vessel B.
[Ans: 2:1]
Answer:
Data : V_A = 2V_B, T_A = 2T_B, P_A = 2P_B PV = Nk_BT

This is the required ratio.

Question 13.
A gas in a cylinder is at pressure P. If the masses of all the molecules are made one third of their original value and their speeds are doubled, then find the resultant pressure. [Ans: 4/3 P]
Answer:
Data : m₂ = m₁/3, v_{rms 2} = 2v_{rms 1} as the speeds of all molecules are doubled

This is the resultant pressure.

Question 14.
Show that rms velocity of an oxygen molecule is \(\sqrt{2}\) times that of a sulfur dioxide molecule at S.T.P.
Answer:

Question 15.
At what temperature will oxygen molecules have same rms speed as helium molecules at S.T.P.? (Molecular masses of oxygen and helium are 32 and 4 respectively)
[Ans: 2184 K]
Answer:
Data : T₂ = 273 K, M₀₁ (oxygen) = 32 × 10^-3 kg/mol, M₀₂ (hydrogen) = 4 × 10^{– 3} kg/mol
The rms speed of oxygen molecules, v₁ = \(\sqrt{\frac{3 R T_{1}}{M_{01}}}\) and that of helium molecules, v₂ = \(\sqrt{\frac{3 R T_{2}}{M_{02}}}\)
When v₁ = v₂,

Question 16.
Compare the rms speed of hydrogen molecules at 127 ºC with rms speed of oxygen molecules at 27 ºC given that molecular masses of hydrogen and oxygen are 2 and 32 respectively. [Ans: 8: 3]
Answer:
Data : M₀₁ (hydrogen) = 2 g/mol,
M₀₂ (oxygen) = 32 g/mol,
T₁ (hydrogen) = 273 + 127 = 400 K,
T₂ (oxygen) = 273 + 27 = 300 K
The rms speed, v_rms = \(\sqrt{\frac{3 R T}{M_{0}}}\),
where M₀ denotes the molar mass

Question 17.
Find kinetic energy of 5000 cc of a gas at S.T.P. given standard pressure is 1.013 × 10⁵ N/m².
[Ans: 7.598 × 10² J]
Answer:
Data : P = 1.013 × 10⁵ N/m², V = 5 litres
= 5 × 10^-3 m³
E = \(\frac{3}{2}\)PV
= \(\frac{3}{2}\)(1.013 × 10⁵ N/m²) (5 × 10^-3 m³)
= 7.5 × 1.013 × 10² J = 7.597 × 10² J
This is the required energy.

Question 18.
Calculate the average molecular kinetic energy
(i) per kmol
(ii) per kg
(iii) per molecule of oxygen at 127 ºC, given that molecular weight of oxygen is 32, R is 8.31 J mol^-1 K^-1 and Avogadro’s number N_A is 6.02 × 10²³ molecules mol^-1.
[Ans: 4.986 × 10⁶J, 1.558 × 10²J 8.282 × 10^-21 J]
Answer:
Data : T = 273 +127 = 400 K, molecular weight = 32 ∴ molar mass = 32 kg/kmol, R = 8.31 Jmol^-1 K^-1, N_A = 6.02 × 10²³ molecules mol^-1

(i) The average molecular kinetic energy per kmol of oxygen = the average kinetic energy per mol of oxygen × 1000
= \(\frac{3}{2}\)RT × 1000 = \(\frac{3}{2}\) (8.31) (400) (10³)\(\frac{\mathrm{J}}{\mathrm{kmol}}\)
= (600)(8.31)(10³) = 4.986 × 10⁶ J/kmol

(ii) The average molecular kinetic energy per kg of

(iii) The average molecular kinetic energy per molecule of oxygen

Question 19.
Calculate the energy radiated in one minute by a blackbody of surface area 100 cm² when it is maintained at 227ºC. (Take Stefen’s constant σ = 5.67 × 10^-8J m^-2s^-1K^-4)
[Ans: 2126.25 J]
Answer:
Data : t = one minute = 60 s, A = 100 cm²
= 100 × 10^-4 m² = 10^-2 m², T = 273 + 227 = 500 K,
σ = 5.67 × 10^-8 W/m².K⁴
The energy radiated, Q = σAT⁴t
= (5.67 × 10^-8)(10^-2)(500)⁴(60) J
= (5.67)(625)(60)(10^-2)J = 2126 J

Question 20.
Energy is emitted from a hole in an electric furnace at the rate of 20 W, when the temperature of the furnace is 727 ºC. What is the area of the hole? (Take Stefan’s constant σ to be 5.7 × 10^-8 J s^-1 m^-2 K^-4) [Ans: 3.509 × 10^-4m²]
Answer:
Data : \(\frac{Q}{t}\) = 20 W, T = 273 + 727 = 1000 K

Question 21.
The emissive power of a sphere of area 0.02 m² is 0.5 kcal s^-1 m^-2. What is the
amount of heat radiated by the spherical surface in 20 second? [Ans: 0.2 kcal]
Answer:
Data : R = 0.5 kcal s^-1m^-2, A = 0.02 m², t = 20 s Q = RAt = (0.5) (0.02) (20) = 0.2 kcal
This is the required quantity.

Question 22.
Compare the rates of emission of heat by a blackbody maintained at 727ºC and at 227ºC, if the black bodies are surrounded
by an enclosure (black) at 27ºC. What would be the ratio of their rates of loss of heat ?
[Ans: 18.23:1]
Answer:
Data : T₁ = 273 + 727 = 1000 K, T₂ = 273 + 227 = 500 K, T₀ = 273 + 27 = 300 K.
(i) The rate of emission of heat, \(\frac{d Q}{d t}\) = σAT⁴.
We assume that the surface area A is the same for the two bodies.

Question 23.
Earth’s mean temperature can be assumed to be 280 K. How will the curve of blackbody radiation look like for this temperature? Find out λ_max. In which part of the electromagnetic spectrum, does this value lie? (Take Wien’s constant b = 2.897 × 10^-3 m K) [Ans: 1.035 × 10^-5m, infrared region]
Answer:
Data : T = 280 K, Wien’s constant b = 2.897 × 10^-3 m.K
λ_maxT = b

This value lies in the infrared region of the electromagnetic spectrum.
The nature of the curve of blackbody radiation will be the same as shown in above, but the maximum will occur at 1.035 × 10^-5 m.

Question 24.
A small-blackened solid copper sphere of radius 2.5 cm is placed in an evacuated chamber. The temperature of the chamber is maintained at 100 ºC. At what rate energy must be supplied to the copper sphere to maintain its temperature at 110 ºC? (Take Stefan’s constant σ to be 5.670 × 10^-8 J s^-1 m^-2 K^-4 , π = 3.1416 and treat the sphere as a blackbody.)
[Ans: 0.9624 W]
Answer:
Data : r = 2.5 cm = 2.5 × 10^-2 m, T₀ = 273 + 100 = 373 K, T = 273 + 110 = 383 K,
σ = 5.67 × 10^-8 J s^-1 m^-2 k^-4
The rate at which energy must be supplied
σA(T⁴ — T₀⁴) = σ 4πr²(T⁴ – T₀⁴)
= (5.67 × 10^-8) (4) (3.142) (2.5 × 10^-2)² (383⁴ – 373⁴)
= (5.67) (4) (3.142) (6.25) (3.83⁴ – 3.73⁴) × 10^-4
= 0.9624W

Question 25.
Find the temperature of a blackbody if its spectrum has a peak at (a) λ_max = 700 nm (visible), (b) λ_max = 3 cm (microwave region) and (c) λ_max = 3 m (short radio waves) (Take Wien’s constant b = 2.897 × 10^-3 m K). [Ans: (a) 4138 K, (b) 0.09657 K, (c) 0.9657 × 10^-3 K]
Answer:
Data:(a) λ_max = 700nm=700 × 10^-9m,
(b) λ_max = 3cm = 3 × 10^-2 m, (c) λ_max = 3 m, b = 2.897 × 10^-3 m.K
λ_maxT = b

12th Physics Digest Chapter 3 Kinetic Theory of Gases and Radiation Intext Questions and Answers

Remember This (Textbook Page No. 60)

Question 1.
Distribution of speeds of molecules of a gas.
Answer:
Maxwell-Boltzmann distribution of molecular speeds is a relation that describes the distribution of speeds among the molecules of a gas at a given temperature.

The root-mean-square speed v_rms gives us a general idea of molecular speeds in a gas at a given temperature. However, not all molecules have the same speed. At any instant, some molecules move slowly and some very rapidly. In classical physics, molecular speeds may be considered to cover the range from 0 to ∞. The molecules constantly collide with each other and with the walls of the container and their speeds change on collisions. Also the number of molecules under consideration is very large statistically. Hence, there is an equilibrium distribution of speeds.

If dN_v represents the number of molecules with speeds between v and v + dv, dN_v remains fairly constant at equilibrium. We consider a gas of total N molecules. Let r\vdv be the probability that a molecule has its speed between v and v + dv. Then, dN_v = Nη_vdv
so that the fraction, i.e., the relative number of molecules with speeds between v and v + dv is dN_v/N = η_vdv
Below figure shows the graph of η_v against v. The area of the strip with height η_v and width dv gives the fraction dN_v/N.

Remember This (Textbook Page No. 64)

Question 1.
If a hot body and a cold body are kept in vacuum, separated from each other, can they exchange heat ? If yes, which mode of transfer of heat causes change in their temperatures ? If not, give reasons.
Answer:
Yes. Radiation.

Remember This (Textbook Page No. 66)

Question 1.
Can a perfect blackbody be realized in practice ?
Answer:
For almost all practical purposes, Fery’s blackbody is very close to a perfect blackbody.

Question 2.
Are good absorbers also good emitters ?
Answer:
Yes.

Use your brain power (Textbook Page No. 68)

Question 1.
Why are the bottom of cooking utensils blackened and tops polished ?
Answer:
The bottoms of cooking utensils are blackened to increase the rate of absorption of radiant energy and tops are polished to increase the reflection of radiation.

Question 2.
A car is left in sunlight with all its windows closed on a hot day. After some time it is observed that the inside of the car is warmer than outside air. Why ?
Answer:
The air inside the car is trapped and hence is a bad conductor of heat.

Question 3.
If surfaces of all bodies are continuously emitting radiant energy, why do they not cool down to 0 K ?
Answer:
Bodies absorb radiant energy from the surroundings.

Can you tell? (Textbook Page No. 71)

Question 1.
λ_max the wavelength corresponding to maximum intensity for the Sun is in the blue-green region of visible spectrum. Why does the Sun then appear yellow to us ?
Answer:
The colour that we perceive depends upon a number of factors such as absorption and scattering by atmosphere (which in turn depends upon the composition of air) and spectral response of the human eye. The colour may be yellow/orange/ red/white.

Balbharti Maharashtra State Board 12th Physics Textbook Solutions Chapter 15 Structure of Atoms and Nuclei Textbook Exercise Questions and Answers.

Maharashtra State Board 12th Physics Solutions Chapter 15 Structure of Atoms and Nuclei

In solving problems, use m_e = 0.00055 u = 0.5110 MeV/c², m_p = 1.00728 u, m_n = 1.00866u, m_H = 1.007825 u, u = 931.5 MeV, e = 1.602 × 10^-19 C, h = 6.626 × 10^-34 Js, ε₀ = 8.854 × 10^-12 SI units and me = 9.109 × 10^-31 kg.

1. Choose the correct option.

i) In which of the following systems will the radius of the first orbit of the electron be the smallest?
(A) hydrogen
(B) singly ionized helium
(C) deuteron
(D) tritium
Answer:
(D) tritium

ii) The radius of the 4th orbit of the electron will be smaller than its 8th orbit by a factor of
(A) 2
(B) 4
(C) 8
(D) 16
Answer:
(B) 4

iii) In the spectrum of hydrogen atom which transition will yield longest wavelength?
(A) n = 2 to n = 1
(B) n = 5 to n = 4
(C) n = 7 to n = 6
(D) n = 8 to n = 7
Answer:
(D) n = 8 to n = 7

iv) Which of the following properties of a nucleus does not depend on its mass number?
(A) radius
(B) mass
(C) volume
(D) density
Answer:
(D) density

v) If the number of nuclei in a radioactive sample at a given time is N, what will be the number at the end of two half-lives?
(A) \(\frac{N}{2}\)
(B) \(\frac{N}{4}\)
(C) \(\frac{3N}{4}\)
(D) \(\frac{N}{8}\)
Answer:
(B) \(\frac{N}{4}\)

2. Answer in brief.

i) State the postulates of Bohr’s atomic model.
Answer:
The postulates of Bohr’s atomic model (for the hydrogen atom) :

1. The electron revolves with a constant speed in acircular orbit around the nucleus. The necessary centripetal force is the Coulomb force of attraction of the positive nuclear charge on the negatively charged electron.
2. The electron can revolve without radiating energy only in certain orbits, called allowed or stable orbits, in which the angular momentum of the electron is equal to an integral multiple of h/2π, where h is Planck’s constant.
3. Energy is radiated by the electron only when it jumps from one of its orbits to another orbit having lower energy. The energy of the quantum of elec-tromagnetic radiation, i.e., the photon, emitted is equal to the energy difference of the two states.

ii) State the difficulties faced by Rutherford’s atomic model.
Answer:
(1) According to Rutherford, the electrons revolve in circular orbits around the atomic nucleus. The circular motion is an accelerated motion. According to the classical electromagnetic theory, an accelerated charge continuously radiates energy. Therefore, an electron during its orbital motion, should go on radiating energy. Due to the loss of energy, the radius of its orbit should go on decreasing. Therefore, the electron should move along a spiral path and finally fall into the nucleus in a very short time, of the order of 10^-16 s in the case of a hydrogen atom. Thus, the atom should be unstable. We exist because atoms are stable.

(2) If the electron moves along such a spiral path, the radius of its orbit would continuously decrease. As a result, the speed and frequency of revolution of the electron would go on increasing. The electron, therefore, would emit radiation of continuously changing frequency, and hence give rise to a con-tinuous spectrum. However, atomic spectrum is a line spectrum.

iii) What are alpha, beta and gamma decays?
Answer:
(a) A radioactive transformation in which an α-particle is emitted is called α-decay.
In an α-decay, the atomic number of the nucleus decreases by 2 and the mass number decreases by 4.
Example : \({ }_{92}^{238} \mathrm{U} \rightarrow{ }_{90}^{234} \mathrm{Th}+{ }_{2}^{4} \alpha\)
Q = [m_u – m_Th – m_α]c²

(b) A radioactive transformation in which a β-particle is emitted is called β-decay.
In a β^–-decay, the atomic number of the nucleus increases by 1 and the mass number remains unchanged.
Example : \({ }_{90}^{234} \mathrm{Th} \rightarrow{ }_{91}^{234} \mathrm{~Pa}+{ }_{-1}^{0} e+\bar{v}_{\mathrm{e}}\)
where \(\bar{v}_{\mathrm{e}}\) is the neutrino emitted to conserve the momentum, energy and spin.
Q = [m_u – m_Th – m_α]c²
In a β⁺-decay, the atomic number of the nucleus decreases by 1 and the mass number remains unchanged.
Example : \(\begin{aligned}
&30 \\
&15
\end{aligned} P \rightarrow{ }_{14}^{30} \mathrm{Si}+{ }_{+1}^{0} e+v_{\mathrm{e}}\)
where v_e is the neutrino emitted to conserve the momentum, energy and spin.
Q = [m_P – m_Si – m_e]c²
[Note : The term fi particle refers to the electron (or positron) emitted by a nucleus.]

A given nucleus does not emit α and β-particles simultaneously. However, on emission of α or β-particles, most nuclei are left in an excited state. A nucleus in an excited state emits a γ-ray photon in a transition to the lower energy state. Hence, α and β-particle emissions are often accompanied by γ-rays.

iv) Define excitation energy, binding energy and ionization energy of an electron in an atom.
Answer:
(1) Excitation energy of an electron in an atom : The energy required to transfer an electron from the ground state to an excited state (a state of higher energy) is called the excitation energy of the electron in that state.

(2) Binding energy of an electron in an atom is defined as the minimum energy that should be provided to an orbital electron to remove it from the atom such that its total energy is zero.

(3) Ionization energy of an electron in an atom is defined as the minimum energy required to remove the least strongly bound electron from a neutral atom such that its total energy is zero.

v) Show that the frequency of the first line in Lyman series is equal to the difference between the limiting frequencies of Lyman and Balmer series.
Answer:
For the first line in the Lyman series,
\(\frac{1}{\lambda_{\mathrm{L} 1}}=R\left(\frac{1}{1^{2}}-\frac{1}{2^{2}}\right)=R\left(1-\frac{1}{4}\right)=\frac{3 R}{4}\)
∴ v_L1 = \(\frac{c}{\lambda_{\mathrm{L} 1}}=\frac{3 R_{c}}{4}\), where v denotes the frequency,
c the speed of light in free space and R the Rydberg constant.
For the limit of the Lyman series,

Hence, the result.

Question 3.
State the postulates of Bohr’s atomic model and derive the expression for the energy of an electron in the atom.
Answer:
The postulates of Bohr’s atomic model (for the hydrogen atom) :
(1) The electron revolves with a constant speed in acircular orbit around the nucleus. The necessary centripetal force is the Coulomb force of attraction of the positive nuclear charge on the negatively charged electron.
(2) The electron can revolve without radiating energy only in certain orbits, called allowed or stable orbits, in which the angular momentum of the electron is equal to an integral multiple of h/2π, where h is Planck’s constant.
(3) Energy is radiated by the electron only when it jumps from one of its orbits to another orbit having lower energy. The energy of the quantum of elec-tromagnetic radiation, i.e., the photon, emitted is equal to the energy difference of the two states.

Consider the electron revolving in the nth orbit around the nucleus of an atom with the atomic number Z. Let m and e be the mass and the charge of the electron, r the radius of the orbit and v the linear speed of the electron.

According to Bohr’s first postulate, centripetal force on the electron = electrostatic force of attraction exerted on the electron by the nucleus
∴ \(\frac{m v^{2}}{r}=\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{Z e^{2}}{r^{2}}\) ……………. (1)
where ε₀ is the permittivity of free space.
∴ Kinetic energy (KE) of the electron
= \(\frac{1}{2} m v^{2}=\frac{Z e^{2}}{8 \pi \varepsilon_{0} r}\) ………….. (2)
The electric potential due to the nucleus of charge +Ze at a point at a distance r from it is
V = \(\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{\mathrm{Ze}}{r}\)
∴ Potential energy (PE) of the electron
= charge on the electron × electric potential
= – e × \(\frac{1}{4 \pi \varepsilon_{0}} \frac{Z e}{r}=-\frac{Z e^{2}}{4 \pi \varepsilon_{0} r}\) …………….. (3)
Hence, the total energy of the electron in the nth orbit is
E = KE + PE = \(\frac{-Z e^{2}}{4 \pi \varepsilon_{0} r}+\frac{Z e^{2}}{8 \pi \varepsilon_{0} r}\)
∴ E = \(-\frac{Z e^{2}}{8 \pi \varepsilon_{0} r}\) ………….. (4)
This shows that the total energy of the electron in the nth orbit of the atom is inversely proportional to the radius of the orbit as Z, ε₀ and e are constants. The radius of the nth orbit of the electron is
r = \(\frac{\varepsilon_{0} h^{2} n^{2}}{\pi m Z e^{2}}\) …………….. (5)
where h is Planck’s constant.
From Eqs. (4) and (5), we get,
E_n = \(-\frac{Z e^{2}}{8 \pi \varepsilon_{0}}\left(\frac{\pi m Z e^{2}}{\varepsilon_{0} h^{2} n^{2}}\right)=-\frac{m Z^{2} e^{4}}{8 \varepsilon_{0}^{2} h^{2} n^{2}}\) ……………… (6)
This gives the expression for the energy of the electron in the nth Bohr orbit. The minus sign in the expression shows that the electron is bound to the nucleus by the electrostatic force of attraction.
As m, Z, e, ε₀ and h are constant, we get
E_n ∝ \(\frac{1}{n^{2}}\)
i.e., the energy of the electron in a stationary energy state is discrete and is inversely proportional to the square of the principal quantum number.
[ Note : Energy levels are most conveniently expressed in electronvolt. Hence, substituting the values of m, e, £0 and h, and dividing by the conversion factor 1.6 × 10^-19 J/eV,
E_n ≅ \(-\frac{13.6 Z^{2}}{n^{2}}\) (in eV)
For hydrogen, Z = 1
∴ E_n ≅ \(-\frac{13.6}{n^{2}}\) (in eV).

Question 4.
Starting from the formula for energy of an electron in the n^th orbit of hydrogen atom, derive the formula for the wavelengths of Lyman and Balmer series spectral lines and determine the shortest wavelengths of lines in both these series.
Answer:
According to Bohr’s third postulate for the model of the hydrogen atom, an atom radiates energy only when an electron jumps from a higher energy state to a lower energy state and the energy of the
quantum of electromagnetic radiation emitted in this process is equal to the energy difference between the two states of the electron. This emission of radiation gives rise to a spectral line.

The energy of the electron in a hydrogen atom,
when it is in an orbit with the principal quantum
number n, is
E_n = \(-\frac{m e^{4}}{8 \varepsilon_{0}^{2} h^{2} n^{2}}\)
where m = mass of electron, e = electronic charge, h = Planck’s constant and = permittivity of free space.

Let E_m be the energy of the electron in a hydrogen atom when it is in an orbit with the principal quantum number m and E, its energy in an orbit with the principal quantum number n, n < m. Then
E_m = \(-\frac{m e^{4}}{8 \varepsilon_{0}^{2} h^{2} m^{2}}\) and E_n = \(-\frac{m e^{4}}{8 \varepsilon_{0}^{2} h^{2} m^{2}}\)
Therefore, the energy radiated when the electron jumps from the higher energy state to the lower energy state is
E_m – E_n = \(\frac{-m e^{4}}{8 \varepsilon_{0}^{2} h^{2} m^{2}}-\left(-\frac{m e^{4}}{8 \varepsilon_{0}^{2} h^{2} n^{2}}\right)\)
= \(\frac{m e^{4}}{8 \varepsilon_{0}^{2} h^{2}}\left(\frac{1}{n^{2}}-\frac{1}{m^{2}}\right)\)
This energy is emitted in the form of a quantum of radiation (photon) with energy hv, where V is the frequency of the radiation.
∴ E_m – E_n = hv
∴ v = \(\frac{E_{m}-E_{n}}{h}=\frac{m e^{4}}{8 \varepsilon_{0}^{2} h^{3}}\left(\frac{1}{n^{2}}-\frac{1}{m^{2}}\right)\)
The wavelength of the radiation is λ = \(\frac{c}{v^{\prime}}\)
where c is the speed of radiation in free space.
The wave number, \(\bar{v}=\frac{1}{\lambda}=\frac{v}{c}\)
\(\bar{v}=\frac{1}{\lambda}=\frac{m e^{4}}{8 \varepsilon_{0}^{2} h^{3} c}\left(\frac{1}{n^{2}}-\frac{1}{m^{2}}\right)=R\left(\frac{1}{n^{2}}-\frac{1}{m^{2}}\right)\)
where \(R\left(=\frac{m e^{4}}{8 \varepsilon_{0}^{2} h^{3} c}\right)\) is a constant called the Ryd berg constant.

This expression gives the wave number of the radiation emitted and hence that of a line in hydrogen spectrum.

For the Lyman series, n = 1,m = 2, 3, 4, ………… ∞
∴ \(\frac{1}{\lambda_{\mathrm{L}}}=R\left(\frac{1}{1^{2}}-\frac{1}{m^{2}}\right)\) and for the shortest wavelength line m this series, \(\frac{1}{\lambda_{\mathrm{Ls}}}=R\left(\frac{1}{1^{2}}\right)\) as m = ∞.
For the Balmer series, n = 2, m = 3, 4, 5, … ∞.
∴ \(\frac{1}{\lambda_{\mathrm{B}}}=R\left(\frac{1}{4}-\frac{1}{m^{2}}\right)\) and for the shortest wavelength line in this series, \(\frac{1}{\lambda_{\mathrm{Bs}}}=R\left(\frac{1}{4}\right)\) as m = ∞
[Note: Johannes Rydberg (1854—1919), Swedish spectroscopist, studied atomic emission spectra and introduced the idea of wave number. The empirical formula \(\bar{v}=\frac{1}{\lambda}=R\left(\frac{1}{n^{2}}-\frac{1}{m^{2}}\right)\) where m and n are simple integers, is due to Rydberg. When we consider the finite mass of the nucleus, we find that R varies slightly from element to element.]

Question 5.
Determine the maximum angular speed of an electron moving in a stable orbit around the nucleus of hydrogen atom.
Answer:
The radius of the ,ith Bohr orbit is
r = \(\frac{\varepsilon_{0} h^{2} n^{2}}{\pi m Z e^{2}}\) ………….. (1)
and the linear speed of an electron in this orbit is
ν = \(\frac{Z e^{2}}{2 \varepsilon_{0} n h}\) …………… (2)
where ε ₀ permittivity of free space, h ≡ Planck’s constant, n ≡ principal quantum number, m ≡ electron mass, e electronic charge and Z ≡ atomic number of the atom.
Since angular speed ω = \(\frac{v}{r}\), then from Eqs. (1) and (2), we get,
ω = \(\frac{v}{r}=\frac{Z e^{2}}{2 \varepsilon_{0} n h} \cdot \frac{\pi m Z e^{2}}{\varepsilon_{0} h^{2} n^{2}}=\frac{\pi m Z^{2} e^{4}}{2 \varepsilon_{0}^{2} h^{3} n^{3}}\) ………………. (3)
which gives the required expression for the angular speed of an electron in the nth Bohr orbit.
From Eq. (3), the frequency of revolution of the electron,
f = \(\frac{\omega}{2 \pi}=\frac{1}{2 \pi} \times \frac{\pi m Z^{2} e^{4}}{2 \varepsilon_{0}^{2} h^{3} n^{3}}=\frac{m Z^{2} e^{4}}{4 \varepsilon_{0}^{2} h^{3} n^{3}}\) …………….. (4)
as required.
[Note : From Eq. (4), the period of revolution of the electron, T = \(\frac{1}{f}=\frac{4 \varepsilon_{0}^{2} h^{3} n^{3}}{m Z e^{4}}\). Hence, f ∝ \(\frac{1}{n^{3}}\) and T ∝ n³].

Obtain the formula for ω and continue as follows :

This is required quantity.

Question 6.
Determine the series limit of Balmer, Paschen and Bracket series, given the limit for Lyman series is 912 Å.
Answer:
Data : λ_L∞ = 912 Å

as n = 5 and m = ∞
From Eqs. (1) and (2), we get,
\(\frac{\lambda_{\mathrm{Pa} \infty}}{\lambda_{\mathrm{L} \infty}}=\frac{R_{\mathrm{H}}}{R_{\mathrm{H}} / 9}\) = 9
∴ λ_Pa∞ = 9λ_L∞ = (9) (912) = 8202 Å
\(\frac{\lambda_{\mathrm{Pf} \infty}}{\lambda_{\mathrm{L} \infty}}=\frac{R_{\mathrm{H}}}{R_{\mathrm{H}} / 25}\) = 25
∴ λ_Pf∞ = 25λ_L∞ = (25) (912) = 22800 Å
This is the series limit of the pfund series.

Question 7.
Describe alpha, beta and gamma decays and write down the formulae for the energies generated in each of these decays.
Answer:
(a) A radioactive transformation in which an α-particle is emitted is called α-decay.
In an α-decay, the atomic number of the nucleus decreases by 2 and the mass number decreases by 4.
Example : \({ }_{92}^{238} \mathrm{U} \rightarrow{ }_{90}^{234} \mathrm{Th}+{ }_{2}^{4} \alpha\)
Q = [m_u – m_Th – m_α]c²

(b) A radioactive transformation in which a β-particle is emitted is called β-decay.
In a β^–-decay, the atomic number of the nucleus increases by 1 and the mass number remains unchanged.
Example : \({ }_{90}^{234} \mathrm{Th} \rightarrow{ }_{91}^{234} \mathrm{~Pa}+{ }_{-1}^{0} e+\bar{v}_{\mathrm{e}}\)
where \(\bar{v}_{\mathrm{e}}\) is the neutrino emitted to conserve the momentum, energy and spin.
Q = [m_u – m_Th – m_α]c²
In a β⁺-decay, the atomic number of the nucleus decreases by 1 and the mass number remains unchanged.
Example : \(\begin{aligned}
&30 \\
&15
\end{aligned} P \rightarrow{ }_{14}^{30} \mathrm{Si}+{ }_{+1}^{0} e+v_{\mathrm{e}}\)
where v_e is the neutrino emitted to conserve the momentum, energy and spin.
Q = [m_P – m_Si – m_e]c²
[Note : The term fi particle refers to the electron (or positron) emitted by a nucleus.]

A given nucleus does not emit α and β-particles simultaneously. However, on emission of α or β-particles, most nuclei are left in an excited state. A nucleus in an excited state emits a γ-ray photon in a transition to the lower energy state. Hence, α and β-particle emissions are often accompanied by γ-rays.

Question 8.
Explain what are nuclear fission and fusion giving an example of each. Write down the formulae for energy generated in each of these processes.
Answer:
Nuclear fission is a nuclear reaction in which a heavy nucleus of an atom, such as that of uranium, splits into two or more fragments of comparable size, either spontaneously or as a result of bombardment of a neutron on the nucleus (induced fission). It is followed by emission of two or three neutrons.
The mass of the original nucleus is more than the sum of the masses of the fragments. This mass difference is released as energy, which can be enormous as in the fission of ²³⁵U.
Nuclear fission was discovered by Lise Meitner, Otto Frisch, Otto Hahn and Fritz Strassmann in 1938.

The products of the fission of ²³⁵U by thermal neutrons are not unique. A variety of fission fragments are produced with mass number A ranging from about 72 to about 138, subject to the conservation of mass-energy, momentum, number of protons (Z) and number of neutrons (N). A few typical fission equations are

A type of nuclear reaction in which lighter atomic nuclei (of low atomic number) fuse to form a heavier nucleus (of higher atomic number) with the’ release of enormous amount of energy is called nuclear fusion.

Very high temperatures, of about 107 K to 108 K, are required to carry out nuclear fusion. Hence, such a reaction is also called a thermonuclear reaction.

Example : The D-T reaction, being used in experimental fusion reactors, fuses a deuteron and a triton nuclei at temperatures of about 10⁸ K.

(2) The value of the energy released in the fusion of two deuteron nuclei and the temperature at which the reaction occurs mentioned in the textbook are probably misprints.]

Question 9.
Describe the principles of a nuclear reactor. What is the difference between a nuclear reactor and a nuclear bomb?
Answer:
In a nuclear reactor fuel rods are used to provide a suitable fissionable material such as \(\begin{gathered}
236 \\
92
\end{gathered}\)U. Control rods are used to start or stop the reactor. Moderators are used to slow down the fast neutrons ejected in a nuclear fission to the appropriate lower speeds. Material used as a coolant removes the energy released in the nuclear reaction by converting it into thermal energy for production of electricity.

In a nuclear reactor, a nuclear fission chain reaction is used in a controlled manner, while in a nuclear bomb, the nuclear fission chain reaction is not controlled, releasing tremendous energy in a very short time interval.
[Note : The first nuclear bomb (atomic bomb) was dropped on Hiroshima in Japan on 06 August 1945. The second bomb was dropped on Nagasaki in Japan on 9 August 1945.]

Question 10.
Calculate the binding energy of an alpha particle given its mass to be 4.00151 u.
Answer:
Data : M = 4.00151 u, = 1.00728 u,
m_n = 1.00866 u, 1 u = 931.5 MeV/c²
The binding energy of an alpha particle
(Zm_p + N_n -M)c²
=(2m_p + 2m_n -M)c²
= [(2)(1.00728u) + 2(1.00866 u) – 4.00151 u]c²
= (2.01456 + 2.01732 – 4.00151)(931.5) MeV
= 28.289655 MeV
= 28.289655 × 10⁶ eV × 1.602 × 10^-19 J
= 4.532002731 × 10^-12 J

Question 11.
An electron in hydrogen atom stays in its second orbit for 10^-8 s. How many revolutions will it make around the nucleus in that time?
Answer:
Data : z = 1, m = 9.1 × 10^-31 kg, e = 1.6 × 10^-19 C, ε₀ = 8.85 × 10^-12 C² / N.m², h = 6.63 × 10 ^-34 J.s, n = 2, t = 10^-8 s
The periodic time of the electron in a hydrogen atom,

Let N be the number of revolutions made by the electron in time t. Then, t = NT.
∴ N = \(\frac{t}{T}=\frac{10^{-8}}{3.898 \times 10^{-16}}\) = 2.565 × ⁷

Question 12.
Determine the binding energy per nucleon of the americium isotope \(_{95}^{244} \mathrm{Am}\) , given the mass of \({ }_{95}^{244} \mathrm{Am}\) to be 244.06428 u.
Answer:
Data : Z = 95, N = 244  – 95 = 149,
m_p = 1.00728 u, m_n = 1.00866 u,
M = 244.06428 u, 1 u = 931.5 MeV/c²
The binding energy per nucleon,

= 7.3209 MeV/nucleon

Question 13.
Calculate the energy released in the nuclear reaction \({ }_{3}^{7} \mathrm{Li}\) + p → 2α given mass of \({ }_{3}^{7} \mathrm{Li}\) atom and of helium atom to be 7.016 u and 4.0026 u respectively.
Answer:
Data: M₁ (\({ }_{3}^{7} \mathrm{Li}\) Li atom)= 7.016 u, M₂ (He atom)
= 4.0026 u, m_p = 1.00728 u, 1 u = 931.5 MeV/c²
∆M = M₁ + m_p – 2M₂
= [7.016 + 1.00728 – 2(4.0026)]u
= 0.01808 u = (0.01808)(931.5) MeV/c²
= 16.84152 MeV/c²
Therefore, the energy released in the nuclear reaction = (∆M) c² = 16.84152 MeV

Question 14.
Complete the following equations describing nuclear decays.

Answer:
(a) \({ }_{88}^{226} \mathrm{Ra} \rightarrow{ }_{2}^{4} \alpha+{ }_{86}^{222} \mathrm{Em}\)
Em (Emanation) ≡ Rn (Radon)
Here, α particle is emitted and radon is formed.

(b) \({ }_{8}^{19} \mathrm{O} \rightarrow e^{-}+{ }_{9}^{19} \mathrm{~F}\)
Here, e^– ≡ \({ }_{-1}^{0} \beta\) is emitted and fluorine is formed.

(c) \(\underset{90}{228} \mathrm{Th} \rightarrow{ }_{2}^{4} \alpha+{ }_{88}^{224} \mathrm{Ra}\)
Here, α particle is emitted and radium is formed.

(d) \({ }_{7}^{12} \mathrm{~N} \rightarrow{ }_{6}^{12} \mathrm{C}+{ }_{1}^{0} \beta\)
\({ }_{1}^{0} \beta\) is e⁺ (positron)
Here, β⁺ is emItted and carbon is formed.

Question 15.
Calculate the energy released in the following reactions, given the masses to be \({ }_{88}^{223} \mathrm{Ra}\) : 223.0185 u, \({ }_{82}^{209} \mathrm{~Pb}\) : 208.9811, \({ }_{6}^{14} C\) : 14.00324, \({ }_{92}^{236} \mathrm{U}\) : 236.0456, \({ }_{56}^{140} \mathrm{Ba}\) : 139.9106, \({ }_{36}^{94} \mathrm{Kr}\) : 93.9341, \({ }_{6}^{11} \mathrm{C}\) : 11.01143, \({ }_{5}^{11} \mathrm{~B}\) : 11.0093. Ignore neutrino energy.

Answer:

(a) \({ }_{88}^{223} \mathrm{Ra} \rightarrow{ }_{82}^{209} \mathrm{~Pb}+{ }_{6}^{14} \mathrm{C}\)
The energy released in this reaction = (∆M) c²
= [223.0185 – (208.9811 + 14.00324)j(931 .5) MeV
= 31.820004 MeV

(b) \({ }_{92}^{236} \mathrm{U} \rightarrow{ }_{56}^{140} \mathrm{Ba}+{ }_{36}^{94} \mathrm{Kr}+2 \mathrm{n}\)
The energy released in this reaction =
(∆M) c² = [236.0456 – (139.9106 + 93.9341 + (2)(1 .00866)1(93 1 .5)MeV
= 171.00477 MeV

(c) \({ }_{6}^{11} \cdot \mathrm{C} \rightarrow{ }_{5}^{11} \mathrm{~B}+e^{+}\) + neutrino
The energy released in this reaction = (∆M) c²
= [11.01143 – (11.0093 + O.00055)](931.5) MeV
= 1.47177 MeV

Question 16.
Sample of carbon obtained from any living organism has a decay rate of 15.3 decays per gram per minute. A sample of carbon obtained from very old charcoal shows a disintegration rate of 12.3 disintegrations per gram per minute. Determine the age of the old sample given the decay constant of carbon to be 3.839 × 10^-12 per second.
Answer:
Data: 15.3 decays per gram per minute (living organism), 12.3 disintegrations per gram per minute (very old charcoal). Hence, we have,
\(\frac{A(t)}{A_{0}}=\frac{12.3}{15.3}\), λ = 3.839 × 10^-12 per second

Question 17.
The half-life of \({ }_{38}^{90} \mathrm{Sr}\) is 28 years. Determine the disintegration rate of its 5 mg sample.
Answer:
Data: T_1/2 = 28 years = 28 × 3.156 × 10⁷ s
=8.837 × 10⁸s, M = 5 mg =5 × 10^-3g
90 grams of \({ }_{38}^{90} \mathrm{Sr}\) contain 6.02 × 10²³ atoms
Hence, here, N = \(\frac{\left(6.02 \times 10^{23}\right)\left(5 \times 10^{-3}\right)}{90}\)
= 3.344 × 10¹⁹ atoms
∴ The disintegration rate = Nλ = N\(\frac{0.693}{T_{1 / 2}}\)
= \(\frac{\left(3.344 \times 10^{19}\right)(0.693)}{8.837 \times 10^{8}}\)
= 2.622 × 10¹⁰ disintegrations per second

Question 18.
What is the amount of \({ }_{27}^{60} \mathrm{Co}\) necessary to provide a radioactive source of strength 10.0 mCi, its half-life being 5.3 years?
Answer:
Data : Activity = 10.0 mCi = 10.0 × 10^-3 Ci
= (10.0 × 10^-3)(3.7 × 10¹⁰) dis/s = 3.7 × 10⁸ dis/s
T_1/2 = 5.3 years = (5.3)(3.156 × 10⁷) s
= 1.673 × 10⁸ s
Decay constant, λ = \(\frac{0.693}{T_{1 / 2}}=\frac{0.693}{1.673 \times 10^{8}} \mathrm{~s}^{-1}\)
=4.142 × 10^-9 s^-1
Activity = Nλ
∴ N = \(\frac{\text { activity }}{\lambda}=\frac{3.7 \times 10^{8}}{4.142 \times 10^{-9}} \text { atoms }\)
= 8.933 × 10¹⁶ atoms
=60 grams of \({ }_{27}^{60} \mathrm{Co}\) contain 6.02 × 10²³ atoms
Mass of 8.933 × 10¹⁶ atoms of \({ }_{27}^{60} \mathrm{Co}\)
= \(\frac{8.933 \times 10^{16}}{6.02 \times 10^{23}} \times 60 \mathrm{~g}\)
= 8.903 × 10^-6 g = 8.903 µg
This is the required amount.

Question 19.
Disintegration rate of a sample is 10¹⁰ per hour at 20 hrs from the start. It reduces to 6.3 × 10⁹ per hour after 30 hours. Calculate its half life and the initial number of radioactive atoms in the sample.
Answer:
Data : A (t₁) = 10¹⁰ per hour, where t₁ = 20 h,
A (t₂) = 6.3 × 10⁹ per hour, where t₂ = 30 h
A(t) = A₀e^-λt ∴ A(t₁) = A₀e^-λt₁ and A(t₂) = A_oe^-λt₂

∴ 1.587 e^10λ ∴ 10λ =2.3031og₁₀(1.587)
∴ λ = (0.2303)(0.2007) = 0.04622 per hour
The half life of the material, T_1/2 = \(\frac{0.693}{\lambda}=\frac{0.693}{0.04622}\)
= 14.99 hours
Now, A₀ = A (t₁)e^λt₁ = 10¹⁰e^{(0.04622)(20)}
= 10¹⁰ e^0.9244
Let x = e^0.9244 ∴ 2.3031og₁₀x = 0.9244
∴ 1og₁₀x = \(\frac{0.9244}{2.303}\) = 0.4014
∴ x = antilog 0.4014 = 2.52
∴ A₀ = 2.52 × 10¹⁰ per hour
Now A₀ = N₀λ ∴ N₀ = \(\frac{A_{0}}{\lambda}=\frac{2.52 \times 10^{10}}{0.04622}\)
= 5.452 × 10¹¹
This is the initial number of radioactive atoms in the sample.

Question 20.
The isotope ⁵⁷Co decays by electron capture to ⁵⁷Fe with a half-life of 272 d. The ⁵⁷Fe nucleus is produced in an excited state, and it almost instantaneously emits gamma rays.
(a) Find the mean lifetime and decay constant for ⁵⁷Co.
(b) If the activity of a radiation source ⁵⁷Co is 2.0 µCi now, how many ⁵⁷Co nuclei does the source contain?
(c) What will be the activity after one year?
Answer:
Data: T_1/2 = 272d = 272 × 24 × 60 × 60s = 2.35 × 10⁷ s,
A₀ = 2.0uCi = 2.0 × 10^-6 × 3.7 × 10¹⁰
= 7.4 × 10⁴ dis/s
t = 1 year = 3.156 × 10⁷ s
(a) T_1/2 = \(\frac{0.693}{\lambda}\) = 0.693 τ ∴ The mean lifetime for
⁵⁷Co = τ = \(\frac{T_{1 / 2}}{0.693}=\frac{2.35 \times 10^{7}}{0.693}\) = 3391 × 10⁷ s
The decay constant for ⁵⁷Co = λ = \(\frac{1}{\tau}\)
= \(\frac{1}{3.391 \times 10^{7} \mathrm{~s}}\)
= 2949 × 10^-8 s^-1

(b)A₀ = N₀A ∴ N₀ = \(\frac{A_{0}}{\lambda}\) = A₀τ
= (7.4 × 10⁴)(3.391 × 10⁷)
= 2.509 × 10¹² nuclei
This is the required number.

(c) A(t) = A₀e^-λt = 2e^{-(2.949 × 10-8)(3.156 × 107)}
= 2e^-0.9307 = 2 / e^0.9307
Let x = e^0.9307 ∴ Iog_ex = 0.9307
∴ 2.303log₁₀x = 0.9307
∴ log₁₀x = \(\frac{0.9307}{2.303}\) = 0.4041
∴ x = antilog 0.4041 = 2.536
∴ A (t) = \(\frac{2}{2.536}\) μCi = 0.7886 μCi

Question 21.
A source contains two species of phosphorous nuclei, \({ }_{15}^{32} \mathrm{P}\) (T_1/2 = 14.3 d) and \({ }_{15}^{32} \mathrm{P}\) (T_1/2 = 25.3 d). At time t = 0, 90% of the decays are from \({ }_{15}^{32} \mathrm{P}\) . How much time has to elapse for only 15% of the decays to be from \({ }_{15}^{32} \mathrm{P}\) ?
Answer:

∴ (0.04846 – 0.02739) t = 2.303 (2.1847 – 0.4771)
∴ t = \(\frac{(2.303)(1.7076)}{0.02107}\) = 186.6 days
This is the required time.

Question 22.
Before the year 1900 the activity per unit mass of atmospheric carbon due to the presence of ¹⁴C averaged about 0.255 Bq per gram of carbon. (a) What fraction of carbon atoms were ¹⁴C? (b) An archaeological specimen containing 500 mg of carbon, shows 174 decays in one hour. What is the age of the specimen, assuming that its activity per unit mass of carbon when the specimen died was equal to the average value of the air? Half-life of 14C is 5730 years?
Answer:
0.693
Data: T_1/2 = 5730y ∴ λ = \(\frac{0.693}{5730 \times 3.156 \times 10^{7}} \mathrm{~s}^{-1}\)
= 3.832 × 10^-12 s^-1, A = 0.255 Bq per gram of carbon in part (a); M = 500 mg = 500 × 10^-3 g,
174 decays in one hour \(\frac{174}{3600}\) dis/s = 0.04833 dis/s in part (b) (per 500 mg].

(a) A = Nλ ∴ N = \(\frac{A}{\lambda}=\frac{0.255}{3.832 \times 10^{-12}}\)
= 6.654 × 10¹⁰
Number of atoms in 1 g of carbon = \(\frac{6.02 \times 10^{23}}{12}\)
=5.017 × 10²²
\(\frac{5.017 \times 10^{22}}{6.654 \times 10^{10}}\) = 0.7539 × 10¹²
∴ 1 ¹⁴C atom per 0.7539 × 10¹² atoms of carbon
∴ 4 ¹⁴C atoms per 3 × 10¹² atoms of carbon

(b) Present activity per gram = \(\)
= 0.09666 dis/s per gram
A₀ = 0.255 dis/s per gram
Now, A(t) = A₀e^-λt

This is the required quantity.

Question 23.
How much mass of ²³⁵U is required to undergo fission each day to provide 3000 MW of thermal power? Average energy per fission is 202.79 MeV
Answer:
Data: Power = 3000 MW = 3 × 10⁹ J/s
∴ Energy to be produced each day
=3 × 10⁹ × 86400 J each day
= 2.592 × 10¹⁴ J each day
Energy per fission = 202.79 MeV
= 202.79 × 10⁶ × 1.6 × 10^-19 J = 3,245 × 10^-11 J
∴ Number of fissions each day
= \(\frac{2.592 \times 10^{14}}{3.245 \times 10^{-11}}\) × 10²⁴ each day
0.235 kg of ²³⁵U contains 6.02 × 10²³ atoms
7988 x 1024
∴ M = \(\left(\frac{7.988 \times 10^{24}}{6.02 \times 10^{23}}\right)\) (o.235) = 3.118 kg
This is the required quantity.

Question 24.
In a periodic table the average atomic mass of magnesium is given as 24.312 u. The average value is based on their relative natural abundance on earth. The three isotopes and their masses are \({ }_{12}^{24} \mathrm{Mg}\) (23.98504 u), \({ }_{12}^{25} \mathrm{Mg}\) (24.98584 u) and \({ }_{12}^{26} \mathrm{Mg}\) (25.98259 u). The natural abundance of \({ }_{12}^{24} \mathrm{Mg}\) is 78.99% by mass. Calculate the abundances of other two isotopes.
[Answer: 9.3% and 11.7%]
Answer:
Data : Average atomic mass of magnesium =

12th Physics Digest Chapter 15 Structure of Atoms and Nuclei Intext Questions and Answers

Use your brain power (Textbook Page No. 336)

Question 1.
Why don’t heavy nuclei decay by emitting a single proton or a single neutron?
Answer:
According to quantum mechanics, the probability for these emissions is extremely low.