CBSE Class 12 Physics Radiation Theory and Examples

This Class 12 Physics page covers the theory of heat radiation with solved examples: modes of heat transfer, nature and spectrum of radiation, interaction of radiation with matter, reflecting, absorbing and transmitting powers, emissive power, black body, Prevost's theory, Kirchhoff's law, Stefan's law, Newton's law of cooling, Wien's displacement law, the solar constant and the change of colour of a body with temperature.

Topics Covered

1. Heat transfer
2. Nature of radiation
3. Spectrum of electromagnetic waves
4. Interaction with matter
5. Definitions
6. Ideal black body
7. Laws of radiation
8. Stefan's law
9. Newton's law of cooling
10. Spectrum of radiations emitted by a black body
11. Solar constant
12. Change of colour of a substance with temperature

1. HEAT TRANSFER

The process of transmission of heat is by way of
(i) conduction (in solids)
(ii) convection (in liquids and gases)
(iii) radiation (through vacuum in the form of electromagnetic waves)

(i) Conduction - The process in which the material takes an active part by molecular action and energy is passed from one particle to another is called conduction. It is predominant in solids.

(ii) Convection - The transfer of energy by actual motion of particles of medium from one place to another is called convection. It is predominant in fluids (liquids and gases).

(iii) Radiation: Quickest way of transmission of heat is known as radiation. In this mode of energy transmission, heat is transferred from one place to another without affecting the intervening medium.

2. NATURE OF RADIATION

(i) Radiation is electromagnetic wave (0 to ∞ wavelength) and travels with velocity of light.
(ii) Radiation travels in straight line.
(iii) Radiation can travel through vacuum.
(iv) Radiation obeys the laws of reflection, refraction, interference, diffraction, polarization etc (all optical phenomena).
(v) Radiation does not affect the medium through which it travels.
(vi) For point source, radiation energy obeys inverse square law i.e.,
(a) For point source intensity at a point is inversely proportional to the square of the distance of point from the source: I ∝ \( \frac{1}{r^{2}} \)
(b) For line source, intensity at a point is inversely proportional to the distance of the point from the source: I ∝ \( \frac{1}{r} \)

3. SPECTRUM OF ELECTROMAGNETIC WAVE OR RADIATION

(i) Maxwell on the basis of his electromagnetic theory proved that all radiations are electromagnetic waves and their sources are vibrations of charged particles in atoms and molecules.
(ii) The spectrum of electromagnetic waves has an infinite spread and the different parts are given different names.

4. INTERACTION WITH MATTER

[Figure: A slab with an incident radiation arrow Q striking its upper surface. Three arrows leave or enter: Q_r is the reflected part going away from the upper surface, Q_a is the part absorbed inside the slab, and Q_t is the part transmitted out of the lower surface.]

(i) Due to incident radiations on the surface of a body following phenomena occur, by which the radiation is divided into three parts
(a) Reflection of radiation (b) Absorption of radiation (c) Transmission of radiation
(ii) If the reflected, absorbed and transmitted parts are Q_r, Q_a and Q_t respectively then from the law of conservation of energy
Q = Q_r + Q_a + Q_t
or \( \frac{Q}{Q} \) = \( \frac{Q_r}{Q} \) + \( \frac{Q_a}{Q} \) + \( \frac{Q_t}{Q} \)
Where Q is the total incident radiation.
1 = r + a + t
(iii) r = \( \frac{Q_r}{Q} \) → reflection coefficient or reflecting power
a = \( \frac{Q_a}{Q} \) → absorption coefficient or absorptive power
t = \( \frac{Q_t}{Q} \) → transmission coefficient or transmitting power
(iv) If a = 0, t = 0, then r = 1. Such a surface is a total or perfect reflector.
(v) If r = 0, t = 0, then a = 1. Such a surface is a total absorber and is called a black body.
(vi) If r = 0, a = 0, then t = 1. The surface is then perfectly transparent and the substance is called diathermanous.

5. DEFINITIONS

(i) Reflecting power or reflecting coefficient (r): Ratio of amount of radiation reflected by the surface to the amount of heat incident over it.
(ii) Transmitting power or transmitting coefficient (t): Ratio of amount of radiation transmitted by the surface to the amount of heat incident over it.
(iii) Absorption power or absorption coefficient (a): Ratio of amount of radiation absorbed by the surface to the amount of heat incident over it.

Note: (a) A radiation incident over a surface is either reflected, transmitted or absorbed, so a + t + r = 1
(b) The reflecting, refracting and transmitting powers of a surface depends on nature of surface as well as on the wavelength of radiation falling on them.

(iv) Emissive power (e): Emissive power is defined as the amount of heat radiated by unit area of the surface in one second at a particular temperature. Its unit is joule/m² or watt-sec/m².
(v) Spectral emissive power (e_λ): Amount of heat radiated by unit area of the body in one second in unit spectral region at a given wavelength. Its unit is watt/m²-Å. Its value depends on temperature, λ (wavelength and nature of surface).
(vi) Total emissive power (e): Emissive power obtained for total range of spectrum is related to total emissive power as follows: e = \( \int_0^{\infty} e_\lambda \, d\lambda \).
(vii) Spectral absorptive power: Like emissive power, the absorptive power of the surface also depends on the wavelength of radiation incident on it. It is defined as the ratio of radiation absorbed by the surface in unit spectral region at a given wavelength to the amount of radiation incident over it.

Note: Absorption power and spectral absorptive power for perfect black body is taken as one.

(viii) Emissivity or relative emissive power (E): Ratio of emissive power of a body at a given temperature to the emissive power of a black body at the same temperature. It has no units.

6. IDEAL BLACK BODY

(i) Total absorber is said to be ideal black body.
(ii) a_λ = 1
(iii) For lamp black and platinum black a_λ = 1, therefore both can be assumed to be ideal black body.
(iv) Ferry's black body is practical example of black body.
(v) For a black body, a = 1, t = 0 and r = 0.
(vi) When black body is heated, it radiates all the radiations absorbed by it.
(vii) The perfect black body is only an ideal conception. The body which completely absorbs the radiation of all wavelengths incident on it and there is no reflection and transmission is called a perfectly black body. Black body emits radiations of all wavelengths on heating and it is a good emitter. The nature of radiations emitted by the black body depends only on the temperature. It does not depend on its nature, size, mass etc.

7. LAWS OF RADIATION

(A) Prevost's theory of heat exchange

(i) Mutual exchange of heat between any object and its surroundings occurs according to their temperatures.
(ii) This process is continuous.
(iii) A body at high temperature emits more radiations and absorbs less. Therefore it experiences cooling. While the body at low temperature absorbs more heat and experiences heating.
(iv) At absolute zero temperature the radiated energy is zero.
(v) The rate of radiation of energy from a body depends upon its area, its temperature and the nature of surface. This does not depend upon temperature difference between body and its surroundings.
(vi) If temperature of body be T_B and temperature of surroundings be T_S.

(B) Kirchhoff's law

(i) The ratio of spectral emissive power (e) to spectral absorptive power (a) is the same for all bodies at a given temperature and is equal to the spectral emissive power of a blackbody at that temperature. Thus
\( \frac{E_\lambda \text{(body)}}{a_\lambda \text{(body)}} \) = E_λ (black body) for different objects -
\( \left(\frac{e_\lambda}{a_\lambda}\right)_1 \) = \( \left(\frac{e_\lambda}{a_\lambda}\right)_2 \) = \( \left(\frac{e_\lambda}{a_\lambda}\right)_3 \) = ...... = E_λ
(ii) Kirchhoff's law tells that if a body has a high emissive power, it should also have high absorptive power to have the E/a ratio same.
(iii) Black body is the best radiator and hence best absorber.
(iv) According to this law, the specific wavelength radiated at higher temperature are then absorbed by the body at low temperature.
(v) Kirchhoff's law successfully explains the concept of Fraunhofer's lines. When sunlight emitted by photosphere passes through chromosphere and is received on earth, then dark lines are obtained in its continuous spectrum. These lines can be explained in form of absorption spectrum. These are said to be Fraunhofer's lines.

8. STEFAN'S LAW

(i) Energy emitted per second by unit area of a black body at absolute temperature T is proportional to fourth power of absolute temperature: E ∝ T⁴ or E = σT⁴.
where,
σ = Stefan's constant
σ = 5.67 × 10⁻⁸ joule/m² second K⁴
(ii) Dimensions of Stefan's constant = [MT⁻³θ⁻⁴]
Net rate of loss of heat by radiation = amount of radiation emitted per second - amount of radiation absorbed per second
∴ ΔE = (E - E₀) = σ(T⁴ - T₀⁴)
where T is the temperature of the body and T₀ is temperature of the surroundings.
If the surface has an area A, the net rate of loss of heat by radiation will be -
\( \frac{dQ}{dt} \) = σA(T⁴ - T₀⁴) J/s
If A is the surface area of the body, e its emissivity, then amount of radiation emitted per second -
\( \frac{dQ}{dt} \) = σAe (T⁴ - T₀⁴) Joule
but dQ = msdθ cal
where m - mass of the body and s - specific heat in cal/kg-°C.
∴ ms \( \frac{d\theta}{dt} \) = \( \frac{\sigma A e}{J} \) (T⁴ - T₀⁴) cal/s
∴ Rate of fall of temperature = \( \frac{d\theta}{dt} \) = \( \frac{\sigma A e}{msJ} \) (T⁴ - T₀⁴) °C/s
(a) ratio of rates of cooling is -
\( \frac{dQ_1/dt}{dQ_2/dt} \) = \( \left(\frac{r_1}{r_2}\right)^{2} \) \( \left(\frac{T_1^{4} - T_0^{4}}{T_2^{4} - T_0^{4}}\right) \)
(b) rate of fall of temperature will be -
\( \frac{d\theta_1/dt}{d\theta_2/dt} \) = \( \left(\frac{r_2}{r_1}\right) \) \( \left(\frac{T_1^{4} - T_0^{4}}{T_2^{4} - T_0^{4}}\right) \)
(iii) Rate fall of temperature, then, is given by
\( \frac{d\theta}{dt} \) = \( \frac{\sigma A e (T^{4} - T_0^{4})}{msJ} \) °C/second
(iv) This law is valid only for energy radiated and not for total energy loss.
(v) This law is true for only ideal black body.

8.1 Special problem

(a) Two metallic spheres of radii r₁ and r₂ are kept at same temperature and same surroundings
(i) which will have higher rate of cooling? (r₁ is greater than r₂)
(ii) which will have higher rate of loss of temperature?

Answer: \( \frac{dQ}{dt} \) = \( \frac{\sigma A (T^{4} - T_0^{4})}{J} \), so the rate of cooling ∝ r²
⇒ sphere with radius r₁ will have higher rate of loss of energy or rate of cooling.
While \( \frac{d\theta}{dt} \) = \( \frac{\sigma A (T^{4} - T_0^{4})}{msJ} \) = \( \frac{\sigma A (T^{4} - T_0^{4})}{VdsJ} \), where V = volume
⇒ \( \frac{d\theta}{dt} \) ∝ \( \frac{1}{r} \) ⇒ sphere with radius r₂ will have higher rate of loss of temperature.
(b) If all of T, T₀, m, s, V, d are same, rate of loss of temperature or cooling will be maximum in the surface with maximum surface area.

EXAMPLES BASED ON STEFAN'S LAW

Ex.1 Two copper spheres A and B of the same surface finish are taken. Sphere B weighs half A. Both are heated to the same temperature and let in a room to cool by radiation only. What is the ratio of initial rate of loss of heat of A and B.
(A) 1.59
(B) 15.9
(C) .159
(D) 159

Answer:
1. It is given that m_B = \( \frac{1}{2} \) m_A. For spheres of the same material, m ∝ r³, so \( \frac{r_A}{r_B} \) = 2^1/3.
2. The rate of loss of heat by radiation is proportional to the surface area, so \( \frac{(dQ/dt)_A}{(dQ/dt)_B} \) = \( \frac{r_A^{2}}{r_B^{2}} \) = (2)^2/3 = 1.59.
3. Hence the correct option is (A).

Teacher's Note:
a) The temperature and surroundings are the same, so only the surface area differs.
b) Mass is proportional to the cube of the radius, but the radiated heat is proportional to the square of the radius.

Ex.2 A black body emits 10 watts per cm² at 327°C. The sun radiates 10⁵ watt per cm². Then what is the temperature of the sun?
(A) 5000K
(B) 6000K
(C) 7000K
(D) 8000K

Answer:
1. T_body = 327 + 273 = 600 K.
2. By Stefan's law, \( \frac{E_{sun}}{E_{body}} \) = \( \left(\frac{T_{sun}}{T_{body}}\right)^{4} \).
3. \( \frac{T_{sun}}{T_{body}} \) = \( \left(\frac{10^{5}}{10}\right)^{1/4} \) = 10.
4. T_sun = 10 × 600 = 6000 K.
5. Hence the correct option is (B).

Teacher's Note:
a) Always convert the Celsius temperature to kelvin before applying Stefan's law.
b) The ratio of the emissive powers is 10⁴, so the ratio of temperatures is 10.

Ex.3 A bulb made of tungsten filament of surface area 0.5 cm² is heated to a temperature 3000K when operated at 220V. The emissivity of the filament is ε = 0.35 and take σ = 5.7 × 10⁻⁸ mks units. Then the wattage of the bulb is ..... (calculate)
(A) 81W
(B) .81W
(C) 81.2W
(D) 8.12 W

Answer:
1. The emissive power in watt/m² is E = εσT⁴.
2. Therefore the power of the bulb is P = E × area (in watts) = εAσT⁴.
3. P = 0.35 × 0.5 × 10⁻⁴ × 5.7 × 10⁻⁸ × (3000)⁴ = 80.8 W.
4. Hence the correct option is (A).

Teacher's Note:
a) Convert the area from cm² to m² (0.5 cm² = 0.5 × 10⁻⁴ m²).
b) The calculated value 80.8 W is closest to 81 W.

Ex.4 The emissive power of a body at temperature T(K) is E. Then the graph between log_eE and log_eT is of the form
(A) graph A
(B) graph B
(C) graph C
(D) graph D

[Figure: Four graphs of log_eE (vertical) against log_eT (horizontal). Option (A): a straight line with negative slope falling from the upper left. Option (B): a curve rising slowly at first and then very steeply, concave upwards. Option (C): a straight line with positive slope rising from the lower left. Option (D): a curve rising steeply at first and then flattening, concave downwards.]

Answer:
1. For a body, the emissive power is given by E = εσT⁴.
2. Taking log on both sides, log_eE = log_e(εσ) + 4 log_eT.
3. Since log_e(εσ) is some constant value, this equation is of the form y = mx + c, which is the equation of a straight line with positive slope (slope 4).
4. Hence the correct option is (C).

Teacher's Note:
a) Comparing with y = mx + c, the slope is 4 and the intercept is log_e(εσ).
b) Any power law becomes a straight line on a log-log graph.

Ex.5 In the above example, if the temperature of the filament falls to 2000K due to a drop of mains voltage, then what will be the wattage of the bulb?
(A) 15W
(B) 16W
(C) 17W
(D) 18W

Answer:
1. The power of the bulb varies as T⁴, so \( \frac{P_2}{P_1} \) = \( \left(\frac{T_2}{T_1}\right)^{4} \).
2. P₂ = P₁ × \( \left(\frac{2}{3}\right)^{4} \) = 80.8 × \( \frac{16}{81} \).
3. P₂ = 15.96 W, that is, about 16 W.
4. Hence the correct option is (B).

Teacher's Note:
a) Use the earlier value of 80.8 W as P₁.
b) A fall of one-third in temperature reduces the power to about one-fifth.

9. NEWTON'S LAW OF COOLING

(1) Rate of cooling of a body -\( \frac{d\theta}{dt} \) is directly proportional to excess of temperature or -\( \frac{d\theta}{dt} \) ∝ (θ - θ₀)
Where θ₀ = temperature of surroundings
θ = temperature of body or -\( \frac{d\theta}{dt} \) = k (θ - θ₀)
(2) Let there be dθ loss of heat due to dQ loss of heat, then \( \frac{dQ}{dt} \) = ms \( \frac{d\theta}{dt} \)
⇒ ms \( \frac{d\theta}{dt} \) = -k (θ - θ₀)
or \( \frac{d\theta}{dt} \) = -\( \frac{k}{ms} \) (θ - θ₀)
(3) If temperature decreases from θ₁ to θ₂ in time t then \( \frac{\theta_2 - \theta_1}{t} \) = -\( \frac{k}{ms} \) \( \left(\frac{\theta_1 + \theta_2}{2} - \theta_0\right) \)
(4) Here, \( \frac{d\theta}{dt} \) ∝ \( \frac{1}{m} \) and \( \frac{d\theta}{dt} \) ∝ \( \frac{1}{s} \)

9.1 Limitation of Newton's Law

[Figure: Graph of dQ/dt (vertical axis) against (θ - θ₀) (horizontal axis): a straight line with positive slope starting near the origin.]

(a) Temperature difference should not exceed 35°C.
(b) Loss of heat should only be by radiation.
(c) This law is an extended form of Stefan-Boltzmann's law, therefore body should behave as black body.
\( \frac{dQ}{dt} \) = -K (θ - θ₀) or \( \frac{d\theta}{dt} \) = -\( \frac{K}{ms} \) (θ - θ₀)

Imp. Formula to find the specific heat of any liquid from this law (on basis of Newton's law of cooling)

[Figure: Graph of θ (vertical axis) against t (horizontal axis) with two falling curves, one for water and one for the liquid. Two levels θ₁ and θ₂ are marked on the vertical axis; the time to fall from θ₁ to θ₂ is t_w for water and t_l for the liquid.]

\( \frac{(W_l + m_l s)(\theta_1 - \theta_2)}{t_l} \) = \( \frac{(W_w + m_w)(\theta_1 - \theta_2)}{t_w} \)
or S = \( \frac{1}{m_l} \) \( \left[\frac{t_l}{t_w} (W_w + m_w) - W_l\right] \)
Suffix l and w stand for liquid and water respectively.
W_w: water equivalent of calorimeter with water
W_l: water equivalent of calorimeter with liquid

EXAMPLES BASED ON NEWTON'S LAW

Ex.6 0.15 kg of water is filled in a calorimeter of copper with water equivalent .01 kg. Its rate of cooling is 88 × 10⁻³ kcal/min. If xanthate oil is filled in place of water at the same temperature and of same volume. Now find the rate of cooling and rate of fall of temperature. (density of xanthate oil = 800 kg/m³, sp heat of xanthate oil = .42 kcal/kg°C)

Answer:
1. Both liquids are cooled under the same conditions, so the rate of cooling is the same: \( \frac{dQ}{dt} \) = 88 × 10⁻³ kcal/min.
2. The volume of 0.15 kg of water is 0.15 × 10⁻³ m³, so the mass of xanthate oil is m = 0.15 × 10⁻³ × 8 × 10² = 0.12 kg. The water equivalent is W = 0.01 kg.
3. The rate of cooling is \( \frac{dQ}{dt} \) = (ms + W) \( \frac{d\theta}{dt} \).
4. \( \frac{d\theta}{dt} \) = \( \frac{1}{ms + W} \) \( \frac{dQ}{dt} \) = \( \frac{0.088}{0.12 \times 0.42 + 0.01} \) = 1.457 °C/min.

Teacher's Note:
a) The rate of cooling (heat lost per minute) stays the same but the fall of temperature changes because the heat capacity changes.
b) Include the water equivalent of the calorimeter in the heat capacity.

Ex.7 Two liquids of same volume are cooled under same conditions from 65°C to 50°C. Time taken are 200 sec and 480 sec. If ratio of their specific heats is 2 : 3 then find the ratio of their densities. (neglect the water equivalent of calorimeter)

Answer:
1. From Newton's law of cooling, \( \frac{m_1 s_1 + w_1}{t_1} \) (θ₁ - θ₂) = \( \frac{m_2 s_2 + w_2}{t_2} \) (θ₁ - θ₂), with w₁ = w₂ = 0.
2. So \( \frac{m_1 s_1}{t_1} \) = \( \frac{m_2 s_2}{t_2} \), that is, \( \frac{V d_1 s_1}{t_1} \) = \( \frac{V d_2 s_2}{t_2} \).
3. \( \frac{d_1}{d_2} \) = \( \frac{t_1 s_2}{t_2 s_1} \) = \( \frac{200}{480} \) × \( \frac{3}{2} \) = \( \frac{5}{8} \).
4. Hence the ratio of the densities d₁ : d₂ is 5 : 8.

Teacher's Note:
a) The same volume means the mass is proportional to the density.
b) The temperature range and the surroundings are the same, so the heat lost per second is the same.

Ex.8 A metal ball cools from 62°C to 50°C in 10 min and to 42°C the next ten minutes. What will be its temperature at the end of next ten minutes?
(A) 3.67°C
(B) 36.7°C
(C) .376°C
(D) 367°C

Answer:
1. Use \( \frac{\theta_1 - \theta_2}{t} \) = K \( \left(\frac{\theta_1 + \theta_2}{2} - \theta_0\right) \).
2. First interval: \( \frac{62 - 50}{10} \) = K \( \left(\frac{62 + 50}{2} - \theta_0\right) \), that is, \( \frac{12}{10} \) = K (56 - θ₀) ... (1)
3. Second interval: \( \frac{50 - 42}{10} \) = K \( \left(\frac{50 + 42}{2} - \theta_0\right) \), that is, \( \frac{8}{10} \) = K (46 - θ₀) ... (2)
4. Dividing (2) by (1): \( \frac{8}{12} \) = \( \frac{46 - \theta_0}{56 - \theta_0} \), which gives θ₀ = 26°C.
5. From (1), K = \( \frac{12}{10 \times 30} \) = 0.04 per min.
6. Next ten minutes, with final temperature θ: \( \frac{42 - \theta}{10} \) = 0.04 \( \left(\frac{42 + \theta}{2} - 26\right) \), so \( \frac{42 - \theta}{12} \) = \( \frac{42 + \theta - 52}{2(56 - 26)} \).
7. Solving, 60 (42 - θ) = 12 (θ - 10), which gives θ = 36.7°C.
8. Hence the correct option is (B).

Teacher's Note:
a) Eliminate the unknown room temperature θ₀ first by dividing the two equations.
b) The average temperature of the interval is used in this form of Newton's law.
c) The printed question says "the next to minutes"; it means the next ten minutes.

Ex.9 A liquid takes 30 seconds to cool from 95°C to 90°C and 70 seconds to cool from 55 to 50°C. Find the room temperature and the time it will take to cool from 50°C to 45°C
(A) 81 sec
(B) 82 sec
(C) 83 sec
(D) 84 sec

Answer:
1. First data: \( \frac{95 - 90}{30} \) = k \( \left(\frac{95 + 90}{2} - \theta_0\right) \) ... (1)
2. Second data: \( \frac{55 - 50}{70} \) = k \( \left(\frac{55 + 50}{2} - \theta_0\right) \) ... (2)
3. Dividing (1) by (2): \( \frac{7}{3} \) = \( \frac{92.5 - \theta_0}{52.5 - \theta_0} \), which gives θ₀ = 22.5°C ... (3)
4. Let the time taken in cooling from 50°C to 45°C be t: \( \frac{50 - 45}{t} \) = k \( \left(\frac{50 + 45}{2} - \theta_0\right) \) ... (4)
5. Dividing (1) by (4) with θ₀ = 22.5°C: \( \frac{t}{30} \) = \( \frac{92.5 - 22.5}{47.5 - 22.5} \) = \( \frac{70}{25} \), so t = 84 s.
6. Hence the room temperature is 22.5°C, the time taken is 84 s, and the correct option is (D).

Teacher's Note:
a) The mean temperature of each interval is the average of its end temperatures.
b) Dividing two equations removes the constant k.
c) Check: k = \( \frac{5}{30 \times 70} \) = \( \frac{1}{420} \) per s, and \( \frac{5}{25k} \) = 84 s.

10. SPECTRUM OF RADIATIONS EMITTED BY BLACK-BODY

(i) To study this spectrum, prism of rock salt is used which absorbs minimum heat radiations otherwise prism of glass etc absorb heat radiations.
(ii) Spectral range of radiation energy emitted by black body depends on its temperature, not on shape, volume, mass etc.
(iii) At different temperature it gives different energy distributions as shown

[Figure: Graph of E_λ (vertical) against λ (horizontal) showing four curves for temperatures 1649 K, 1449 K, 1259 K and 1095 K. Each curve rises to a peak and falls; the peak is highest and nearest to the vertical axis for 1649 K, and lowest and farthest from the axis for 1095 K.]

[Figure: Graph of E_λ (vertical) against frequency ν (horizontal) showing three curves for T₁, T₂ and T₃ with T₃ greater than T₂ greater than T₁. The peaks lie at frequencies ν_m, ν_m2 and ν_m3, with the peak of the T₃ curve highest and at the highest frequency.]

(iv) We can conclude from the figure λ_m5 is less than λ_m4 is less than λ_m3 is less than λ_m2 is less than λ_m1.
(a) Some energy E_λ is always obtained at any temperature and at each wavelength.
(b) Maximum energy E_λmax is obtained at a particular wavelength λ_m at a constant temperature.
(c) Value of λ_m decreases with increase in temperature. This is called Wien's displacement Law.
(d) According to Wien's law, λ_m ∝ \( \frac{1}{T} \) or λ_mT = b, where b = Wien's constant, b = 0.293 cm K or b = 2.93 × 10⁻³ m K.
(e) At any temperature the area between curve and λ axis gives the total energy emitted at that temperature. This is in accordance with the fact that E ∝ T⁴.
(f) (E_λ)_max ∝ T⁵
(g) Wien's law in form of frequency can be written as ν_m ∝ T or \( \frac{\nu_m}{T} \) = \( \frac{c}{b} \), where c is velocity of light
⇒ \( \frac{\nu_{m1}}{T_1} \) = \( \frac{\nu_{m2}}{T_2} \)
(h) Graph of λ_m against T is a curve with λ_m ∝ \( \frac{1}{T} \).

[Figure: Graph of λ_m (vertical) against T (horizontal): a falling hyperbola-shaped curve approaching both axes, marked λ_m ∝ 1/T.]

EXAMPLES BASED ON WIEN'S LAW

Ex.10 A black body at 2000K emits radiation with λ_m = 1250nm. If for the radiation coming from the star SIRIUS λ_m is 71nm, then the temperature of this star is ....
(A) 3521.1K
(B) 352.11K
(C) 35.21K
(D) 35, 211K

Answer:
1. By Wien's displacement law, λ_mT is constant, so \( \frac{T_2}{T_1} \) = \( \frac{(\lambda_m)_1}{(\lambda_m)_2} \).
2. T₂ = \( \frac{2000 \times 1250 \times 10^{-9}}{71 \times 10^{-9}} \) = 35211 K.
3. Hence the temperature of the star is about 35,211 K, and the correct option is (D).

Teacher's Note:
a) A star with a smaller peak wavelength is hotter, so the answer must be larger than 2000 K.
b) Both wavelengths are in nanometres, so the conversion factors cancel.

Ex.11 When the temperature of the body is increased, the frequency corresponding to maximum emission, changes such that :
(A) \( \frac{\Delta V_m}{V_m} \) = \( \frac{\Delta T}{T} \)
(B) V_mΔV_m = TΔT
(C) ΔV_m = \( \frac{\Delta T}{T} \)
(D) ΔV_m = TΔT

Answer:
1. By Wien's law, the frequency corresponding to maximum emission is given by \( \frac{V_m}{T} \) = b', so V_m = b'T.
2. Hence ΔV_m = b'ΔT, and therefore \( \frac{\Delta V_m}{V_m} \) = \( \frac{\Delta T}{T} \).
3. Hence the correct option is (A).

Teacher's Note:
a) The peak frequency is directly proportional to the temperature.
b) The fractional change in the peak frequency equals the fractional change in temperature.

Ex.12 At 1600K maximum radiation is emitted at a wavelength of 2µm. Then the corresponding wavelength at 2000K will be -
(A) 1.6µm
(B) 16µm
(C) 160µm
(D) .16µm

Answer:
1. By Wien's law, λ_m1T₁ = λ_m2T₂.
2. λ_m2 = \( \frac{\lambda_{m1} T_1}{T_2} \) = \( \frac{2 \times 10^{-6} \times 1600}{2000} \) = 1.6 × 10⁻⁶ m = 1.6 µm.
3. Hence the correct option is (A).

Teacher's Note:
a) A higher temperature must give a shorter peak wavelength, so the answer is below 2 µm.
b) Keep the unit micrometre throughout to avoid errors.

11. SOLAR CONSTANT

The amount of normal incident heat radiations per minute on a unit area placed at average distance between sun and earth is said to be solar constant denoted by S.
S = \( \frac{\sigma T^{4} \, 4\pi R^{2}}{4\pi d^{2}} \) or S = \( \frac{\sigma T^{4} r^{2}}{d^{2}} \)
Unit: Joule/m²-second or watt/m²
S = 1.94 calorie/cm²-minute
R = radius of sun
≈ 2 calorie/cm²-minute = 1358 watt/m²
d = distance between sun and earth

EXAMPLES BASED ON SOLAR CONSTANT

Ex.13 Solar constant for earth is 2 cal/minute/cm². If the distance of the planet mercury from the sun is 0.40 times the distance of the earth from the sun, the Solar constant for the planet mercury in cal/minute/cm² will be -
(A) 12.5
(B) 25
(C) 0.32
(D) 2

Answer:
1. For earth, S = 2 = \( \left(\frac{R_s}{d}\right)^{2} \) σT⁴. For mercury, S' = \( \left(\frac{R_s}{d'}\right)^{2} \) σT⁴.
2. \( \frac{S'}{S} \) = \( \left(\frac{d}{d'}\right)^{2} \), with d' = 0.4 d.
3. S' = \( \frac{2}{(0.4)^{2}} \) = \( \frac{2}{0.16} \) = 12.5 cal/cm²/min.
4. Hence the correct option is (A).

Teacher's Note:
a) The solar constant follows the inverse square law with distance from the sun.
b) A planet closer to the sun receives more energy per unit area.

Ex.14 Earth receives energy at the rate of 2 cal/minute/cm² from sun. If angular diameter of sun is 32' and it is assumed as a black body, then the temperature of the surface of sun is
(A) 5723K
(B) 5773K
(C) 5800K
(D) 5925K

Answer:
1. S = \( \left(\frac{R_s}{d}\right)^{2} \) σT⁴, with S = \( \frac{2 \times 4.2 \times 10^{4}}{60} \) = 1400 J/m²/s.
2. \( \frac{R_s}{d} \) = \( \frac{1}{2} \) angular diameter of sun = 16' = \( \frac{16}{60} \) degree = \( \frac{16}{60} \) × \( \frac{\pi}{180} \) rad = 4.65 × 10⁻³ rad.
3. T⁴ = \( \frac{S}{\sigma \left(\frac{R_s}{d}\right)^{2}} \) = \( \frac{1400}{5.7 \times 10^{-8} \times (4.65 \times 10^{-3})^{2}} \) = 1.13 × 10¹⁵.
4. T ≈ 5803 K, that is, about 5800 K.
5. Hence the correct option is (C).

Teacher's Note:
a) The angle subtended by the radius of the Sun is half the angular diameter.
b) The printed working gives 5806 K with rounding; the result is about 5800 K in either case.
c) Convert 1 cal to 4.2 J and 1 cm² to 10⁻⁴ m².

12. CHANGE OF COLOUR OF A SUBSTANCE WITH TEMPERATURE

When a body is heated continuously, the maximum emitted wavelengths shift to right of electromagnetic spectrum. Change of colour will be as follows -