Black Body Radiation and Wien Displacement Law Questions

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Very Short Answers Questions [1 Mark Questions]

Ques. A black body and a yellow body at room temperature are thrown into a furnace at a very high temperature. Select the correct statement.

1. Initially, both are equally bright, and finally yellow is brighter
2. Initially black is brighter and finally both are equally bright
3. Initially yellow is brighter and finally, both are equally bright
4. Initially yellow is brighter and finally black is brighter.

Ans. The correct answer is c. Initially yellow is brighter and finally, both are equally bright.

Explanation: Initially both body absorbs heat, but the black body absorbs almost all heat of all wavelengths, and the yellow body reflects a yellow wavelength of light. So yellow body is initially brighter.

When the temperature of both bodies becomes equal to the temperature of the furnace, both bodies radiate the energy of all wavelengths. Hence finally both the bodies are equally bright.

Ques. As the wavelength of the radiation declines, the strength of the black body radiations_____

1. Decreases
2. Increases
3. First increases then decreases
4. First decreases then increases

Ans. The correct answer is c. First increases then decreases

Explanation: The wavelength of the emitted radiation decreases as the body gets hotter in the case of black body radiation. Though, the intensity up to a particular wavelength first increases and then starts decreasing.

Ques. The relationship at which the maximum value of monochromatic emissive power occurs, that is, the relation between the wavelength and temperature of a black body, λ_maxT = constant is termed as

1. Lambert’s law
2. Wien’s law
3. Planck’s law
4. Kirchhoff’s law

Ans. The correct answer is b. Wien’s law

Explanation: Wien’s law relates the wavelength of the maximum intensity of the emission to the absolute temperature.

Ques. Two spheres A and B of the same material have radius 2 m and 3 m, and temperatures 6000 K and 2000 K respectively. Then the energy radiated by sphere A is

1. Less than that of sphere B
2. Greater than that of sphere B
3. Two times that of sphere B
4. Equal to that of sphere B

Ans. The correct answer is b. Greater than that of sphere B

Explanation: According to Stefan’s law, the rate at which energy radiated from an object is proportional to the fourth power of absolute temperature. i.e. 

E/t = σAeT⁴

Where

• σ is Stefan’s Boltzmann constant
• A = 4πr² = surface area
• e= emissivity
• T = absolute temperature

⇒ E/t = σ(4πr²)eT⁴

Since the material is the same, therefore

E ∝ r²T⁴

⇒ E_A/E_B = (r_A/r_B)² (T_A/T_B)⁴

⇒ E_A/E_B = (2/3)²(6000/2000)⁴ = 34 = (4 x 81)/9 = 36

⇒ E_A = 36E_B

Hence, the energy radiated by sphere A is greater than that of sphere B.

Ques. A surface for which emissivity is constant at all temperatures and throughout the entire range of wavelength is called

1. Grey
2. Opaque
3. Diathemanous
4. Specular

Ans. The correct answer is a. Grey

Explanation: The emissivity of the Grey surface is almost constant. Grey surface radiates much more because of constant wavelength throughout its entire surface.

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Short Answers Questions [2 Marks Questions]

Ques. State Wien's displacement law.

Ans. According to Wien displacement law, The wavelength of maximum intensity (λm) of emission of black body radiation is inversely proportional to the absolute temperature (T) of the black body. i.e. 

λ_max ∝ 1/T

⇒ λ_max = b/T

Where b is Wien’s displacement constant = 2.898 x 10^-3 mK (meter Kelvin)

Ques. Four similar pieces of copper Black, Yellow, White, and Rough were heated at the same temperature and then left in the environment to cool. Also, these pieces were painted with different colors of paints. Which among the following paints will give fast cooling?

Ans. The emissivity of black paint is highest i.e. 1. Thus, when painted black the emitted radiant energy will be maximum. Hence, among the given pieces of copper, the black-colored copper will give fast cooling.

Ques. The wavelength of maximum solar emission is observed to be approximately 0.475 μm. What is the surface temperature of the sun (assumed as a blackbody)?

Ans. From Wien’s displacement law, we have

λ_m = b/T

Where

• λ_m is the wavelength of the maximum intensity of emission
• b = Wien’s displacement constant = 2.898 x 10^-3 mK
• T is the absolute temperature

Given the wavelength of maximum solar emission, λ_m = 0.475 μm = 0.475 x 10^-6 m

Therefore, the surface temperature of the sun

T = b/λ_m = (2.898 x 10^-3)/(0.475 x 10^-6) = 6.101 x 10³ K = 6101 K

⇒ T = 6101 - 273 = 5828 ℃

Ques. A furnace emits radiation at 2000 K. Treating it as a black body, calculate the wavelength at which emission is maximum.

Ans. From Wien’s displacement law,

Maximum wavelength, λ_m = b/T

⇒ λ_m = (2.898 x 10^-3)/(2000) = 1.449 x 10^-6 m = 1.449 µm

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Long Answers Questions [3 Marks Questions]

Ques. What are the processes of heat transfer?

Ans. The transfer of heat from one body to another body by the following processes

• Conduction: The process by which heat is transferred from the hot part of a body to the cold part through the transfer of energy from one particle to another particle of the body without the actual movement of the particles from their mean positions is called conduction.
• Convection: The process of heat transfer from one part of a fluid to another part by actual movement of the particles of the fluid is called convection.
• Radiation: The process of transfer of heat from one place to another place without heating the intervening medium is called radiation.

Ques. A body at 1227 ℃ emits radiation with maximum intensity at a wavelength of 5000 Å. If the temperature of the body is increased by 1000 ℃, then what is the wavelength of maximum intensity?

Ans. According to Wien’s displacement law, the wavelength of the maximum intensity of emission is inversely proportional to the absolute temperature. i.e.

λ_max ∝ 1/T

⇒ λ₁/λ₂ = T₂/T₁

⇒ λ₂ = λ₁T₁/T₂

Given

• λ₁ = 5000 Å
• T₁ = 1227 ℃ = 1227 + 273 = 1500 K
• T₂ = T₁ + 1000 ℃ = 1227 +1000 ℃ = 2227 ℃ = 2227 + 273 = 2500 K

⇒ λ₂ = (5000 x 1500)/2500

⇒ λ₂ = 3000 Å

Ques. A body at 1527 ℃ emits radiation with maximum intensity at a wavelength of 6000 Å. If the temperature of the body is increased by 2000 ℃, then what is the wavelength of maximum intensity?

Ans. According to Wien’s displacement law, the wavelength of the maximum intensity of emission is inversely proportional to the absolute temperature. i.e.

λ_max ∝ 1/T

⇒ λ₁/λ₂ = T₂/T₁

⇒ λ₂ = λ₁T₁/T₂

Given

• λ₁ = 6000 Å
• T₁ = 1527 ℃ = 1527 + 273 = 1800 K
• T₂ = T₁ + 2000 ℃ = 1527 + 2000 ℃ = 3527 ℃ = 3527 + 273 = 3800 K

⇒ λ₂ = (6000 x 1800)/3800

⇒ λ₂ = 2842 Å

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Very Long Answers Questions [5 Mark Questions]

Ques. The intensity of radiation emitted by the sun has its maximum value at a wavelength of 510 nm and that emitted by the north star has the maximum value at 350 nm. If these stars behave like black bodies, then what is the ratio of the surface temperature of the sun and the north star?

Ans. Given

• The wavelength of maximum intensity of radiation of the sun, λ₁ = 510 nm
• The wavelength of maximum intensity of radiation of the north star, λ₂ = 350 nm

According to Wien’s displacement law, the wavelength of the maximum intensity of emission is inversely proportional to absolute temperature. i.e.

λ_max ∝ 1/T

If T₁ is the surface temperature of the sun and T₂ is the surface temperature of the north star, then

λ₁/λ₂ = T₂/T₁

⇒ T₁/T₂ = λ₂/λ₁

⇒ T₁/T₂ = 350/510 = 0.69

Therefore, the ratio of the surface temperature of the sun and the north star is 0.69

Ques. What is Kirchhoff's law of radiation?

Ans. According to Kirchhoff’s law of radiation, at a given temperature, the ratio of the spectral emissive power of a body to its spectral absorptive power is constant and is equal to the spectral emissive power of a black body at the same temperature.

i.e. E/a = E_b

Where

• E = spectral emissive power
• a = spectral absorptive power
• E_b = spectral emissive power of a black body

But E/E_b = e (emissivity)

⇒ e = a

Therefore, at any given temperature, the emissivity of a body is equal to the spectral absorptive power.

Ques. The temperature and surface area of a body are 227 ℃ and 0.15 m² respectively. If its transmitting power is negligible and its reflecting power is 0.5, then what is the thermal power of the body?

Ans. We have, absorptance (a) + reflectance (r) + transmittance (t) = 1

Given

• Reflecting power or reflectance, r = 0.5
• Transmitting power or transmittance, t = 0

Absorptance, a = 1 - 0.5 - 0 = 0.5

From Kirchhoff’s law of radiation, a = e = 0.5

From Stefan’s law, the thermal power of the body is given by

P = AeσT⁴

Where

• A = surface area of the body = 0.15 m² (given)
• σ = Stefan-Boltzmann constant = 5.67 x 10^-8 W m^-2 K^-4
• T = absolute temperature

Given temperature of the body, T = 227 ℃ = 227 + 273 = 500 K

Therefore, P = 0.15 x 0.5 x 5.67 x 10^-8 x 5004 = 265.78 W

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