Question

A black body is at a temperature of 2880 K. The energy of radiation emitted by this body with wavelength between 499 nm and 500 nm is U₁, between 999 nm and 1000 nm is U₂ and between 1499 nm and 1500 nm is U₃. The Wien's constant, b = 2.88×10⁶ nm-K. Then,

• (A) U₁ = 0
• (B) U₃ = 0
• (C) U₁>U₂
• (D) U₂>U₁

Answer 1

The Correct Option is D

Let λ_m be the wavelength corresponding to the maximum energy emitted by the black body at 2880k. 

Using Wien's displacement law, 

λ_mT = b

⇒ λ_m × 2880K = 2.88 × 10⁶ nm-k

⇒ λ_m = (2.88 × 10⁶ nm-k)/2880K

⇒ λ_m = 1000 nm

Thus, the maximum energy emitted by the star at temperature 2880K is corresponding to the wavelength of 1000 nm. 

Graph of energy versus wavelength of the black body is shown below:

In the above graph, λ corresponds to the point A. This clearly shows that U is the point where energy is maximum and U > U3 or U1Hence, the correct option is (D).

Answer 2

Wien’s Displacement Law

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

λ_m ∝ 1/T

⇒ λ_m = b/T

Where

• λ_m = wavelength of maximum intensity of radiation
• T = absolute temperature
• b = Wien’s constant or Wien’s displacement constant

Wien’s Constant

Aa physical constant that establishes a relationship between the thermodynamic temperature and wavelength of a black body is known as Wien’s constant. 

• It is a combination of temperature and the wavelength of the black body, which becomes shorter as the temperature rises and the wavelength approaches its maximum.
• The value of Wien’s displacement constant is 2.898 x 10^-3 meter kelvin (m K).