The spectral distribution of radiation from a black body at T₁ = 3000 K has a maximum at wavelength λmax. The body cools down to a temperature T₂. If the wavelength corresponding to the maximum of the spectral distribution at T₂ is 1.2 times of the original wavelength λmax, then the temperature T₂ is ________ K (round off to the nearest integer).

The correct answer is 2500.

According to Wien's displacement law, the wavelength corresponding to the maximum intensity of radiation emitted by a black body is inversely proportional to its absolute temperature. Mathematically, this relationship is expressed as:

$${\lambda }_{\text{max}}·T=\text{constant}$$

This implies that for two different states of a black body at absolute temperatures T₁ and T₂ with corresponding peak wavelengths λ₁ and λ₂, we can write the relation as:

$${\lambda }_1{T}_1={\lambda }_2{T}_2$$

From the problem, we are given the following values:
- Initial temperature: T₁ = 3000 K
- Final wavelength: λ₂ = 1.2 λ₁

Substituting these values into the Wien's displacement law equation, we get:

$${\lambda }_1\times 3000=1.2{\lambda }_1\times T_2$$

Since the peak wavelength λ₁ is present on both sides, it cancels out of the equation:

$$3000=1.2\times T_2$$

Now, we solve for the final temperature T₂:

$$T_2=\frac{3000}{1.2}$$

$$T_2=2500\text{K}$$

Thus, the temperature T₂ is 2500 K.