Question

When the temperature of a black body increases, it is observed that the wavelength corresponding to maximum energy changes from $0.26\text{mm}$ to $0.13\text{mm}$. The ratio of the emissive powers of the body at the respective temperature is

• A. $\frac{16}{1}$
• B. $\frac{4}{1}$
• C. $\frac{1}{4}$
• D. $\frac{1}{16}$

Answer 1

The correct option is D $\frac{1}{16}$

Given that,
Let at temperature $T_1$ and $T_2$ wavelength is ${\lambda }_1=0.26\text{mm}$ and ${\lambda }_2=0.13\text{mm}$ respectively
According to Wien's displacement law, and
${\lambda }_1{T}_1={\lambda }_2{T}_2$
$\frac{{\lambda }_1}{{\lambda }_2}=\frac{T_2}{T_1}=\frac{0.26}{0.13}=2$
$T_2=2T_1$

By <!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Stefan-Boltzman's law, emissive power, $E=\sigma T^4$
$E_1=\sigma {T_1}^4;E_2=\sigma {T_2}^4$
$\frac{E_1}{E_2}=\frac{\sigma {T_1}^4}{\sigma {T_2}^4}={\left(\frac{1}{2}\right)}^4=\frac{1}{16}$
i.e. (d) is the correct option.

Why this question? Botton line:- As the temperature is increased, the wavelength corresponding to the highest intensity decreases. $i.e.{\lambda }_{m}T=$ constant