Question

The absolute temperature of a body $A$ is four times that of another body $B$. For two bodies, the difference in wave lengths at which energy radiated is maximum is $3.0$ $\mu$m. Then the wavelength at which the body $B$ radiates maximum energy in micrometre is :

• A. $2$
• B. $2.5$
• C. $4.0$
• D. $4.5$

Answer 1

The correct option is C $4.0$

From Wien's displacement law, ${\lambda }_{m}T=$ constant
$⟹\frac{{\lambda }_B}{{\lambda }_A}=\frac{T_A}{T_B}$
Given : $T_A=4T_B$
$⟹$ $\frac{{\lambda }_B}{{\lambda }_A}=\frac{4T_B}{T_B}=4$
We get ${\lambda }_B=4{\lambda }_A$
Also, we have ${\lambda }_B-{\lambda }_A=3\mu m$
$\therefore$ $4{\lambda }_A-{\lambda }_A=3\mu m$
$⟹{\lambda }_A=1\mu m$
So, we get ${\lambda }_B=4{\lambda }_A=4\times 1=4.0\mu m$