Q. Solar radiation emitted by sun correspond to that emitted by black body at a temperature of 6000 K. Maximum intensity is emitted at wavelength of 4800$\mathring{\mathrm{A}}$. If the sun was to cool down from 6000 K to 3000 K, then the peak intensity of emitted radiation would occur at a wavelength
A) 4800$\mathring{\mathrm{A}}$
B) 9600$\mathring{\mathrm{A}}$
C) 2400$\mathring{\mathrm{A}}$
D) 19200$\mathring{\mathrm{A}}$
Explanation
(b) According to Wien’s displacement law
$\therefore$ $\lambda_{m} = \frac{b}{T}$ (b=constant)
$\therefore$ $\frac{\lambda_{1}}{\lambda_{2}} = \frac{T_{2}}{T_{1}}$
$\Rightarrow$ $\lambda_{2} = \frac{\lambda_{1}T_{1}}{T_{2}}$
Given, $\lambda_{1} = 4800\mathring{\mathrm{A}},\ \ T_{1} = 6000\ K,\ {\ \ \ \ \ T}_{2} = 3000\ K$
$\therefore$ $\lambda_{2} = \frac{4800 \times 6000}{3000} = 9600\mathring{\mathrm{A}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.