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$U_1$, between 999 nm and 1000 nm is $U_2$ and between 1499 nm and 1500 nm is $U_3$. The Wien constant, b = 2.88 $\times 10^6$ nm-K. Then,

• A. $U_1 =0$
• B. $U_3 =0$
• C. $U_1 > U_2$
• D. $U_2 > U_1$

Answer 1

Answer: (D)

$U_2 > U_1$

Answer 2

Wien's displacement law is
$\, \, \, \, \, \, \, \, \, \lambda_m T =b \, \, \, \, \, \, \, \, \, \, \, \,$ (b = Wien's constant)
$\therefore \, \, \, \, \, \, \, \lambda_m =\frac{b}{T} =\frac{2.88 \times 10^6 nm-K}{2880 K}$
$\therefore \, \, \, \, \, \, \, \, \lambda = 1000 nm$
Energy distribution with wavelength will be as follows :
From the graph it is clear that
$\, \, \, \, \, \, \, \, \, \, \, \, U_2 > U_1$ (In fact U_2 is maximum)