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 isU1, between 999 nm and 1000 nm is U2 and between 1499 nm and 1500 nm is U3. The Wien constant, b = 2.88 × 106 nm-K. Then,

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 isU1, between 999 nm and 1000 nm is U2 and between 1499 nm and 1500 nm is U3. The Wien constant, b = 2.88 × 106 nm-K. Then, (A) U1 =0 (B) U3 =0 (C) U1 > U2 (D) U2 > U1

Tardigrade Admin · Accepted Answer

Wien's displacement law is λm T =b (b = Wien's constant) ∴ λm =(b/T) =(2.88 × 106 nm-K/2880 K) ∴ λ = 1000 nm Energy distribution with wavelength will be as follows: From the graph it is clear that U2 > U1 (In fact U2 is maximum)

Q. 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 $\times 1 0^{6}$ nm-K. Then,

$U_{1} = 0$

$U_{3} = 0$

$U_{1} > U_{2}$

$U_{2} > U_{1}$

Wien's displacement law is
$λ_{m} T = b$ (b = Wien's constant)
$∴ λ_{m} = \frac{b}{T} = \frac{2.88 \times 1 0 ^{6} nm - K}{2880 K}$
$∴ λ = 1000 nm$
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)

Q. 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 isU1​, between 999 nm and 1000 nm is U2​ and between 1499 nm and 1500 nm is U3​. The Wien constant, b = 2.88 ×106 nm-K. Then,

U1​=0

U3​=0

U1​>U2​

U2​>U1​