A black body is at a temperature of 5760K. The energy of radiation emitted by the body at wavelength 250nm is U _ 1 , at wavelength 500 nm is U _ 2 and that at 1000nm is U _ 3 . Wien's constant, b = 2.88 × 10 ^ 6 nm K.

NEET 2016

Exam: NEET
Year: 2016
Subject: Physics
Chapter: Thermal Properties of Matter

NEET 2016 · Physics · Thermal Properties of Matter

A black body is at a temperature of 5760K. The energy of radiation...

A black body is at a temperature of

\text{5760}\text{K.}

The energy of radiation emitted by the body at wavelength

\text{250}\text{nm}

{U}_{1}

, at wavelength

\text{500} \text{nm}

{U}_{2}

and that at

\text{1000}\text{nm}

{U}_{3}

. Wien's constant,

b = 2.88 \times {10}^{6}\text{nm K.}

Which of the following is correct? A black body is at a temperature of 5760K. The energy of radiation emitted by the body at wavelength 250nm is U _ 1 ,

Options

A ${U}_{1} = 0$
B ${U}_{3} = 0$
C ${U}_{1} > {U}_{2}$
D(\mathrm{U}_3 < \mathrm{U}_2 > \mathrm{U}_1)

Correct answer

D. (\mathrm{U}_3 < \mathrm{U}_2 > \mathrm{U}_1)

Step-by-step solution

Given

Temperature $= 5760K$

Wien's constant $b = 2 . 88 \times {10}^{6}nmK$

According to Wien's displacement law, maximum amount of emitted radiation corresponding to ${\lambda}_{m} = \dfrac{b}{T}$

${\lambda}_{m} = \dfrac{2 . 88 \times {10}^{6}nmK}{5760K} = 500nm$

A black body is at a temperature of 5760K. The energy of radiation emitted by the body at wavelength 250nm is U _ 1 ,

From the graph, ${U}_{1} {U}_{3}$

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