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Standard XII

Physics

Wien's Displacement Law

A black body ...

Question

A black body is at a temperature of $5760 K$ . The energy of radiation emitted by the body at wavelength $250 n m$ is $U_{1}$ , at wavelength $500 n m$ is $U_{2}$ and that at $1000 n m$ is $U_{3}$ . Wien's constant, $b = 2.88 \times 10^{6} n m K$ . Which of the following is correct?

U_{1} = 0

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U_{3} = 0

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U_{1} > U_{2}

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U_{2} > U_{1}

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Solution

The correct option is D $U_{2} > U_{1}$
Maximum amount of emitted radiation corresponding to $λ_{m} = \frac{b}{T}$

$\Rightarrow λ_{m} = \frac{2.88 \times 10^{6} n m K}{5760 K} = 500 n m$

From the graph we can see that $U_{1} < U_{2} > U_{3}$

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A black body is at a temperature of 5760K. The energy of radiation emitted by the body at wavelength 250nm is U1, at wavelength 500nm is U2 and that at 1000nm is U3. Wien's constant, b=2.88×106nmK. Which of the following is correct?

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A black body is at a temperature of $5760 K$ . The energy of radiation emitted by the body at wavelength $250 n m$ is $U_{1}$ , at wavelength $500 n m$ is $U_{2}$ and that at $1000 n m$ is $U_{3}$ . Wien's constant, $b = 2.88 \times 10^{6} n m K$ . Which of the following is correct?

The correct option is D $U_{2} > U_{1}$
Maximum amount of emitted radiation corresponding to $λ_{m} = \frac{b}{T}$

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From the graph we can see that $U_{1} < U_{2} > U_{3}$