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Standard XIII

Physics

Wien's Displacement Law

A black body ...

Question

A black body is at rest at a temperature of $2880 K$ . The energy of radiation emitted by this object with wavelength between $499 n-m$ and $500 n-m$ is $U_{1}$ , between $999$ and $1000 nm$ is $U_{2}$ and between $1499 nm$ and $1500 nm$ is $U_{3}$ . The Wien's constant $b = 2.88 \times 10^{6} nm-K$ . Then

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 $U_{2} > U_{1}$
Given, Temperature of black body $(T) = 2880 K$
From Wien's law
$λ_{m a x} = \frac{2.88 \times 10^{6}}{2880} = 1000 nm$

Energy distribution with wavelength is as shown in the figure.
From the figure it is clear that $U_{2}$ is corresponding to $λ_{m a x}$ (approx)
$∴ U_{2} > U_{1}$

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Similar questions

Q. A black body is at rest at a temperature of

2880 K

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499 n-m

and

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b = 2.88 \times 10^{6} n m K

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499

nm and

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and between

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and between

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and

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Wien's Displacement Law

Standard XIII Physics

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A black body is at rest at a temperature of 2880 K. The energy of radiation emitted by this object with wavelength between 499 n-m and 500 n-m is U1, between 999 and 1000 nm is U2 and between 1499 nm and 1500 nm is U3. The Wien's constant b=2.88×106 nm-K. Then

The correct option is D U2>U1Given, Temperature of black body (T)=2880 K From Wien's law λmax=2.88×1062880=1000 nm Energy distribution with wavelength is as shown in the figure. From the figure it is clear that U2 is corresponding to λmax(approx) ∴U2>U1