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Standard XII

Physics

Blackbody

A black body ...

Question

A black body is at a temperature of $2880 K$ . The energy of radiation emitted by this body with wavelength between $499 n m$ and $500 n m$ is $U_{1}$ , between $999 n m$ and $1000 n m$ , is $U_{2}$ and between $1499 n m$ and $1500 n m$ is $U_{3}$ . Wien's constant, $b = 2.88 \times 10^{6} n m - K$ . 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 $U_{2} > U_{1}$
Wien's displacement law is
$λ_{m} T = b (b = W i e n^{'} s c o n s t a n t)$
$λ_{m} = \frac{b}{T} = \frac{2.88 \times 10^{6} n m - K}{2880 K}$
$λ = 1000 n m$
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).

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A black body is at a temperature of 2880 K. The

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A black body is at a temperature of 2880 K. The energy of radiation emitted by this object with wave length between 250 nm to 255 nm is U₁, between 500 nm to 505 nm is U₂and between 1500 nm to 1505 nm is U₃. Then for this situation mark the correct statement

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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 U1, between 999 nm and 1000 nm, is U2 and between 1499 nm and 1500 nm is U3. Wien's constant, b=2.88×106nm−K. Then

The correct option is D U2>U1Wien's displacement law isλmT=b(b=Wien′s constant)λm=bT=2.88×106nm−K2880 Kλ=1000 nmEnergy distribution with wavelength will be as follows:From the graph it is clear thatU2>U1 (in fact U2 is maximum).