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Byju's Answer

Standard XII

Physics

Wien's Displacement Law

A black body ...

Question

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

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 A $U_{2} > U_{1}$
According to wein's displacement law we know that $λ_{m a x} T = c o n s t a n t$
So, $λ_{m a x} T = b$ ----(suppose)
So, $λ_{m a x} = \frac{b}{T}$ $= \frac{2.88 \times 10^{6} n m K}{2880 K} = 1000 n m$
Hence, $U_{2}$ is maximum since all the wavelength ranging from zero to infinity,so $U_{1}$ and $U_{2}$ are not zero.

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Similar questions

Q. A black body is at a temperature of

2800

K. The energy of radiation emitted by this object with wavelength between

499

nm and

500

nm is

U_{1}

, between

999

nm and

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and between

1499

nm and

1500

nm is

U_{3}

. (Wiens constant

b = 2.80 \times 10^{6} n m K

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U_{2}

and between

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U_{3}

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

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499

nm and

500

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U_{1}

, between

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and between

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A black body is at a temperature of 2880 K. The energy of radiation emitted by this body between wave lengths 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’s constant b = 2.88 $\times$ 106 nm K. Then

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

energy of radiation emitted by this object 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 Wein's constant

b = 2.88 x $10^{6}$ nm K. Then

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A black body is at a temperature of 2880 K. The energy of radiation emitted by this body between wavelengths 499 nm and 500 nm is U1; between 999 nm and 1000 nm is U2; and between 1499 nm and 1500 nm is U3. Then, (Wein's constant b=2.88×106nmK)

The correct option is A U2>U1According to wein's displacement law we know that λmaxT=constantSo, λmaxT=b----(suppose)So, λmax=bT=2.88×106nmK2880K=1000nmHence, U2 is maximum since all the wavelength ranging from zero to infinity,so U1 and U2 are not zero.

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