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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 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

U₁=0

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U₃=0

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U₁>U₂

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U₂>U₁

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Solution

The correct option is D
U₂>U₁

Wein's displacement law is $λ_{m} T = b$

$\Rightarrow$ $λ_{m} = \frac{b}{T} = \frac{2.88 \times 10^{6}}{2880}$ = 1000 nm.

Energy distribution with wavelength will be as follows

From the graph it is clear that $U_{2} > U_{1}$ .

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Similar questions

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

Q. 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

U_{1}

, between

999 n m

and

1000 n m

, is

U_{2}

and between

1499 n m

and

1500 n m

U_{3}

. Wien's constant,

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

. Then

Q. 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

U_{1}

, between

999

and

1000 nm

U_{2}

and between

1499 nm

and

1500 nm

U_{3}

. The Wien's constant

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

. Then

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

Q. 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 999nm and 1000nm is

U_{2}

and between

1499

nm and

1500

nm is

U_{3}

. The Wein constant is

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

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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 U1, between 999 nm and 1000 nm is U2 and between 1499 nm and 1500 nm is U3. The Wein's constant b = 2.88 x 106 nm K. Then

The correct option is D U2>U1 Wein's displacement law is λmT=b ⇒ λm = bT = 2.88 × 1062880 = 1000 nm. Energy distribution with wavelength will be as follows From the graph it is clear that U2 > U1.

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

The correct option is D
U₂>U₁

Wein's displacement law is $λ_{m} T = b$

$\Rightarrow$ $λ_{m} = \frac{b}{T} = \frac{2.88 \times 10^{6}}{2880}$ = 1000 nm.

Energy distribution with wavelength will be as follows

From the graph it is clear that $U_{2} > U_{1}$ .