You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XI

Physics

Kirchoff's 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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

U_{3} = 0

No worries! We‘ve got your back. Try BYJU‘S free classes today!

U_{1} > U_{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

U_{2} > U_{1}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

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)
$∴ U_{2} > U_{1}$

Suggest Corrections

Similar questions

Q. 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

)

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

1000

nm is

U_{2}

and between

1499

nm and

1500

nm is

U_{3}

. (Wiens constant

b = 2.80 \times 10^{6} n m 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 a temperature of

2880 K

. The radiation energy emitted by this object with wavelength in 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}

. If Wein constant is

2.8 \times 10^{6}

nmK, 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

Join BYJU'S Learning Program

Explore more

Kirchoff's Law

Standard XI Physics

Join BYJU'S Learning Program

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