(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

1)The peak emission from a black body at a certain temperature occurs at a wavelength of 9000 Angstrom. On increasing its temperature the total radiation emitted is increased 81 times. At the initial temperature, when the peak radiation from the black body is incident on a metal surface, it does not cause any photoemission from the surface. After the increase of temperature, the peak radiation from the black body causes photoemission. To bring these photo electrons to rest, a potential equivalent to the excitation energy between n=2 and n=3 Bohr levels of Hydrogen atom is required. Find the work function of the metal.

2. Relevant equations

According to Wein’s displacement law,

[lambda1(max.) x T1] = 2.898 x 10^(-3) metre-Kelvin

According to Stefan’s law,

Energy of incident radiation(E2) = (sigma) x T^4

3. The attempt at a solution

I have doubt only in the first part of the problem. So I haven’t worked out the entire problem.

I solved it in the following way:

Let T1 and T2 be the initial and final temperatures respectively. Let lambda1(max.) and lambda2(max.) be the initial and final wavelengths respectively. Let E1 and E2 be the initial and final energy of the radiations respectively.

According to Stefan’s law,

E is proportional to T^4

Hence,

E1/E2 = [T1/T2]^4

Therefore,

T2 = 3T1 {since given that E2/E1 = 81}

According to Wein’s displacement law,

[lambda1(max.) x T1] = 2.898 x 10^(-3) metre-Kelvin

Hence, T1 = 3220 Kelvin

Hence, T2 = 3T1 = 9660 Kelvin

According to Stefan’s law,

Energy of incident radiation(E2) = (sigma) x T^4

= 5.67 x 10^(-8) x (9660 x 10^(-10))^4

= 4.94 x 10^(-32) Joules

The solution given in my book is as follows:

According to Stefan’s law,

E is proportional to T^4

Hence,

E1/E2 = [T1/T2]^4

Therefore,

T2 = 3T1 {since given that E2/E1 = 81}

According to Wein’s displacement law,

[lambda(max.) x T] = constant

i.e. [lambda1(max.) x T1] = [lambda2(max.) x T2]

lambda2(max.) = 3000 Angstrom

Energy of incident radiation(E2) = hc/ lambda2(max.) = 6.63 x 10^(-19) Joules

Why do I get 2 different answers?

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# Black body radiation problem

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