The peak emission from a black body at a certain temprature occurs at a wavelength of `9000 Å`. On increase its temperature , the total radiation emmited is increased its `81` times. At the intial 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 from the black body caused photoemission. To bring these photoelectrons to rest , a potential equivalent to the excitation energy between `n = 2 and n = 3` bohr levels of hydrogen atoms is required. Find the work function of the metal.

Answer

`2.25 e V`

Explanation

Let `T_(1)` be the initial tamprature and `T_(2)` be the increased tamprature of the black body.
According to Stefan's law.
`((T_(2))/(T_(2))) = 81 = (3)^(4)`
`T_(2) = 3 T_(1)`
Also `lambda_(1) T_(1) = lambda_(2) T_(2)`
` lambda_(2) = lambda_(1 xx T_(1))/(T_(2)) = (9000 xx T_(1))/(3 T_(1)) = 3000 Å`
Now `(hv)/(lambda_(2)) - W = eV_(0) or (hv)/(lambda_(2)) - W =13.6 ((1)/(2^(2)) - (1)/(3^(2)))`
Solving we get , `W = 2.25 e V`