The peak emission from a black body at a certain temperature occurs at a wavelength of 6200 A. On increasing its temperature, the total radiation emitted is increased 16 times. These radiations are allowed to fall on a metal surface. Photoelectrons emitted by the peak radiation at higher temperature can be bought to rest by applying potential equivalent to the excitation potential corresponding to the transition for the level n=4 to n=2 in the Bohr's hydrogen atom. The work function of the metal is given by `alpha/100 eV` where alpha is the numerical constant. Find the value of alpha.

Answer

`0145`

Explanation

Radiation alpha T^(4)
So `T_(2)=2R_(1)` and By wein's displacement law `lambda prop 1/T`
So `lambda_(2)=lambda_(1)/2=3000A`
By Einstein's photo electric equation
`(hc)/lambda-eV_(s)+phi`
`f=(hc)/lambda-eV_(s) =(hc)/(3100A)-(13.6eV) 1^(2)(1/2^(2)-1/4^(2))=4.00-2.55`
`phi=1.45 eV`
`=1.45=(alpha/100) or alpha=145`