Question

two separate monochromatic light beams A and B of the same intensity are falling normally on a unit area of a metallic surface.Their wavelengths are lambda A n lambda B respectively. Assuming that all the incident light is used in ejecting the photoelectrons,find the ratio of number of photoelectrons from beam A to that form B,in terms of their wavelenths

Answer 1

The ratio of number of photoelectrons from beam A to that from B is λa/λb

Explanation :

we know that intensity of the light is given by

I = N/λ

where N = number of photo electron ejected

λ = wavelength

Given for light beam A

wavelength = λa

hence intensity

Ia = Na/λa

similarly for light beam B

wavelength = λb

hence intensity

Ib = Nb/λb

since both the light beam have the same intensity, hence

Ia = Ib

=> Na/λa = Nb/λb

=> Na/Nb = λa/λb

Hence the ratio of number of photoelectrons from beam A to that from B is λa/λb