Two separate monochromatic light beams `A` and `B` of the same intensity (energy per unit area per unit time) are falling normally on a unit area of a metallic surface. Their wavelength are `lambda_(A)` and `lambda_(B)` respectively. Assuming that all the the incident light is used in ejecting the photoelectrons, the ratio of the number of photoelectrons from beam `A` to that from `B` is

Options

A. `(lambda_(A)/lambda_(B))`
B. `(lambda_(B)/lambda_(A))`
C. `(lambda_(A)/lambda_(B))^(2)`
D. `(lambda_(B)/lambda_(A))^(2)`

Answer

A

Explanation

The number of photo electron depends on the Number of photons Number of photon `=I/(hc//lambda)=(lambda.I)/(hc) prop lambda`
Ratio of no. of photo electrons =`lambda_(A)/lambda_(B)`