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//lamda) = (lamda' I)/(hc) prop lamda`
Ratio of no. of photo electrons `= (lamda_(A))/(lamda_(B))`