In a photoelectric experiment the relation between applied potential difference between cathode and anode V and the photoelectric current cathode and anode V and the photoelectric current l and was found to be shown in graph below. If planck's constant `h=6.6xx10^(-34)`Js, the frequency of incident radiation would e nearly (in `s^(-1)`)

Options

A. `0.436xx10^(18)`
B. `0.436xx10^(17)`
C. `0.775xx10^(15)`
D. `0.775xx10^(16)`

Answer

C

Explanation

From the graph, stopping potential, `V_(0)=-3.2V`
for photoelectric effect,
`eV_(0)hv implies v=(eV_(0))/(h)`
`v=(1.6xx10^(-19)xx3.2)/(6.6xx10^(-34))`
`=0.775xx10^(15)Hz`