When a metal surface is exposed to certain frequency of electromagnetic radiation. The kinetic energy of electron ejected from metal surface is 0.20 eV. If its work function `(W_(0))` is 4.80 eV, the approximate frequency of radiation falling on the metal surface in Hz is

Options

A. `1.98xx10^(15)`
B. `1.21xx10^(16)`
C. `1.21xx10^(15)`
D. `1.98xx10^(16)`

Answer

C

Explanation

Given, work function `(omega_(0))` i.e. energy for ejection = 4.80 eV and kinetic energy of the ejected electron.
(KE) = 0.20 eV.
According to photoelectric current relation.
Total energy `(E_(T))=omega_(0)t+KE=4.80+0.20`
`E_(T)=5.0eV`
Also, `1eV=1.6021xx10^(-19)J`
`therefore" "5eV=8.010xx10^(-19)J`
or `E_(T)=8.010xx10^(-19)J`
and frequency (v) = `(E_(T)("Total energy"))/(h("Planks constant"))`
(where, `h = 6.6260xx10m^(2)kg//s`)
`therefore" "V=E_(T)/h=(8.010xx10^(-19))/(6.626xx10^(-34))`
Frequency (V) = `1.21xx10^(15)` Hz.