The maximum kinetic energy of a photoelectron liberated from the surface of lithium with work function 2.35 eV by electromagnetic radiation whose electric component varies with time as:
`E = a [ 1 + cos (2pi f_(1)t) ] cos 2pi f_(2) t ` ( where a is a constant ) is` (f_(1) = 3.6 xx 10^(15) Hz, and f_(2) = 1.2 xx10^(15) `  Hz and planck's constant `h= 6.6 xx 10 ^(-34)` Js)

Options

A. `2.64` eV
B. `7.55` eV
C. `12.53` eV
D. `17.45` eV

Answer

D

Explanation

Here, work function `W_(0) = 2.35 eV` and the electric component of electromagnetic radiation
`E = a Pi + cos (2pif_(1)t)cos (2pif_(2)t)`
`rArr E = [a cos ( 2pif_(2)t) + a cos (2pif_(1) t) cos (2pif_(2)t)]`
`(cos A cos B = 1/2 cos (A+B)- cos (A-B))`
`rArr E = a cos (2pif_(2) t) + a/2 cos 2pi ( f_(1) + f_(2))t-a/2 cos 2 pi ( f_(1) - f_(2))t`
So, the electric component has 2 sub- components with frequencies are,
`f_(2) (f_(1) + f_(2)) and (f_(1) - f_(2))`
So, for maximum kinetic energy of photoelectron, whe take photon of maximum frequency. Hence,
`E_(max)=(hc_(max))/e = (6.6xx10^(-34)xx(3.6xx10xx1.2xx10^(15)))/(1.6xx10^(-19))`
`=19.8 eV`
Hence, the maximum kinetic energy,
`KE_(max) =E_(max) - W_(0) = 19.8 - 2.35 = 17.45 eV`
Hence, the correct option is (d).