The work function of a metal is `4.0 eV`. If the metal is irradiated with radiation of wavelength `200` nm, then the maximum kinetic energy of the photoelectrons would be about.

Options

A. `1:2`
B. `1:3`
C. `1:4`
D. `2 :1`

Answer

A

Explanation

The maximum energy of photoelectron
`K_(max_1) = h upsilon - phi_1 =3 eV -2.75 eV =0.25 eV`
and `K_(max_2)= h upsilon - phi_(2) = 3 eV -2 eV =1 eV`
Momenta of photoelectron is given by
`p = sqrt(2mK)`
`therefore (P_(max_1))/(P_(max_2)) = sqrt((K_(max_1))/(K_(max_2)))=sqrt((0.25eV)/(1.eV))= sqrt(1/4)=1/2`