When the energy of the incident radiation is increased by `20 %` , kinetic energy of the photoelectrons emitted from a metal surface increased from `0.5 eV to 0.8 eV`. The work function of the metal is

Options

A. `0.65 eV`
B. `1.0 eV`
C. `1.3 eV`
D. `1.5 eV`

Answer

B

Explanation

According to Einstein.s photoelectric equation ,
The kinetic energy of emitted photoelectrons is
`K = hupsilon - phi-0`
where `hupsilon` is the energy of incident radiation and `phi_0` is work function of the metal.
`0.5 eV = hupsilon_0 " "...(i) and 0.8 eV =1.2 hupsilon -phi_0 " " ....(ii)`
On solving eqns. (i) and (ii) , we get `phi_0 =1.0 eV`