When a metallic surface is illuminated by a light of frequency `8xx10^(14)`Hz, photoelectron of maximum energy 0.5 eV is emitted. When the same surface is illuminated by light of frequency `12xx10^(14)` Hz, photoelectron of maximum energy 2 eV is emitted. The work function is

Options

A. 0.5 eV
B. 2.85 eV
C. 2.5 eV
D. 3.5 eV

Answer

C

Explanation

`8xx10^(14)j=phi_0+0.5`
`12xx10^(14)h=phi_0+2`
Dividing, we get `(12)/(8)=(phi_0+2)/(phi_0+0.5)`
`(3)/(2)=(phi_0+2)/(phi_0+0.5)`
`3phi_0+1.5=2phi+4`
or `phi_0=2.5eV`