When a metallic surface is illuminated by light of frequency `8xx10^(14)Hz` a photoelectron of energy 0.5eV 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. 1.5 eV
C. 2.5 eV
D. 3.5 eV

Answer

C

Explanation

`8xx10^(14)h= phi_(0)+0*5`
`12xx10^(14)h=phi_(0)+2`
Then `(12)/(8)=(phi_(0)+2)/(phi_(0)+0*5)`
Solving `phi_(0)=2*5eV`