When the electromagnetic radiations of frequencies `4 xx 10^(15) Hz` and `6 xx 10^(15) Hz` fall on the same metal, in different experiments, the ratio of maximum kinetic energy of electrons liberated is `1 : 3`. The threshold frequency for the metal is

Options

A. `2 xx 10 ^(15) Hz`
B. `1 xx 10 ^(15) Hz`
C. `3 xx 10 ^(15) Hz`
D. `1.67 xx 10 ^(15) Hz`

Answer

C

Explanation

According to Einstein.s photoelectric equation
`hupsilon=hupsilon_0 +KE_(max)`
For first experiment , `hxx 4 xx 10^(15) = hupsilon_(0) +x " "..(i)`
For second experiment , `h xx 6 xx 10^(15) = hupsilon_(0) + 3x " "...(ii)`
From eqs. (i) and (ii) , we get `upsilon_0 3 xx 10^(15) Hz`