The work functions of metals `A` and `B` are in the ratio `1 : 2`. If light of frequencies `f` and `2 f` are incident on the surfaces of `A` and `B` respectively , the ratio of the maximum kinetic energy of photoelectrons emitted is ( `f` is greater than threshold frequency of `A , 2f` is greater than threshold frequency of `B`)

Options

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

Answer

B

Explanation

`K_("max") =hf -phi_0implies hf -phi_0implieshf =phi_A+K_A " "....(i)`
and `2hf =phi_B +K_B =2phi_A+K_B " "...(ii)`
[as `phi_A/phi_B=1/2` ]
From (i) and (ii)
`implies (phi_A+K_A)/(2phi_A+K_B)=1/2implies K_A/K_B=1/2`