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

2

Explanation

`((K.E)_(max))/((K.E)_(max))=(hf-varphi)/(h2f-2varphi)=(1)/(2)`