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

`(KE_(1))/(KE_(2))=(E_(1)-W_(1))/(E_(2)-W_(2))=`
`(hv_(1)-w_(1))/(hv_(2)-w_(2))=(hf-w)/(2hf-2w)=(1)/(2)`