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

`E = W_0 + K_(max)` …(i)
`rArr hf = W_A + K_A` …(ii)
and `2hf = W_B + K_B + K_B ( because W_A/W_B = 1/2)`
dividing equation (i) by (ii)
`1/2 = (W_A + K_A)/(2W_A + K_B) rArr K_A/K_B = 1/2`