A and B are two metals with threshold frequencies `1.8 xx 10^(14)Hz` and `2.2 xx 10^(14)Hz`. Two identical photons of energy 0.825 eV each are incident on them. Then, photoelectrons are emitted by (Taking, `h = 6.6 xx 10^(-34)J-s`)

Options

A. B alone
B. A alone
C. Neither A nor B
D. Both A and B

Answer

B

Explanation

Threshold energy of A is
`E_A = hv_A`
`= 6.6 xx 10^(-34) xx 1.8 xx 10^14`
`11.88 xx 10^(-20) J`
`= (11.88 xx 10^(-20))/(1.6 xx 10^(-19)) eV`
`= 0.74 eV`
Similarly, `E_B = 0.91 eV`
Since, the incident photons have energy greater than `E_A` but less than `E_B`.So, photoelectrons will be emitted from metal A only.