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`)

Answer

B

Explanation

Threshold energy of A is
`E_(A)=hv_(A)`
`=6.6xx10^(-34)xx1.8 xx10^(14)`
`11.88xx10^(-20)J`
`=(11.88xx10^(-20))/(1.6 xx 10^(-19))eV`
`=0.74 eV`
Similarl, `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.