## In This post you will get answer of Q.

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

Solution:

## $phi_{0A}=frac{hupsilon_{0}}{e}eV=frac{left(6.6times10^{-34}right)timesleft(1.8times10^{14}right)}{1.6times10^{-19}}eV$

$=0.74,eV$

$phi_{0B}=frac{left(6.6times10^{-34}right)timesleft(2.2times10^{14}right)}{1.6times10^{-19}}eV$

$=0.91,eV$

Since the incident energy $0.825 ,eV$ is greater than $0.74, eV $ and less than $0.91 ,eV$, so photoelectrons are emitted from metal $A$ only