Question

$A$ and $B$ are two metals with threshold frequencies $1.8 \times 10^{14}\, Hz$ and $2.2 \times 10^{4}\, Hz$ Two identical photons of energy $0.825 \,eV$ each are incident on them. Then photoelectrons are emitted by (Take $\left.h=6.6 \times 10^{-34} \,J - s \right)$

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

Answer 1

Answer: (B)

Threshold energy of $A$ is
$E_A=hv_A $
$=6.6 \times 10^{-34} \times 1.8 \times 10^{14}$
$=11.88 \times 10^{-20} \,J$
$=\frac{11.88 \times 10^{-20}}{1.6 \times 10^{-19}} \,eV$
$ =0.74 \,e V$
Similarly, $E_{B}=0.91 \,eV$
As the incident photons have energy greater than $E_{A}$ but less than $E_{B}$.
So, photoelectrons will be emitted from metal $A$ only