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 $0.825eV$ each are incident on them. Then photoelectrons are emitted by (Take $h=6.6\times {10}^{-34}J-s$)

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

Answer 1

The correct option is B $A$ alone

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