Q. $A$ and $B\ $are two metals with threshold frequencies $1.8 \times 10^{14}$Hz and $2.2 \times 10^{14}$Hz. Two idential photons of energy 0.825 eV 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 $B$
Explanation
(b) Threshold energy of A is
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ E_{A} = hv_{A}$
$= \ 6.6 \times 10^{- 34} \times 1.8 \times 10^{14}$
$= \ 11.88 \times 10^{- 20}J$
$= \ \frac{11.88 \times 10^{- 20}}{1.6 \times 10^{- 19}}\ eV$
$= \ 0.74\ eV$
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.
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.