Question

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

• 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

Given- ${\nu }_{0A}=1.8\times {10}^{14}Hz,{\nu }_{0B}=2.2\times {10}^{14}Hz,E=0.825eV=0.825\times 1.6\times {10}^{-19}J$ ,

Frequency of incident photon ,

$E=h\nu$ ,

or $\nu =E/h=\frac{0.825\times 1.6\times {10}^{-19}}{6.6\times {10}^{-34}}=2\times {10}^{14}Hz$ ,

The photoelectrons will be emitted from the metal surface , if frequency of incident photon is greater than threshold frequency .

As $\nu >{\nu }_{0A}$ and $\nu <{\nu }_{0B}$ ,

Therefore, photoelectrons will be emitted from A alone .