Physics - Modern Physics Question with Solution | TestHub

$\begin{array}{rl}{\text{ Threshold energy of A E }}_{\mathrm{A}}={\mathrm{hv}}_{\mathrm{A}}& \\ =& 6.6\times 10^{-34}\times 1.8\times 10^{14}\\ =& 11.88\times 10^{-20}\mathrm{J}\\ =& \frac{11.88\times 10^{-20}}{1.6\times 10^{-19}}\mathrm{eV}=0.74\mathrm{eV}\end{array}$ Similarly, ${\mathrm{E}}_{\mathrm{B}}=0.91\mathrm{eV}$ As the incident photons have energy greater than ${\mathrm{E}}_{\mathrm{A}}$ but less than ${\mathrm{E}}_{\mathrm{B}}$ So, photoelectrons will be emitted from metal A only.