Chemistry - Atomic Structure Question with Solution | TestHub

Minimum energy required for the emission of photoelectrons$=4.95\mathrm{ }\times {10}^{-19}\mathrm{J}$

Let the critical frequency or threshold frequency of the photon to eject the electron is$\mathrm{ }\mathrm{v}\mathrm{ }{\mathrm{s}\mathrm{e}\mathrm{c}}^{-1}$.

We know,$\mathrm{E}\mathrm{ }=\mathrm{ }\mathrm{h}\mathrm{\upsilon }\mathrm{ }\mathrm{ }\mathrm{o}\mathrm{r}\mathrm{ }\mathrm{\upsilon }\mathrm{ }=\mathrm{ }\frac{\mathrm{E}}{\mathrm{h}}\mathrm{ }=\mathrm{ }\frac{4.95\times {10}^{-19}}{6.6\times {10}^{-34}}\mathrm{ }=\mathrm{ }7.5\mathrm{ }\times {10}^{14}{\mathrm{s}}^{-1}$