Physics - Modern Physics Question with Solution | TestHub

As we know, energy of photo electrons,

$\mathit{\text{E}}=\mathit{\text{hν}}=\frac{\mathit{\text{hc}}}{\lambda }$

$\mathit{\text{E}}=\frac{\left(\text{6.6}\times 10^{-34}\right)\left(3\times 10^8\right)}{\left(5000\times 10^{-10}\right)}$

$=\frac{\text{19.8}\times 10^{-19}}{5} \text{joule}$

$=\frac{\text{19.8}\times 10^{-19}}{5\times \text{1.6}\times 10^{-19}} \text{eV}$

$=\frac{\text{19.8}}{8}=\text{2.48 eV}$

Now,$\mathit{\text{hν}}=\varphi +\frac{1}{2}{\mathit{\text{mv}}}_{\text{max}}^2$

$\Rightarrow \frac{1}{2}{\mathit{\text{mv}}}_{\text{max}}^2=\mathit{\text{hν}}-\varphi$

$=\text{2.48}-\text{1.9}=\text{0.58 eV}$