Question

Radiation of wavelength $\lambda$ is incident on a photocell. The fastest emitted electron has a speed $v$. If the wavelength is changed to $\frac{3\lambda }{4}$, the speed of the fastest emitted electron will be :

Answer 1

We know,

$\frac{h_c}{\lambda }-\varphi =\frac{1}{2}{mv}^2⟶\left(i\right)$

$\frac{4h_c}{3\lambda }-\varphi =\frac{1}{2}{mv}^2⟶\left(ii\right)$

$\left(ii\right)$ - $\left(i\right)$ gives

$\frac{h_c}{3\lambda }=\frac{1}{2}m(v^2-v^1)$

$\Rightarrow v^1=\sqrt{v^2+\frac{2h_c}{3{\lambda }_m}}⟶\left(iii\right)$

also from $\left(i\right)$ $\frac{h_c}{\lambda }=\varphi +\frac{{mv}^2}{2}$

$\Rightarrow \frac{{2h}_c}{{\lambda }_m}=\frac{2\varphi }{m}{v}^2$

$\Rightarrow \frac{{2h}_c}{{3\lambda }_m}=\frac{2\varphi }{3m}+\frac{r^2}{3}>\frac{v^2}{3}$ $⟶\left(iv\right)$

$v>\sqrt{\frac{{4v}^2}{3}}$