In a photoelectric experiment ultraviolet light of wavelength 280 nm is used with lithium cathode having work-function $\mathrm{\varphi }=2.5 \mathrm{eV}$. If the wavelength of incident light is switched to 400 nm, find out the change in the stopping potential. $\left(\mathrm{h}=6.63\times {10}^{-34} \mathrm{Js}, \mathrm{and} \mathrm{c}=3\times {10}^8 {\mathrm{ms}}^{-1}\right)$

By Einstein’s photoelectric equation, we have

       $\mathrm{hf}=\mathrm{\varphi }+(\mathrm{KE}{)}_{max}$

where, f is the frequency, $\mathrm{\varphi }$ is the work-function of metal and (KE)_max is the maximum kinetic energy of emitted photoelectron.

$\begin{array}{l}\Rightarrow \frac{\mathrm{hc}}{\mathrm{\lambda }}=\mathrm{\varphi }+{\mathrm{eV}}_0\\ \Rightarrow {\mathrm{eV}}_0=\frac{\mathrm{hc}}{\mathrm{\lambda }}-\mathrm{\varphi }\\ \text{ Here, }\mathrm{\lambda }=280\mathrm{nm}\text{ and }\mathrm{\varphi }=2.5\mathrm{eV}\\ \Rightarrow \mathrm{e}{\left({\mathrm{V}}_0\right)}_1=\frac{1240}{280}-2.5\\ \Rightarrow \mathrm{e}{\left({\mathrm{V}}_0\right)}_1=1.93\mathrm{eV}\\ \Rightarrow {\left({\mathrm{V}}_0\right)}_1=1.93\mathrm{V}\end{array}$

Similarly, stopping potential for the light of wavelength of 400 nm,

$\begin{array}{l} \mathrm{e}{\left({\mathrm{V}}_0\right)}_2=\frac{1240}{400}-2.5\\ \Rightarrow {\left({\mathrm{V}}_0\right)}_2=0.6\mathrm{V}\end{array}$

$\therefore$ Change in stopping potential,

           $\begin{array}{l}\mathrm{\Delta V}={\left({\mathrm{V}}_0\right)}_1-{\left({\mathrm{V}}_0\right)}_2\\ =1.93-0.6=1.33\simeq 1.3\mathrm{V}\end{array}$