Question

what is stopping potential? how is it related to maximum velocity of ejected electron? find the stopping potential if the maximum kinetic energy of emitted photoelectrons from plate is 0.5825 eV.

Answer 1

Dear Student,

If we increase the negative potential of the collector, the photoelectric current decreases rapidly. At a certain critical value of the negative potential of the collector, the photoelectric current becomes zero. This potential is called stopping potential or cut-off potential.

The stopping potential is related to the maximum kinetic energy of the ejected electrons by
$$\begin{gathered} {\mathrm{eV}}_0={\mathrm{K}}_{\max } \\ \mathrm{where},{\mathrm{K}}_{\max }=\mathrm{maximum}\mathrm{kinetic}\mathrm{energy} \\ {\mathrm{V}}_0=\mathrm{stopping}\mathrm{potential} \\ \mathrm{e}=\mathrm{charge}\mathrm{of}\mathrm{electron} \\ {\mathrm{eV}}_0=\frac{1}{2}{\mathrm{mv}}_{\max }^2 \\ \mathrm{where},{\mathrm{v}}_{\max }=\mathrm{maximum}\mathrm{velocity}\mathrm{of}\mathrm{ejected}\mathrm{electron} \end{gathered}$$
$$\begin{gathered} {\mathrm{eV}}_0=\mathrm{kinetic}\mathrm{energy} \\ \mathrm{Kinetic}\mathrm{energy},\mathrm{K}=0.5825\mathrm{eV} \\ 1\mathrm{eV}=1.6\times {10}^{-19}\mathrm{J} \\ \mathrm{K}=0.5825\times 1.6\times {10}^{-19}\mathrm{J} \\ {\mathrm{V}}_0=\frac{\mathrm{K}}{\mathrm{e}} \\ {\mathrm{V}}_0=\frac{0.5825\times 1.6\times {10}^{-19}}{1.6\times {10}^{-19}} \\ {\mathrm{V}}_0=0.5825\mathrm{V} \end{gathered}$$