If the wavelength of the light incident on a photoelectric cell is reduced from  ${\lambda }_1\stackrel{\circ }{A}$to ${\lambda }_2\stackrel{\circ }{A}$ then calculate the change in the cut-off potential

Let the work function of the surface be${\varphi }_0$. If be the frequency of the light falling on the st face, then according to Einstein's photoelectic equation, the maximum kinetic energy $K_{\max }$ the emitted photoelectron is given by

$K_{\max }=hv-{\varphi }_0=\frac{hc}{\lambda }-{\varphi }_0$

Since, we know that the maximum kinetic energy of the photoelectrons emitted and the stopping potential are related to each other as $K_{\max }=eV$

$\Rightarrow e{V}_s=\frac{hc}{\lambda }-{\varphi }_0$

$\Rightarrow V_s=\frac{hc}{e\lambda }-\frac{{\varphi }_0}{e}$

Now, $\Delta V_s=V_{s2}-V_{s1}$

$\Rightarrow \Delta V_s=\left(\frac{hc}{e{\lambda }_2}-\frac{{\varphi }_0}{e}\right)-\left(\frac{hc}{e{\lambda }_1}-\frac{{\varphi }_0}{e}\right)$

$\Rightarrow \Delta V_s=\frac{hc}{e}\left(\frac{1}{{\lambda }_2}-\frac{1}{{\lambda }_1}\right)$

$\Rightarrow \Delta V_s=\frac{hc}{e}\left(\frac{{\lambda }_1-{\lambda }_2}{{\lambda }_1{\lambda }_2}\right)$