The stopping potential for electrons emitted from a photosensitive surface illuminated by light of wavelength

(2)

From the photoelectric effect equation

$\frac{h c}{\lambda}=\phi+\operatorname{ev}_{s}$

so $\mathrm{ev}_{\mathrm{s}_{1}}=\frac{\mathrm{hc}}{\lambda_{1}}-\phi$........

$\mathrm{ev}_{83}=\frac{\mathrm{hc}}{1}-\phi \ldots . . .(\mathrm{ii})$

Subtract equation (i) from equation (ii) ev $_{s_{1}}-\mathrm{ev}_{s_{2}}=\frac{h c}{\lambda_{1}}-\frac{h c}{\lambda_{2}}$

$v_{s_{1}}-v_{s_{2}}=\frac{h c}{e}\left(\frac{1}{\lambda_{1}}-\frac{1}{\lambda_{2}}\right)$

$(0.710-1.43)=1240\left(\frac{1}{491}-\frac{1}{\lambda_{2}}\right)$

$\frac{-0.72}{1240}=\frac{1}{491}-\frac{1}{\lambda_{2}}$

$\frac{1}{\lambda_{2}}=\frac{1}{491}+\frac{0.72}{1240}$

$\frac{1}{\lambda_{2}}=0.00203+0.00058$

$\frac{1}{\lambda_{2}}=0.00261$

$\lambda_{2}=383.14$

$\lambda_{2} \simeq 382 \mathrm{~nm}$