# when light of 3000 armstrong is incident on sodium cathode ,the stopping potential is 1.85V and when light of 4000 armstrong is incident then the stopping potential becomes 0.82v.calculate :plank's constant ,work function of sodium,threshold wavelength for sodium.

$\mathrm{Einstein}\mathrm{Photoelectric}\mathrm{equation}:\phantom{\rule{0ex}{0ex}}\mathrm{E}=\mathrm{W}+\mathrm{KE}\phantom{\rule{0ex}{0ex}}\frac{\mathrm{hc}}{\mathrm{\lambda }}=\frac{\mathrm{hc}}{{\mathrm{\lambda }}_{0}}+\mathrm{KE}\phantom{\rule{0ex}{0ex}}\mathrm{Case}\mathrm{I}:\phantom{\rule{0ex}{0ex}}\frac{\mathrm{hc}}{3000×{10}^{-10}}=\frac{\mathrm{hc}}{{\mathrm{\lambda }}_{0}}+1.85\mathrm{eV}\phantom{\rule{0ex}{0ex}}\frac{\mathrm{hc}}{3000×{10}^{-10}}=\frac{\mathrm{hc}}{{\mathrm{\lambda }}_{0}}+\left(1.85×1.6×{10}^{-19}\right)\mathrm{J}\phantom{\rule{0ex}{0ex}}\frac{\mathrm{hc}}{3000×{10}^{-10}}=\frac{\mathrm{hc}}{{\mathrm{\lambda }}_{0}}+2.96×{10}^{-19}......\left(1\right)\phantom{\rule{0ex}{0ex}}\mathrm{Case}\mathrm{II}:\phantom{\rule{0ex}{0ex}}\frac{\mathrm{hc}}{4000×{10}^{-10}}=\frac{\mathrm{hc}}{{\mathrm{\lambda }}_{0}}+0.82\mathrm{eV}\phantom{\rule{0ex}{0ex}}\frac{\mathrm{hc}}{4000×{10}^{-10}}=\frac{\mathrm{hc}}{{\mathrm{\lambda }}_{0}}+\left(0.82×1.6×{10}^{-19}\right)\mathrm{J}\phantom{\rule{0ex}{0ex}}\frac{\mathrm{hc}}{4000×{10}^{-10}}=\frac{\mathrm{hc}}{{\mathrm{\lambda }}_{0}}+1.312×{10}^{-19}......\left(2\right)\phantom{\rule{0ex}{0ex}}\left(1\right)-\left(2\right):\phantom{\rule{0ex}{0ex}}\mathrm{hc}\left(\frac{1}{3000×{10}^{-10}}-\frac{1}{4000×{10}^{-10}}\right)=1.648×{10}^{-19}\phantom{\rule{0ex}{0ex}}\mathrm{hc}\left(8.33×{10}^{5}\right)=1.648×{10}^{-19}\phantom{\rule{0ex}{0ex}}\mathrm{h}=\frac{1.648×{10}^{-19}}{8.33×{10}^{5}×3×{10}^{8}}\phantom{\rule{0ex}{0ex}}\mathbf{=}\mathbf{6}\mathbf{.}\mathbf{6}\mathbf{×}{\mathbf{10}}^{\mathbf{-}\mathbf{34}}\mathbf{}{\mathbf{m}}^{\mathbf{2}}\mathbf{kg}\mathbf{/}\mathbf{s}\phantom{\rule{0ex}{0ex}}\mathrm{Work}\mathrm{function}:\phantom{\rule{0ex}{0ex}}\mathrm{W}=\mathrm{E}-\mathrm{KE}\phantom{\rule{0ex}{0ex}}=\frac{\mathrm{hc}}{3000×{10}^{-10}}-2.96×{10}^{-19}\phantom{\rule{0ex}{0ex}}=\frac{\left(6.6×{10}^{-34}\right)×\left(3×{10}^{8}\right)}{3000×{10}^{-10}}-2.96×{10}^{-19}\phantom{\rule{0ex}{0ex}}\mathbf{=}\mathbf{3}\mathbf{.}\mathbf{632}\mathbf{×}{\mathbf{10}}^{\mathbf{-}\mathbf{19}\mathbf{}}\mathbf{}\mathbf{J}\phantom{\rule{0ex}{0ex}}\mathrm{Threshold}\mathrm{wavelength}:\phantom{\rule{0ex}{0ex}}{\mathrm{\lambda }}_{\mathrm{o}}=\frac{\mathrm{hc}}{\mathrm{W}}\phantom{\rule{0ex}{0ex}}=\frac{\left(6.6×{10}^{-34}\right)×\left(3×{10}^{8}\right)}{3.632×{10}^{-19}}\phantom{\rule{0ex}{0ex}}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{5}\mathbf{.}\mathbf{452}\mathbf{×}{\mathbf{10}}^{\mathbf{-}\mathbf{10}}\mathbf{}\mathbf{m}\phantom{\rule{0ex}{0ex}}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{5452}\mathbf{}\mathbf{A}$

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