Question

Consider the situation of the previous problem. Consider the faster electron emitted parallel to the large metal plate. Find the displacement of this electron parallel to its initial velocity before it strikes the large metal plate.

Answer 1

Electric field of the metal plate,
$$\begin{gathered} E=\frac{\sigma }{{\epsilon }_0}=\frac{1\times {10}^{-9}}{8.85\times {10}^{-12}} \\ =113\mathrm{V}/\mathrm{m} \\ \mathrm{Acceleration}, \\ a=\frac{q\mathrm{E}}{m}, \\ \text{w}\mathrm{here}q=\mathrm{charge}\mathrm{on}\mathrm{electron} \\ E=\mathrm{electric}\mathrm{field} \\ m=\mathrm{mass}\mathrm{of}\mathrm{electron} \\ a=\frac{1.6\times {10}^{-19}\times 113}{9.1\times {10}^{-31}}=19.87\times {10}^{12} \\ t=\sqrt{\frac{2\mathrm{y}}{a}}=\sqrt{\frac{2\times 20\times {10}^{-2}}{19.87\times {10}^{12}}} \\ =1.41\times {10}^{-7}\mathrm{s} \\ \mathrm{From}\mathrm{Einstein}\text{'}s\mathrm{photoelectric}\mathrm{equation}, \\ \mathrm{K}.\mathrm{E}.=\frac{hc}{\mathrm{\lambda }}-\mathrm{W}=1.2\mathrm{eV} \\ =1.2\times 1.6\times {10}^{-19}\mathrm{J} \\ [\because \mathrm{in}\mathrm{problem}31:\mathrm{KE}=1.2\mathrm{eV}] \\ \therefore \mathrm{Velocity},v=\sqrt{\frac{2\mathrm{KE}}{m}} \\ =\sqrt{{\displaystyle \frac{2\times 1.2\times 1.6\times {10}^{-19}}{4.1\times {10}^{-31}}}}\sqrt{{\displaystyle \frac{2\times 1.2\times 1.6\times {10}^{-19}}{4.1\times {10}^{-31}}}} \\ =0.665\times {10}^{-6}\mathrm{m}/\mathrm{s} \end{gathered}$$

∴ Horizontal displacement,
$$\begin{gathered} S=v\times t \\ S=0.665\times {10}^{-6}\times 1.4\times {10}^{-7} \\ S=0.092\mathrm{m}=9.2\mathrm{cm} \end{gathered}$$