• # question_answer                 In a photo-emissive cell, with exciting wavelength $\lambda$, the fastest electron has speed v. If the exciting wavelength is changed to $3\lambda /4$, the speed of the fastest emitted electron will be:     A)                 $v\,{{(3/4)}^{1/2}}$ B)                 $v{{(4/3)}^{1/2}}$                                      C)                 less than $v{{(4/3)}^{1/2}}$         D)                 greater than $v{{(4/3)}^{1/2}}$

Einstein?s photoelectric equation is given                 ${{E}_{k}}=E-W$                 but         ${{E}_{k}}=\frac{1}{2}m{{v}^{2}}$ and $E=\frac{hc}{\lambda }$                 $\therefore \frac{1}{2}m{{v}^{2}}=\frac{hc}{\lambda }-W$                 Suppose v? be the new speed, when $\lambda$ is changed to $\frac{3\lambda }{4}$,                 $\therefore \frac{1}{2}mv{{'}^{2}}=\frac{hc}{(3\lambda /4)}-W$                 $or\frac{1}{2}mv{{'}^{2}}=\frac{4}{3}\frac{hc}{\lambda }-W$                 Dividing Eq. (ii) by Eq. (i), we get                 $\frac{v{{'}^{2}}}{{{v}^{2}}}=\frac{\frac{4}{3}\frac{hc}{\lambda }-W}{\frac{hc}{\lambda }-W}$                 $=\frac{\frac{4}{3}\frac{hc}{\lambda }-\frac{4}{3}W+\frac{1}{3}W}{\frac{hc}{\lambda }-W}$                 $=\frac{4}{3}+\frac{W}{3\left( \frac{hc}{\lambda }-W \right)}>\frac{4}{3}$                 $\therefore \frac{v'}{v}>\sqrt{\frac{4}{3}}\,\,or\sqrt{\frac{4}{3}}\,v$