Chemistry - Atomic Structure Question with Solution | TestHub

The kinetic energy of the ejected electron is given by the equation $$\begin{array}{rlrl}hv& =h{v}_0+\frac{1}{2}m{v}^2\because v=\frac{c}{\lambda }& & \\ & \text{ or }\frac{hc}{\lambda }& =\frac{hc}{{\lambda }_0}+\frac{1}{2}m{v}^2& \\ \frac{1}{2}m{v}^2& =\frac{hc}{\lambda }-\frac{hc}{{\lambda }_0}& & \\ & =hc\left(\frac{{\lambda }_0-\lambda }{\lambda {\lambda }_0}\right)& & \\ \therefore & v^2& =\frac{2hc}{m}\left(\frac{{\lambda }_0-\lambda )}{\lambda {\lambda }_0}\right)& \\ & & v& =\sqrt{\frac{2hc}{m}\left(\frac{{\lambda }_0-\lambda }{\lambda {\lambda }_0}\right)}\end{array}$$