Question

When a certain metal is radiated with a radiation 4.5x10¹⁶ Hzfrequency. The electron ejected has three times the kinetic energy as the kinetic energy of an electron emitted when the same metal is radiated with a radiation of 2.5x10¹⁶Hz. Calculate the threshold frequency.

Answer 1

$$\begin{gathered} KE=h\nu -h{\nu }_0=h(\nu -v_0) \\ K{E}_1=h({\nu }_1-v_0)-----\left(1\right) \\ K{E}_2=h({\nu }_2-v_0)-----\left(2\right) \\ Dividingequationiibyi \\ \frac{K{E}_2}{K{E}_1}=\frac{h({\nu }_2-v_0)}{h({\nu }_1-v_0)} \\ =\frac{({\nu }_2-v_0)}{({\nu }_1-v_0)} \\ 3=\frac{({\nu }_2-v_0)}{({\nu }_1-v_0)} \\ 3{\nu }_1-3{\nu }_0={\nu }_2-v_0 \\ 3{\nu }_1-{\nu }_2=3{\nu }_0-{\nu }_0 \\ 3\times 2.5\times {10}^{16}-4.5\times {10}^{16}=2{\nu }_0 \\ {\nu }_0=\frac{3\times {10}^{16}}{2}=1.5\times {10}^{16}se{c}^{-1} \end{gathered}$$