Light of frequency 5 · 0 × 10^14 Hz is incident on a metal surface. If the maximum speed of photoelectrons emitted is 6.63 × 10^5 ms^-1, calculate…

\[\begin{aligned} K . E & =h v-h v_{0} \\ v_{0} & =\frac{h v-K . E}{h} \\ & =\frac{\left.\left(6.63 \times 10^{-34} \times 5 \times 10^{14}\right)-\left(\dfrac{1}{2} \times 9.1 \times 10^{-31}\right) \times(6.63)^{2} \times 10^{10}\right)}{6.63 \times 10^{-34}} \\ & =\frac{\left(5 \times 10^{-20}\right)-\left(3 \times 10^{-20}\right)}{10^{-34}} \\ & =2 \times 10^{14} \mathrm{~Hz} \end{aligned} \]