A light of wavelength 400 nm is incident on metal surface whose work function is 3.0 × 10^-19 J. Calculate the speed of the fastest photoelectrons…

$\displaystyle h \nu=\phi_{0}+K_{\max }$ $\displaystyle \frac{h c}{\lambda}=\phi_{0}+\frac{1}{2} m v_{\max }^{2}$ $\displaystyle \frac{6.63 \times 10^{-34} \times 3 \times 10^{8}}{4 \times 10^{-7}}=3 \times 10^{-19}+\frac{1}{2} m v_{\max }^{2}$ $\displaystyle \frac{19.89}{4} \times 10^{-19}=3 \times 10^{-19}+\frac{1}{2} m v_{\max }^{2}$ $\displaystyle \frac{4 \times 10^{-19}}{9 \times 10^{-31}}=v_{\max }^{2}$ $\displaystyle v_{\max }=\frac{2}{3} \times 10^{6} \mathrm{~m} / \mathrm{s}$