The threshold wavelength for a metal is $\displaystyle 3315 \AA$. What should be the wavelength of light incident on the metal surface so that the maximum kinetic energy of the emitted photoelectrons be $\displaystyle 1.25$ eV ?

$\displaystyle \lambda_o = 3315\ \text{Å}$$\displaystyle \mathrm{K_{max}} = 1.25\ \text{eV} = 1.25 \times 1.6 \times 10^{-19}\ J$ $\displaystyle = 2 \times 10^{-19}\ J$ $\displaystyle = 0.2 \times 10^{-18}\ J$Work function $\displaystyle = \phi_o = \dfrac{hc}{\lambda_o}$$\displaystyle = \dfrac{6.63 \times 10^{-34} \times 3 \times 10^{8}}{3315 \times 10^{-10}} = 0.6 \times 10^{-18}\ \text{J}$Using $\displaystyle h\nu = \phi_o + K_{max}$$\displaystyle \dfrac{hc}{\lambda} = \phi_o + K_{max}$$\displaystyle = (0.6 \times 10^{-18} + 0.2 \times 10^{-18})\,J$$\displaystyle = 0.8 \times 10^{-18}\ J$$\displaystyle \lambda = \dfrac{hc}{0.8 \times 10^{-18}} = \dfrac{6.63 \times 10^{-34} \times 3 \times 10^{8}}{0.8 \times 10^{-18}}$$\displaystyle = 24.86 \times 10^{-8}\ \text{m}$$\displaystyle = 2486 \times 10^{-10}\ \text{m} = 2486\ \text{Å}$