Light of wavelength 500 nm is incident on a metal with work function 2.28 eV. The de Broglie wavelength of the emitted electron is

 According to Einstein's photoelectric equation, the maximum kinetic energy of the emitted electron is

$K_{\max }=\frac{hc}{\lambda }-{\varphi }_0$

where $\lambda$ is the wavelength of incident light and ${\varphi }_0$ is the work function.

$\text{ Here, }\lambda =500\mathrm{nm},hc=1240\mathrm{eVnm}$

$\text{ and }{\varphi }_0=2.28\mathrm{eV}$

$\therefore K_{\max }=\frac{1240\mathrm{eVnm}}{500\mathrm{nm}}-2.28\mathrm{eV}$

$=2.48\mathrm{eV}-2.28\mathrm{eV}=0.2\mathrm{eV}$

The de Broglie wavelength of the emitted electron is 

${\lambda }_{\min }=\frac{h}{\sqrt{2m{K}_{\max }}}$

where h is the Planck's constant and m is the mass of the electron.

$\text{ As }h=6.6\times {10}^{-34} \mathrm{J} \mathrm{s}, m=9\times {10}^{-31} \mathrm{kg}$

$\text{ and }{K}_{\max }=0.2\mathrm{eV}=0.2\times 1.6\times {10}^{-19} \mathrm{J}$

$\therefore {\lambda }_{\min }=\frac{6.6\times {10}^{-34} \mathrm{J} \mathrm{s}}{\sqrt{2\left(9\times {10}^{-31} \mathrm{kg}\right)\left(0.2\times 1.6\times {10}^{-19} \mathrm{J}\right)}}$

$=\frac{6.6}{2.4}\times {10}^{-9} \mathrm{m}=2.8\times {10}^{-9} \mathrm{m}$

$\text{ So, }\lambda \ge 2.8\times {10}^{-9} \mathrm{m}$