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

\(K_{\max }=\frac{h c}{\lambda}-\phi_{0}\)

where \(\lambda\) is the wavelength of incident light and \(\phi _0\) is the work function.

Here, \(\lambda=500\, \mathrm{nm}, h c=1240\, \mathrm{eV} \,\mathrm{nm}\) and \(\phi_{0}=2.28\, \mathrm{eV}\)

\(\therefore \)  \(K_{\max } =\frac{1240\, \mathrm{eV} \mathrm{nm}}{500\, \mathrm{nm}}-2.28 \,\mathrm{eV}\)

\(=2.48\, \mathrm{eV}-2.28\, \mathrm{eV}=0.2\, \mathrm{eV}\)

The de Broglie wavelength of the emitted electron is

\(\lambda_{\min }=\frac{h}{\sqrt{2 m K_{\max }}}\)

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

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

and \(K_{\max }=0.2\,\mathrm{eV}=0.2 \times 1.6 \times 10^{-19}\, \mathrm{J}\)

\(\therefore\) \( \lambda_{\min } =\frac{6.6 \times 10^{-34} \,\mathrm{Js}}{\sqrt{2\left(9 \times 10^{-31} \,\mathrm{kg}\right)\left(0.2 \times 1.6 \times 10^{-19} \,\mathrm{J}\right)}}\)

\(=\frac{6.6}{2.4} \times 10^{-9}\, \mathrm{m}=2.8 \times 10^{-9}\, \mathrm{m}\)

So, \(\lambda \geq 2.8 \times 10^{-9}\, \mathrm{m}\)