Answer
$545\;nm$
Work Step by Step
If $f_{0}$ be cutoff frequency, then the work function ($\phi$) of the metallic sodium is
$\phi=hf_{0}$
or, $\phi=\frac{ch}{\lambda_{0}}$, where $\lambda_{0}$ the corresponding cutoff wavelength
or, $\lambda_{0}=\frac{ch}{\phi}$
Given $\phi=2.28\;eV=2.28\times1.6\times 10^{-19}\;J$
$\therefore\;\lambda_{0}=\frac{3\times 10^{8}\times 6.63\times 10^{-34}}{2.28\times1.6\times 10^{-19}}\;m$
or, $\lambda_{0}\approx5.45\times 10^{-7}\;m$
or, $\lambda_{0}=545\;nm$
Therefore, the cutoff wavelength for photoelectric emission from sodium is $545\;nm$
