Answer
$5.48\times 10^{-7}$ $m$
Work Step by Step
We have already calculated the work function $(\phi)$ of sodium in the previous part of the question:
$\phi=2.27$ $eV$ $=2.27\times1.6\times 10^{-19}$ $J$ $=3.632\times 10^{-19}$ $J$
If $f_{0}$ be cutoff frequency, then the work function ($\phi$) of sodium is
$\phi=hf_{0}$
or, $\phi=\frac{ch}{\lambda_{0}}$, where $\lambda_{0}$ the corresponding cutoff wavelength.
$\therefore \lambda_{0}=\frac{ch}{\phi}$
or, $\lambda_{0}=\frac{3\times 10^{8}\times 6.63\times 10^{-34}}{3.632\times 10^{-19}}$ $m$
or, $\lambda_{0}\approx 5.48\times 10^{-7}$ $m$
Therefore, the cutoff wavelength for sodium is $5.48\times 10^{-7}$ $m$
