Answer
$234$ $nm$
Work Step by Step
If $f_{0}$ be cutoff frequency, then the work function ($\phi$) of platinum metal can be written as
$\phi=hf_{0}$
or, $\phi=\frac{ch}{\lambda_{0}}$, ..................$(1)$
where $\lambda_{0}$ the corresponding cutoff wavelength.
The value of $\phi$ is given: $\phi=5.32$ $eV$ $=5.32\times1.6\times 10^{-19}$ $J$ $=8.512\times 10^{-19}$ $J$
From eq. $1$ we get.
$\lambda_{0}=\frac{ch}{\phi}$
or, $\lambda_{0}=\frac{3\times 10^{8}\times 6.63\times 10^{-34}}{8.512\times 10^{-19}}$ $m$
or, $\lambda_{0}\approx2.34\times 10^{-7}$ $m$
or, $\lambda_{0}=234$ $nm$
$\therefore$ The longest wavelength of incident sunlight that can eject an electron from the platinum is $234$ $nm$
