Answer
0.82 nm.
Work Step by Step
At such a low energy, the electrons are nonrelativistic.
The maximum kinetic energy of the photoelectrons is given by Eq. 27–5b. Use the nonrelativistic equation for momentum and KE.
$$KE_{max}= \frac{hc}{\lambda}-W_0=\frac{p^2}{2m}$$
Now use the de Broglie relation.
$$ \frac{hc}{\lambda}-W_0=\frac{h^2}{2m\lambda_e^2}$$
The shortest electron wavelength corresponds to the largest momentum and kinetic energy.
$$\lambda_e=\frac{h}{\sqrt{2m(\frac{hc}{\lambda}-W_0)}}$$
$$=\frac{6.626 \times10^{-34}J s}{\sqrt{2(9.11\times10^{-31}kg)(\frac{1240 eV nm}{280nm}-2.2eV)(1.60\times10^{-19}J/eV)}}$$
$$=8.2\times10^{-10}m$$