Answer
See explanation.
Work Step by Step
a. Find the photon energy from the wavelength.
$$E =hf=\frac{hc}{\lambda }=\frac{1.24\times10^{-6}eV\cdot m}{5.00\times10^{-9}m}=248 eV$$
b. Find the electron energy from the wavelength.
For a nonrelativistic electron, $E=\frac{p^2}{2m}=\frac{h^2}{2m\lambda^2}$.
$$E=\frac{(6.626\times10^{-34}J \cdot s)^2}{2(9.11\times10^{-31}kg)( 5.00\times10^{-9}m)^2}$$
$$E= 9.64\times10^{-21}J=0.0602eV
$$
The answer confirms that the electron is nonrelativistic, since its energy is much less than its rest energy.
