Step 1: Find the wavelength of an electron and photon.
The momentum of an elementary particle is given by de Broglie relation:$\lambda =\frac{h}{p}\Rightarrow p=\frac{h}{\lambda }$
It is clear that momentum depends only on the wavelength of the particle. Since the wavelengths of an electron and a photon are
equal, both have an equal momentum.
$\therefore p=\frac{6.63\times {10}^{-34}}{1\times {10}^{-9}}=6.63\times {10}^{-23}kgm{s}^{-1}$

(b)
Hint: \(E=\frac{hc}{\lambda}\)

Step 1: Find the energy of the photon.
The energy of a photon is given by the relation:
$E=\frac{hc}{\lambda }$
Where, Speed of light, c = 3 x ${10}^3$ m/s

$\begin{array}{l}\therefore E=\frac{6.63\times {10}^{-34}\times 3\times {10}^8}{1\times {10}^{-9}\times 1.6\times {10}^{-19}}\\ =1243.1eV=1.243keV\end{array}$

Therefore, the energy photon is 1.243 keV.

(c)
Hint: \(K=\frac{p^{2}}{2m}\)

Step 1: Find the kinetic energy of the electron.
The kinetic energy (K) of an electron having momentum p, is given by the relation:

$K=\frac{1}{2}\frac{p^2}{m}$

Where
m = Mass of the electron= 9.1 x ${10}^{-31}$ kg
p = \(6.63\times10^{-25}\) kg m ${\mathrm{s}}^{-1}$

$\begin{array}{l}\therefore K=\frac{1}{2}\times \frac{(6.63\times {10}^{-25}{)}^2}{9.1\times {10}^{-31}}=2.415\times {10}^{-10}J\\ =\frac{2.415\times {10}^{-10}}{1.6\times {10}^{-19}}=1.51eV\end{array}$

Hence, the kinetic energy of the electron is 1.51 eV.