An electron and a photon each have a wavelength of $1.00 \; nm$. Find
$(a)$ their momenta,
$(b)$ the energy of the photon,and
$(c)$ the kinetic energy of the electron.

(N/A) The momentum $p$ for both a photon and an electron is given by the de Broglie relation: $p = h / \lambda$.
Given $\lambda = 1.00 \; nm = 1.00 \times 10^{-9} \; m$ and $h = 6.63 \times 10^{-34} \; J \cdot s$.
$p = (6.63 \times 10^{-34}) / (1.00 \times 10^{-9}) = 6.63 \times 10^{-25} \; kg \cdot m/s$.
$(b)$ The energy of a photon is given by $E = hc / \lambda$.
$E = (6.63 \times 10^{-34} \; J \cdot s \times 3.00 \times 10^8 \; m/s) / (1.00 \times 10^{-9} \; m) = 1.989 \times 10^{-16} \; J$.
Converting to $eV$: $E = (1.989 \times 10^{-16} \; J) / (1.602 \times 10^{-19} \; J/eV) \approx 1.24 \; keV$.
$(c)$ The kinetic energy $K$ of the electron is given by $K = p^2 / (2m_e)$.
Using $p = 6.63 \times 10^{-25} \; kg \cdot m/s$ and $m_e = 9.11 \times 10^{-31} \; kg$:
$K = (6.63 \times 10^{-25})^2 / (2 \times 9.11 \times 10^{-31}) = 4.39569 \times 10^{-49} / 1.822 \times 10^{-30} \approx 2.41 \times 10^{-19} \; J$.
Converting to $eV$: $K = (2.41 \times 10^{-19} \; J) / (1.602 \times 10^{-19} \; J/eV) \approx 1.51 \; eV$.