The wavelength of light from the spectral emission line of sodium is $589 \; nm$. Find the kinetic energy at which
$(a)$ an electron,and
$(b)$ a neutron,would have the same de Broglie wavelength.

(N/A) Given,wavelength $\lambda = 589 \; nm = 589 \times 10^{-9} \; m$.
The momentum $p$ associated with this wavelength is given by $p = \frac{h}{\lambda}$,where $h = 6.626 \times 10^{-34} \; J \cdot s$.
$p = \frac{6.626 \times 10^{-34}}{589 \times 10^{-9}} \approx 1.125 \times 10^{-27} \; kg \cdot m/s$.
$(a)$ For an electron,mass $m_e = 9.11 \times 10^{-31} \; kg$. The kinetic energy $K_e = \frac{p^2}{2m_e}$.
$K_e = \frac{(1.125 \times 10^{-27})^2}{2 \times 9.11 \times 10^{-31}} \approx 6.95 \times 10^{-25} \; J$.
$(b)$ For a neutron,mass $m_n = 1.675 \times 10^{-27} \; kg$. The kinetic energy $K_n = \frac{p^2}{2m_n}$.
$K_n = \frac{(1.125 \times 10^{-27})^2}{2 \times 1.675 \times 10^{-27}} \approx 3.78 \times 10^{-28} \; J$.