Question 11.14: 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.
(a) Hint: \(\frac12m_ev^2\) Step 1: Relate velocity with de Broglie wavelength. \(~~~~~~~~~~~~~\lambda=\frac{h}{m_ev}\\ ~~~~~~\Rightarrow~v^2=\frac{h^2}{\lambda^2m_e~^2}~~~~~.......(i)\)
Step 2: Find the kinetic energy of the electron
As \(K=\frac12m_ev^2 ~~~.........(ii)\)
Substituting equation (i) in equation (ii), we get \(K=\frac12\frac{m_eh^2}{\lambda^2m_e^2}=\frac{h^2}{2\lambda^2m_e}\\ ~~~~=\frac{(6.6\times10^{-34})^2}{2\times(589\times10^{-9})^2\times9.1\times10^{-31}}\\ ~~~~\approx6.9\times10^{-25}~\text J\\ ~~~~=\frac{6.9\times^{-25}}{1.6\times10^{-19}}\\ ~~~~=4.31\times10^{-6}~\text{ev}\\ ~~~~=4.31~\mu ev\)
(b) Hint: Kneutron\(=\frac{h^2}{2h^2m_n}\) Step: Find the kinetic energy of the neutron. \(K_n=\frac{h^2}{2\lambda^2m_n}\\ ~~~~~=\frac{(6.6\times10^{-34})^2}{2\times(589\times10^{-9})^2\times1.66\times10^{-27}}\\ ~~~~~=3.78\times10^{-28}\text J\\ ~~~~~=\frac{3.78\times10^{-28}}{1.6\times10^{-19}}=2.36\times10^{-9}\text{ ev}\\ ~~~~~=2.36\text{ nev}\)
