Question

What is the energy, momentum and wavelength of the photon emitted by a hydrogen atom when an electron makes a transition from $n=2$ to $n=1$? Given that ionization potential is $13.6eV$.

Answer 1

$16.32\times {10}^{-19}J$, $5.44\times {10}^{-27}$ $kg$ $m/sec$, $1218\mathring{A}$.
$E_1=-13.6eV$
$E_2=\frac{-13.6}{4}eV$
$\Delta E=\frac{3}{4}\times 13.6eV$
$=0.75\times 13.6\times 1.6\times {10}^{-19}J=1.632\times {10}^{-18}J$
$\frac{hc}{\lambda }=1.632\times {10}^{-18}$
$\lambda =\frac{6.626\times {10}^{-34}\times 3\times {10}^8}{1.632\times {10}^{-18}}=1218\times {10}^{-10}m=1218\mathring{A}$.
$\lambda =\frac{h}{p}$
$\therefore$ $p=\frac{h}{\lambda }=\frac{6.626\times {10}^{-34}}{1218\times {10}^{-10}}=5.44\times {10}^{-27}kg-m/sec$.