The acceleration of an electron due to the mutual attraction between the electron and a proton when they are 1.6 $\stackrel{0}{\mathrm{A}}$ apart is, $\left(\begin{array}{l}\text{ take, }\frac{1}{4\pi {\epsilon }_0}=9\times {10}^9{\mathrm{Nm}}^2{\mathrm{C}}^{-2},m_e\simeq 9\times {10}^{-31} \mathrm{kg},\\ e=1.6\times {10}^{-19}\end{array}\right)$            

Force due to mutual attraction between the electron and proton. (when,  $r=1.6 \stackrel{0}{\mathrm{A}}=1.6\times {10}^{-10} \mathrm{m}$)  is given as

$F=9\times {10}^9\times \frac{e^2}{r^2}$

$=9\times {10}^9\times \frac{{\left(1.6\times {10}^{-19}\right)}^2}{{\left(1.6\times {10}^{-10}\right)}^2}=9\times {10}^{-9} \mathrm{N}$

$\therefore$   Acceleration of electron

$=\frac{F}{m_e}=\frac{9\times {10}^{-9}}{9\times {10}^{-31}}={10}^{22} \mathrm{m}/{\mathrm{s}}^2$