The acceleration of an electron due to the mutual attraction between the electron and a proton when they are $1.6 \mathrm{\AA }$ apart is,
$\left(m_{\mathrm{e}}\simeq 9\times {10}^{-31} \mathrm{kg}, e=1.6\times {10}^{-19} \mathrm{C}\right)$ (Take $\frac{1}{4\pi {\epsilon }_0}=9\times {10}^9 \mathrm{N} {\mathrm{m}}^2 {\mathrm{C}}^{-2}$)

• A

  ${10}^{24} \mathrm{m} {\mathrm{s}}^{-2}$
• B

  ${10}^{23} \mathrm{m} {\mathrm{s}}^{-2}$
• Correct answer
  C

  ${10}^{22} \mathrm{m} {\mathrm{s}}^{-2}$
• D

  ${10}^{25} \mathrm{m} {\mathrm{s}}^{-2}$

Hence, the correct answer is option: 'C'.