In a hydrogen atom, an electron of mass 9 . 1 × 10 - 31   kg revolves about a proton in circular…

The centripetal force is equal to the electrostatic force, when
an electron revolves around a proton.

$\frac{m_e{v}^2}{r}=\frac{k{q}_1{q}_2}{r^2}$

Rewrite the above equation.

$m_e{v}^{2}r=k{q}_1{q}_2$

$m_e(r\omega {)}^{2}r=kqq$

$\omega =\sqrt{\frac{k{q}^2}{m_e{r}^3}}$

Substitute the given values in the above equation.

$\omega =\sqrt{\frac{9\times {10}^9\times {\left(1.6\times {10}^{-19}\right)}^2}{9.1\times {10}^{-31}\times {\left(0.53\times {10}^{-10}\right)}^3}}$

$\omega =4.1\times {10}^{16} {\mathrm{s}}^{-1}$

The radial acceleration of the electron can be calculated as,

$a_r=r{\omega }^2$

$a_r=0.53\times {10}^{-10}\times {\left(4.1\times {10}^{16}\right)}^2$

$a_r=8.9\times {10}^{22} \mathrm{m} {\mathrm{s}}^{-2}$

$a_r\approx 9\times {10}^{22} \mathrm{m} {\mathrm{s}}^{-2}$