The acceleration of an electron due to the mutual attraction between the electron and a proton when they are…

Recall the formula of electrostatic force interns of charge on electron and septation between them, $F=K\frac{e^2}{r^2}$, now use Newton's second law, $ma=K\frac{e^2}{r^2}$

$\Rightarrow a=K\frac{e^2}{m{r}^2}$, 

so,  the acceleration, $a=9\times {10}^9\frac{{\left(1.6\times {10}^{-19}\right)}^2}{{\left(1.6\times {10}^{-10}\right)}^2\left(9\times {10}^{-31}\right)}$

$\Rightarrow a={10}^{-29}\times {10}^{51}={10}^{22} \mathrm{m} {\mathrm{s}}^{-2}$