Question
Find the maximum Coulomb force that can act on the electron due to the nucleus in a hydrogen atom.

Answer

$\text{F}=\frac{\text{q}_1\text{q}_2}{4\pi\in_0\text{r}^2}$[Smallest dist. Between the electron and nucleus in the radius of first Bohrs orbit]
$=\frac{\big(1.6\times10^{-19}\big)\times\big(1.6\times10^{-19}\big)\times9\times10^9}{\big(0.53\times10^{-10}\big)^2}$
$=82.02\times10^{-9}=8.202\times10^{-8}=8.2\times10^{-8}\text{N}$

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