Question

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

Answer 1

Since F = $\frac{q_1{q}_2}{4\pi {ϵ}_0{r}^2}$

[Smallest distance between the electron and nucleus in the radius of first Bohr's Orbit]

$\frac{(1.6\times {10}^{-19})\times (1.6\times {10}^{-19}\times 9\times {10}^9}{(0.53\times {10}^{-10}{)}^2}$

$\frac{1.6\times 1.6\times 9\times {10}^9}{(0.53{)}^3}=82.02\times {10}^{-9}$

= 8.2 $\times {10}^{-8}N$