Physics - Modern Physics Question with Solution | TestHub

According to Bohr's theory of the hydrogen atom, the radius of the $n$-th orbit is proportional to the square of the principal quantum number, $r\propto n^2$.

The electrostatic force of attraction acting on the electron is given by Coulomb's law:

$F=\frac{1}{4\pi {ϵ}_0}\frac{e^2}{r^2}$

Substituting the proportionality of $r$ into the force equation, we get:

$F\propto \frac{1}{(n^2{)}^2}$

$F\propto n^{-4}$

Alternatively, using the centripetal force expression $F=\frac{m{v}^2}{r}$, and knowing that velocity $v\propto n^{-1}$ and $r\propto n^2$, we obtain:

$F\propto \frac{(n^{-1}{)}^2}{n^2}=n^{-4}$

Answer: $n^{-4}$