Physics - Modern Physics Question with Solution | TestHub

The potential energy is given by $V(r)=-\frac{k{e}^2}{3r^3}$.

According to Bohr's theory, the angular momentum is quantized:

$L=mvr=n\hslash$.

The centripetal force is provided by the derivative of the potential energy: $\frac{m{v}^2}{r}=-\frac{dV}{dr}=\frac{3k{e}^2}{3r^4}=\frac{k{e}^2}{r^4}$.

Solving for $v$ and substituting into the angular momentum equation, we find that $r\propto n^2$.

The energy $E=\frac{1}{2}m{v}^2+V=\frac{1}{2}\frac{k{e}^2}{r^3}-\frac{k{e}^2}{3r^3}=\frac{k{e}^2}{6r^3}$.

Thus, $E\propto \frac{1}{r^3}\propto \frac{1}{n^6}$ and $E\propto \frac{1}{L^3}$. Therefore, both (1) and (2) are correct.