Physics - Modern Physics Question with Solution | TestHub

The magnitude of total energy of an electron in the $n^{th}$ Bohr orbit is given by $E\propto \frac{1}{n^2}$.

The angular momentum of the electron in the $n^{th}$ Bohr orbit is given by $L=\frac{nh}{2\pi }$.

This implies $L\propto n$.

Substituting $n\propto L$ in the expression for energy, we get $E\propto \frac{1}{L^2}$.

Thus, $E\propto L^{-2}$.

Answer: $E\propto L^{-2}$