Physics - Modern Physics Question with Solution | TestHub

The time period of revolution in Bohr's orbit is given by $T=\frac{2\pi r}{v}$,

where $r\propto n^2$ and $v\propto \frac{1}{n}$.

Therefore, $T\propto n^3$.

The ground state corresponds to $n=1$ and the first excited state corresponds to $n=2$. Thus, the ratio of the times is

$\frac{T_1}{T_2}=\frac{1^3}{2^3}=\frac{1}{8}$.

The ratio is $1:8$.