Question

An electron in a hydrogen atom makes a transition $n_1\to$ $n_2$, where $n_1$ and $n_2$ are the principal quantum numbers of the two states. Assume Bohr's model is valid in this case. The frequency of the orbital motion of the electron in the initial state is $1/27^{\text{th }}$ of that in the final state. Then the ratio of $n_1$ to $n_2$ is

Answer 1

Answer: 3

We know that frequency of orbital motion is given as
$f\propto \frac{1}{n^3};$ Given $f_1=\frac{1}{27}{f}_2$
$\Rightarrow \left(\frac{n_2^3}{n_1^3}\right)=\frac{f_1}{f_2}$
$\Rightarrow \frac{n_2}{n_1}={\left(\frac{1}{27}\right)}^{1/3}=\frac{1}{3}$