Question

An electron in a hydrogen atom undergoes a transition from an orbit with quantum number $𝑛_i$ to another with quantum number $𝑛_f. 𝑉_i$ and $𝑉_f$ are respectively the initial and final potential energies of the electron. If $\frac{V_{i}}{V_{f}}=6.25$, then the smallest possible $𝑛_f$ is

Answer 1

$| V | \propto \frac{1}{ n ^{2}} $
$\frac{\left| V _{ i }\right|}{\left| V _{ f }\right|}=\frac{ n _{ f }^{2}}{ n _{ i }^{2}}=6.25$
$\frac{ n _{ f }^{2}}{ n _{ i }^{2}}=6.25 $
$\frac{ n _{ f }}{ n _{ i }}=\sqrt{6.25}=2.5 $
$ n _{ f }=2.5 n _{ i }$
Smallest value of $n_{i}=2$ for which $n_{f}=5$