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

$\frac{V_i}{V_f}=6.25$

We have

$E\propto \frac{Z^2}{n^2}$

$V=2E$

$V_i=\frac{\frac{1}{n_i^2}}{\frac{1}{n_f^2}}=\frac{n_f^2}{n_i^2}=6.25$

$\frac{n_f}{n_i}=2.5$

$n_f=\frac{5}{2}{n}_i$

$n_f=5$ when $n_i=2$