Q44 / 54 · JEE Advanced 2017 (Paper 1) · Physics

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

Correct answer: 5

Solution

$U = - \frac{Z^{2}}{n^{2}} E_{0}$

$Z = 1$

$U = - \frac{E_{0}}{n^{2}}$

$\frac{U_{i}}{U_{f}} = \frac{n_{f}^{2}}{n_{i}^{2}} = 6.25$

Taking $n_{i} = 2$

$n_{f} = 5$

Hence, the correct answer is 5.