Modern Physics Ques 53
- 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
(2017 Adv.)
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Solution:
- Potential energy of hydrogen atom $(Z=1)$ in $n$th orbit (in $eV$ )
$$ \begin{aligned} PE & =-\frac{27.2}{n^{2}} \\ \frac{v _f}{v _i} & =\frac{-\frac{27.2}{n _f^{2}}}{-\frac{27.2}{n _i^{2}}}=\frac{1}{6.25} \\
BETA
