Q. Hydrogen atom from excited state comes to the ground state by emitting a photon of wavelength $\lambda$. If $R$ is the Rydberg constant, the principal quantum number $n$ of the excited state is
A) $\sqrt{\frac{\lambda R}{\lambda R - 1}}$
B) $\sqrt{\frac{\lambda}{\lambda R - 1}}$
C) $\sqrt{\frac{\lambda R^{2}}{\lambda R - 1}}$
D) $\sqrt{\frac{\lambda R}{\lambda - 1}}$
Explanation
(a) According to Rydberg’s formula
$\frac{1}{\lambda} = R\left( \frac{1}{n_{f}^{2}} - \frac{1}{n_{i}^{2}} \right)$
Here, $n_{f} = 1,n_{i} = n$
$\therefore\frac{1}{\lambda} = R\left( \frac{1}{1^{2}} - \frac{1}{n^{2}} \right) \Rightarrow \frac{1}{\lambda} = R\left( 1 - \frac{1}{n^{2}} \right)\ \ \ \ldots(i)$
Multiplying equation (i) by $\lambda$ on both sides,
$1 = \lambda R\left( 1 - \frac{1}{n^{2}} \right) \Rightarrow \frac{1}{\lambda R} = 1 - \frac{1}{n^{2}}$
$\Rightarrow \frac{1}{n^{2}} = 1 - \frac{1}{\lambda R} \Rightarrow \frac{1}{n^{2}} = \frac{\lambda R - 1}{\lambda R} \Rightarrow n = \sqrt{\frac{\lambda R}{\lambda R - 1}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.