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) Here, $n_{f}$ $= \ 1$, $n_{i} = n$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{1}{\lambda}$ = $R\left( \frac{1}{1^{2}} - \frac{1}{n^{2}} \right)$
$\ \Rightarrow \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{1}{\lambda}$ = $R\left( 1 - \frac{1}{n^{2}} \right)$ $..(i)$
or$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{1}{\lambda R}$ =1$- \frac{1}{n^{2}}$ or $\frac{1}{n^{2}} = 1 - \frac{1}{\lambda R}$
or $n$ = $\sqrt{\frac{\lambda R}{\lambda R - 1}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.