Search any question & find its solution
Question:
Answered & Verified by Expert
An excited hydrogen atom emits a photon of wavelength ' $\lambda$ ' in returning to ground state. The quantum number ' $n$ ' of the excited state is ( $\mathrm{R}=$ Rydberg's constant)
Options:
Solution:
1290 Upvotes
Verified Answer
The correct answer is:
$\sqrt{\frac{\lambda \mathrm{R}}{(\lambda \mathrm{R}-1)}}$
Using Rydberg's formula
$\frac{1}{\lambda}=\mathrm{R}\left[\frac{1}{1^2}-\frac{1}{\mathrm{n}^2}\right] \quad \ldots(\because \mathrm{m}=1)$
$\therefore \quad \frac{\mathrm{n}^2-1}{\mathrm{n}^2}=\frac{1}{\lambda \mathrm{R}} \Rightarrow 1-\frac{1}{\mathrm{n}^2}=\frac{1}{\lambda \mathrm{R}} \Rightarrow \frac{\lambda \mathrm{R}-1}{\lambda \mathrm{R}}=\frac{1}{\mathrm{n}^2}$
$\therefore \quad \mathrm{n}=\sqrt{\frac{\lambda \mathrm{R}}{\lambda \mathrm{R}-1}}$
$\frac{1}{\lambda}=\mathrm{R}\left[\frac{1}{1^2}-\frac{1}{\mathrm{n}^2}\right] \quad \ldots(\because \mathrm{m}=1)$
$\therefore \quad \frac{\mathrm{n}^2-1}{\mathrm{n}^2}=\frac{1}{\lambda \mathrm{R}} \Rightarrow 1-\frac{1}{\mathrm{n}^2}=\frac{1}{\lambda \mathrm{R}} \Rightarrow \frac{\lambda \mathrm{R}-1}{\lambda \mathrm{R}}=\frac{1}{\mathrm{n}^2}$
$\therefore \quad \mathrm{n}=\sqrt{\frac{\lambda \mathrm{R}}{\lambda \mathrm{R}-1}}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.