Question

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 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. $\sqrt{\frac{\lambda R}{\lambda -1}}$

Answer 1

Answer: (A)

$\sqrt{\frac{\lambda R}{\lambda R+1}}$

$\begin{array}{c}\text{We Know, wavelength emitted by photon}\\ \text{during its transition}\end{array}$

$\frac{1}{\lambda }=R{z}^2(\frac{1}{n_1^2}-\frac{1}{n_2^2})$

$\begin{array}{c}z=1\text{(hydrogen)}\\ n_1=n\\ n_2=1\text{(ground state)}\\ \frac{1}{\lambda }=R(1{)}^2(\frac{1}{n^2}-\frac{1}{1})\\ \frac{1}{\lambda }=R(\frac{1}{n^2}-1)\\ \frac{1}{n^2}=\frac{1}{\lambda R}+1\\ n=\sqrt{\frac{\lambda R}{\lambda R+1}}\end{array}$