Question

An excited electron of $H$-atoms emits a photon of wavelength $\lambda$ and returns in the ground state, the principal quantum number of excited state is given by:

• A. $\sqrt{\lambda R(\lambda R-1)}$
• B. $\sqrt{\frac{\lambda R}{(\lambda R-1)}}$
• C. $\frac{1}{\sqrt{\lambda R(\lambda R-1)}}$
• D. $\sqrt{\frac{(\lambda R-1)}{\lambda R}}$

Answer 1

Answer: (B)

$\sqrt{\frac{\lambda R}{(\lambda R-1)}}$

we know that, $\frac{1}{\lambda }=R{Z}^2(\frac{1}{n_L^2}-\frac{1}{n_H^2})$

The hydrogen atom returns to the ground state

$n_L^2=1$ and Z for hydrogen atom=1

Substituting in the equation, we get

$\frac{1}{\lambda }=R{Z}^2(\frac{1}{1^2}-\frac{1}{n_H^2})$

$\frac{1}{\lambda }=R.1.(1-\frac{1}{n^2})$

$\frac{1}{\lambda }=R(1-\frac{1}{n^2})$

$\frac{1}{\lambda R}=R(1-\frac{1}{n^2})$

$\frac{1}{\lambda R}=1-\frac{1}{n^2}$

$\frac{1}{n^2}=1-\frac{1}{\lambda R}$

$\frac{1}{n^2}=\frac{\lambda R-1}{\lambda R}$

$n=\sqrt{\frac{\lambda R}{\lambda R-1}}$

So, the correct option is $B$