Question

Hydrogen atom from excited state comes to the ground stage by emitting a photon of wavelength λ. 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. $|\frac{\lambda R^{2}}{\lambda R-1}$
• D. $\sqrt{\frac{\lambda R}{\lambda-1}}$

Hint:

Always set $n_1=1$ for transitions to the ground state.

Answer 1

The Correct Option is A

Step 1: Rydberg Formula
$\frac{1}{\lambda} = R (\frac{1}{n_f^2} - \frac{1}{n_i^2})$.
Step 2: Ground State
For ground state, $n_f = 1$ and $n_i = n$. So, $\frac{1}{\lambda} = R (1 - \frac{1}{n^2})$.
Step 3: Solve for n
$\frac{1}{\lambda R} = 1 - \frac{1}{n^2} \Rightarrow \frac{1}{n^2} = 1 - \frac{1}{\lambda R} = \frac{\lambda R - 1}{\lambda R}$. Taking the reciprocal and square root: $n = \sqrt{\frac{\lambda R}{\lambda R - 1}}$.
Final Answer: (a)