Question

The wavelength of a certain line in Balmer series is observed to be 4103 $\stackrel{o}{A}$. Find the value of state from which the de-excitation took place.
(Rydberg’s constant = $109678c{m}^{-1}$)

• A. n = 4
• B. n = 3
• C. n = 1
• D. n = 6

Answer 1

The correct option is D n = 6

Given: $\lambda =4103\stackrel{o}{A}$ = $4103\times {10}^{-8}cm$
We know,
$\frac{1}{\lambda }=R_H[\frac{1}{n_1^2}-\frac{1}{n^2}]\frac{1}{\lambda }=R_H[\frac{1}{2^2}-\frac{1}{n^2}]\frac{1}{n^2}=\frac{1}{4}-\frac{1}{\lambda \times R_H}=\frac{1}{4}-\frac{1}{4103\times {10}^{-8}\times 109678}=0.0277\therefore n^2=\frac{1}{0.0277}=36$
or $n=6$