Question

The first emission line in the atomic spectrum of hydrogen in the Balmer Series appears at:

• A. $\frac{9R_H}{400}c{m}^{-1}$
• B. $\frac{7R_H}{144}c{m}^{-1}$
• C. $\frac{3R_H}{4}c{m}^{-1}$
• D. $\frac{5R_H}{36}c{m}^{-1}$

Answer 1

Answer: (D)

$\frac{5R_H}{36}c{m}^{-1}$

$\frac{1}{\lambda }=R_H{Z}^2[\frac{1}{n_1^2}-\frac{1}{n_2^2}]$
For Balmer Series' first emission line,
$\frac{1}{\lambda }=R_H\times 1[\frac{1}{2^2}-\frac{1}{3^2}]=\frac{R_{H}5}{36}c{m}^{-1}$