Question

The shortest wavelength in H spectrum of Lyman series when $R_H=109678c{m}^{-1}$ is

• A. 1002.7 Å
• B. 1215.67 Å
• C. 1127.30 Å
• D. 911.7 Å

Answer 1

The correct option is D

911.7 Å

From line spectrum of hydrogen, we know that
For lyman series, $n_1=1$
For shortest 'wavelength' of lyman series, the energy difference in two levels showing transition should be maximum (i.e., $n_2=\infty$)

$\frac{1}{\lambda }=R_H[\frac{1}{1^2}-\frac{1}{\infty }]=\lambda =\frac{1}{109678}$
$=109678\Rightarrow \lambda =\frac{1}{109678}911.7\times {10}^{-8}=911.7\mathring{A}$