Question

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

• A. $1002.7\mathring{A}$
• B. $1215.67\mathring{A}$
• C. $1127.30\mathring{A}$
• D. $911.7\mathring{A}$
• E. $1234.7\mathring{A}$

Answer 1

Answer: (D)

Rydberg's formula is

$\frac{1}{\lambda }=R_H{Z}^2\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)$

For hydrogen,

$Z=1$ and for lyman series, $n_1=1$ and $n_2=\infty$

(for shortest wavelength)

On substituting values, we get

$\frac{1}{\lambda }=109678\times (1{)}^2\times \left[\frac{1}{(1{)}^2}-\frac{1}{(\infty {)}^2}\right]$

$\frac{1}{\lambda }=109678c{m}^{-1}$

or $\lambda =\frac{1}{109678}c{m}^{-1}$

$=9.117\times 10^{-6}cm$

$=911.7\times 10^{-8}cm$

$=911.7\mathring{A}$