Question

The shortest wavelength of the line in hydrogen atomic spectrum of Lyman series when $R_H = 109678 \, cm^{-1}$ 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] $

$\frac{1}{\lambda}=109678\, cm ^{-1}$

or $\lambda =\frac{1}{109678}\, cm ^{-1} $

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

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

$=911.7\,\mathring{A} $