Q. The shortest wavelength in hydrogen spectrum of Lyman series when $R_{H} = 109678\ cm^{- 1}$ is

A) $1002.7\ \mathring{\mathrm{A}}$
B) $1215.67\ \mathring{\mathrm{A}}$
C) $1127.30\ \mathring{\mathrm{A}}$
D) $911.7\ \mathring{\mathrm{A}}$
Explanation

(d) The wavelength of different spectral lines of Lyman series is given by
$\frac{1}{\lambda} = R_{H}\left\lbrack \frac{1}{1^{2}} - \frac{1}{n^{2}} \right\rbrack$; where $n = 2,\ 3,\ 4,\ldots..$
For shortest wavelength, $n = \infty$
$\therefore\frac{1}{\lambda} = \frac{R_{H}}{1}$
Or $\lambda = \frac{1}{R_{H}} = \frac{1}{109678cm^{- 1}}$
$= 9.117 \times 10^{- 6}cm = 9.117 \times 10^{- 8}m$
$= 911.7 \times 10^{- 10}m = 911.7\mathring{\mathrm{A}}$

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

Ph.D. in Physics with 19+ years of teaching experience.

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