Q. The wavelength of a spectral line in Lyman series, when electron jumps back from 2nd orbit, is

A) 1162 Å
B) 1216 Å
C) 1362 Å
D) 1176 Å
Explanation

(b)
$\frac{1}{\lambda} = R_{H}\left\lbrack \frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}} \right\rbrack$
For Lyman series, $n_{1} = 1,\ n_{2} = 2\$
$\frac{1}{\lambda} = 10,9678\ \left\lbrack \frac{1}{(1)^{2}} - \frac{1}{(2)^{2}} \right\rbrack$
$= \frac{10,9678 \times 3}{4}$
$\lambda = 1216\ \mathring{\mathrm{A}}$

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

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