Q. If $\lambda_{1}$ and $\ \lambda_{2}$are the wavelengths of the first members of the Lyman and Paschen series respectively, then $\ \lambda_{1}:\ \lambda_{2}$is

A) 1:3
B) 1:30
C) 7:50
D) 7:108
Explanation

(d) For first line of Lyman series,
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ n_{1}$=1 and $n_{2}$=2
$\therefore\ \ \ \ \ \ \ \ \ \ \frac{1}{\lambda_{1}} = R\left( \frac{1}{1^{2}} - \frac{1}{2^{2}} \right) = R\left( 1 - \frac{1}{4} \right) = \frac{3R}{4}$
For first line of Paschen Series
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ n_{1}$=3 and $n_{2}$=4
$\therefore\ \ \ \ \ \ \ \ \ \frac{1}{\lambda_{2}} = R\left( \frac{1}{3^{2}} - \frac{1}{4^{2}} \right) = R\left( \frac{1}{9} - \frac{1}{16} \right) = \frac{7R}{144}$
$\therefore\ \ \ \ \ \frac{\lambda_{1}}{\lambda_{2}} = \ \frac{7R}{144}\ \times \ \frac{4}{3R} = \frac{7}{108}$

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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