Q. The first emission line in the electronic spectrum of hydrogen in the Balmer series appears at $cm^{- 1}$
A) $\frac{9R}{400}cm^{- 1}$
B) $\frac{7R}{144}cm^{- 1}$
C) $\frac{3R}{4}cm^{- 1}$
D) $\frac{5R}{36}cm^{- 1}$
Explanation
(d)
$\overline{v} = R\left( \frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}} \right)$
For Balmer series
$n_{1} = 2,\ n_{2} = 3,\ 4,5,\ldots\infty$
For first emission line $n_{2} = 3$
$\therefore\ \overline{v} = R\left( \frac{1}{2^{2}} - \frac{1}{3^{2}} \right)$
$\ \ \ \ \ \ \ \ = R\left( \frac{1}{4} - \frac{1}{9} \right) = R\left( \frac{5}{36} \right)$
$\ \ \ \ \overline{v} = \frac{5R}{36}cm^{- 1}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.