Question

The first emission line of Balmer series for $H$-spectrum has the wave no. equal to:

• A. $\frac{9R_H}{400}c{m}^{-1}$
• B. $\frac{7R_H}{144}c{m}^{-1}$
• C. $\frac{3R_H}{4}c{m}^{-1}$
• D. $\frac{5R_H}{36}c{m}^{-1}$

Answer 1

The correct option is D $\frac{5R_H}{36}c{m}^{-1}$

$\overline{v}=\frac{1}{\lambda }=R_H[\frac{1}{n_1^2}-\frac{1}{n_2^2}]$; $n_1=2$ for Balmer series and $n_2=3$ for first line or $H_{\alpha }$ line of Balmer series.

$\overline{v}=\frac{1}{\lambda }=R_H[\frac{1}{{\left(2\right)}^2}-\frac{1}{{\left(3\right)}^2}]$

$\overline{v}=\frac{1}{\lambda }=\frac{5R_H}{36}$

Hence, the correct option is $D$