Question

The first emission line in the $H$-atom spectrum in the Balmer series appears at:

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

Answer 1

The correct option is A $\frac{5R}{36}{cm}^{-1}$

The first emission line in the H-atom spectrum in the Balmer series appears at
The first emission line in the H-atom spectrum in the Balmer series corresponds to the transition $3\to 2$
$\overline{\nu }=R[\frac{1}{2^2}-\frac{1}{3^2}]c{m}^{-1}$
$\overline{\nu }=R[\frac{1}{4}-\frac{1}{9}]c{m}^{-1}$
$\overline{\nu }=R\left[\frac{9-4}{36}\right]c{m}^{-1}$
$\overline{\nu }=R\left[\frac{5}{36}\right]c{m}^{-1}$