Question

The wavelength of $H_{\alpha }$ line of the Balmer series is $6500\stackrel{o}{A}$. The wavelength of the $H_{\beta }$ line of Balmer series is:

• A. $4814.8\stackrel{o}{A}$
• B. $6500.5\stackrel{o}{A}$
• C. $1625.3\stackrel{o}{A}$
• D. $5428.7\stackrel{o}{A}$

Answer 1

The correct option is A $4814.8\stackrel{o}{A}$

$H_{\alpha }$ line of Balmer series is produced due to the transition from n = 3 to n = 2
$H_{\beta }$ line of Balmer series is produced due to the transition from n = 4 to n = 2
Therefore,
$\frac{1}{{\lambda }_{\alpha }}=R_H[\frac{1}{2^2}-\frac{1}{3^2}]$ ...(1)
$\frac{1}{{\lambda }_{\beta }}=R_H[\frac{1}{2^2}-\frac{1}{4^2}]$ ...(2)
Where $R_H$ is the Rydberg constant.
From the equation (1) and (2) we get,
$\frac{{\lambda }_{\beta }}{{\lambda }_{\alpha }}=\frac{5\times 16}{36\times 3}$
${\lambda }_{\beta }=\frac{5\times 16}{36\times 3}\times 6500\stackrel{o}{A}$
$\Rightarrow {\lambda }_{\beta }=4814.8\stackrel{o}{A}$