Question

The wavelength of $\beta$-line of the Balmer series is $4815\stackrel{o}{A}$ . What is the wavelength of $\alpha$-line of the Balmer series of the same atom?

Answer 1

The wavelength of the visible line in blue- green region corresponds to $H_{\beta }$, the balmer's formula is,

$\frac{1}{\lambda }=R(\frac{1}{2^2}-\frac{1}{n^2})$

$R$=Rydberg's constant=$1.097\times {10}^7{m}^{-1}$

For $H_{\beta }$ $n=3$

$\frac{1}{\lambda }=1.097\times {10}^7{m}^{-1}(\frac{1}{4}-\frac{1}{9})$

$\lambda =4.9\times {10}^{-7}m$

$\lambda =4900\stackrel{o}{A}$