Question

The wavelength of the first member of the Balmer series of hydrogen is $6563\times {10}^{-10}m$. Calculate the wavelength of its second member.

Answer 1

$\frac{1}{{\lambda }_1}=R_H[\frac{1}{2^2}-\frac{1}{3^2}]$ and $\frac{1}{{\lambda }_2}=R_H[\frac{1}{2^2}-\frac{1}{4^2}]$
$\frac{{\lambda }_2}{{\lambda }_1}=\frac{5}{36}\times \frac{16}{3}=\frac{20}{27}$
${\lambda }_2=\frac{20}{27}\times 6563\times {10}^{-10}=4861\times {10}^{-10}m$.