Question

Wavelength of first line of Balmer series of hydrogen atom is $6560\stackrel{o}{A}$. Find wavelength of third line of this series.

Answer 1

Wavelength of Balmer series $\frac{1}{\lambda }=R(\frac{1}{2^2}-\frac{1}{n^2})$ where $n=3,4,5,......$
First line : $n=3$
$\therefore$ $\frac{1}{{\lambda }_1}=R(\frac{1}{2^2}-\frac{1}{3^2})=\frac{5R}{36}$
$⟹{\lambda }_1=\frac{36}{5R}$ ....(1)
Third line : $n=5$
$\therefore$ $\frac{1}{{\lambda }_3}=R(\frac{1}{2^2}-\frac{1}{5^2})=\frac{21R}{100}$
$⟹{\lambda }_3=\frac{100}{21R}$ ....(2)
$⟹\frac{{\lambda }_3}{{\lambda }_1}=\frac{100}{21}\times \frac{5}{36}=0.66$
Or ${\lambda }_3=0.66\times {\lambda }_1=0.66\times 6560=4340A^o$