Question

In the hydrogen spectrum, the ratio of the wavelengths for Lyman-alpha radiation to Balmer-alpha radiation is

• A. $5/27$
• B. $5/48$
• C. $27/5$
• D. $1/3$

Answer 1

Answer: (C)

Lyman-alpha line sometimes denoted as $Ly-\alpha$ line is obtained when an electron falls from $n=2$ to $n=1$ orbital in an one electron ion or atom.
Similarly, Balmer-alpha radiation is emitted when an electron makes a transition from $n=3$ to $n=2$ quantum state.
Now, using $\frac{1}{\lambda }=R\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)$
We have, $\frac{1}{{\lambda }_{Ly-a}}=R\left(\frac{1}{1^2}-\frac{1}{2^2}=\frac{3R}{4}\right)$
and $\frac{1}{{\lambda }_{Ba-\alpha }}=R\left(\frac{1}{4}-\frac{1}{9}\right)=\frac{5}{36}R$
Hence, $\frac{{\lambda }_{Ly-\alpha }}{{\lambda }_{Ba-\alpha }}=\frac{\frac{3}{4}R}{\frac{5}{36}R}=\frac{27}{5}$