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Q.
In the hydrogen spectrum, the ratio of the wavelengths for Lyman-alpha radiation to Balmer-alpha radiation is
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Solution:
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, $\frac{\lambda_{Ly-\alpha}}{\lambda_{Ba-\alpha}}=\frac{\frac{3}{4}R}{\frac{5}{36}R} =\frac{27}{5}$