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Q. If $\lambda_{1}$ and $\lambda_{2}$ are the wavelengths of the first members of the Lyman and Paschen series respectively, then $\lambda_{1}: \lambda_{2}$ is

KCETKCET 2006Atoms

Solution:

For first line of Lyman series,
$n_{1}=1$ and $n_{2}=2$
$\therefore \frac{1}{\lambda_{1}}=R\left(\frac{1}{1^{2}}-\frac{1}{2^{2}}\right)$
$=R\left(1-\frac{1}{4}\right)=\frac{3 R}{4}$
For first line of Paschen series $n_{1}=3$ and $n_{2}=4$
$\therefore \frac{1}{\lambda_{2}}=R\left(\frac{1}{3^{2}}-\frac{1}{4^{2}}\right)$
$=R\left(\frac{1}{9}-\frac{1}{16}\right)=\frac{7 R}{144}$
$\frac{\lambda_{1}}{\lambda_{2}}=\frac{7 R}{144} \times \frac{4}{3 R}$
$=\frac{7}{108}$