If $\lambda_{1}$ and $\lambda_{2}$ are the wavelengths of the third member of Lyman and first
member of the Paschen series respectively, then the value of $\lambda_{1}: \lambda_{2}$ is :
Correct Option: , 3
(3)
For Lyman series
$n_{1}=1, \quad n_{2}=4$
$\frac{1}{\lambda_{1}}=R z^{2}\left(\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right)$
$\frac{1}{\lambda_{1}}=\mathrm{Rz}^{2}\left(\frac{1}{1_{1}^{2}}-\frac{1}{4^{2}}\right)$
$\frac{1}{\lambda_{1}}=\frac{15 \mathrm{Rz}^{2}}{16}$
$\lambda_{1}=\frac{16}{15 \mathrm{Rz}^{2}}$
For paschen series
$n_{1}=3, \quad n_{2}=4$
$\frac{1}{\lambda_{2}}=\mathrm{Rz}^{2}\left(\frac{1}{3^{2}}-\frac{1}{4^{2}}\right)$
$\frac{1}{\lambda_{2}}=R z^{2}\left(\frac{16-9}{9 \times 16}\right)$
$\frac{1}{\lambda_{2}}=\mathrm{Rz}^{2}\left(\frac{7}{9 \times 16}\right)$
$\lambda_{2}=\frac{9 \times 16}{7 \mathrm{Rz}^{2}}$
So,
$\frac{\lambda_{1}}{\lambda_{2}}=\frac{\frac{16}{15 \mathrm{Rx}^{2}}}{\frac{9 \times 16}{7 \mathrm{Rx}^{2}}}$
$=\frac{16 \times 7}{15 \times 9 \times 16}$
$=\frac{7}{135}$