Physics - Modern Physics Question with Solution | TestHub

Using Rydberg's formula, $\frac{1}{\lambda }=R\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)$

For the first line of the Lyman series, $n_1=1$ and $n_2=2$.

$\frac{1}{{\lambda }_1}=R\left(\frac{1}{1^2}-\frac{1}{2^2}\right)=\frac{3R}{4}$

${\lambda }_1=\frac{4}{3R}$

For the first line of the Paschen series, $n_1=3$ and $n_2=4$.

$\frac{1}{{\lambda }_2}=R\left(\frac{1}{3^2}-\frac{1}{4^2}\right)=R\left(\frac{1}{9}-\frac{1}{16}\right)=\frac{7R}{144}$

${\lambda }_2=\frac{144}{7R}$

Taking the ratio of ${\lambda }_2$ to ${\lambda }_1$:

$\frac{{\lambda }_2}{{\lambda }_1}=\frac{\frac{144}{7R}}{\frac{4}{3R}}=\frac{144}{7}\times \frac{3}{4}=\frac{108}{7}$

The ratio is $108:7$.

Answer: $108:7$