Physics - Modern Physics Question with Solution | TestHub

For last Balmer series,
$\frac{1}{{\lambda }_b}=R \left[\frac{1}{2^2}-\frac{1}{{\infty }^2}\right]$
${\lambda }_b=\frac{4}{R}$
For last Lyman series,
$\frac{1}{{\lambda }_l}=R \left[\frac{1}{1^2}-\frac{1}{{\infty }^2}\right]$
${\lambda }_l=\frac{1}{R}$
$\frac{{\lambda }_b}{{\lambda }_l}=\frac{\frac{4}{R}}{\frac{1}{R}}$
$\frac{{\lambda }_b}{{\lambda }_l}=4$

Hence this is the correct answer.