Question

Sol. For last line of Balmer : n₁ = 2 & n₂ = $\infty$
$\frac{1}{{\lambda }_B}=R{Z}^2\left[\frac{{\displaystyle 1}}{n_1^2}-\frac{{\displaystyle 1}}{n_2^2}\right]=R(1{)}^2\left[\frac{{\displaystyle 1}}{2^2}-\frac{{\displaystyle 1}}{{\infty }^2}\right]$
${\lambda }_B=\frac{4}{R} ....\left(1\right)$
For last line of Lyman series : n₁ = 1 & n₂ = $\infty$
$\frac{1}{{\lambda }_L}=R{Z}^2\left[\frac{1}{n_1^2}-\frac{{\displaystyle 1}}{{\displaystyle n_2^2}}\right]=R(1{)}^2\left[\frac{{\displaystyle 1}}{{\displaystyle 1^2}}-\frac{1}{{\infty }^2}\right]$
${\lambda }_L = 1/R$    ..... (2)
$\boxed{\frac{{\lambda }_B}{{\lambda }_L}=\frac{(4/R)}{(1/R)}=4}$