Wavelength of spectral line in hydrogen atom $(z=1)$ given as
$\frac{1}{\lambda}=R\left[\frac{1}{n_1^2}-\frac{1}{n_2^2}\right] \ldots \text { (i) }[\because Z=1]$
For last line of Lyman series,
$n_1=1 \text { and } n_2=\infty$
$\therefore$ From Eq. (i), we have
$ \frac{1}{\lambda_L}=R\left(\frac{1}{1^2}-\frac{1}{\infty}\right)=R$
$ \Rightarrow \lambda_L=\frac{1}{R} \cdots \cdots(i i)$
Similarly, for last line of Balmer series, $n_1=2$ and $n_2=\infty$
$\therefore$ From Eq. (i), we get
$ \frac{1}{\lambda_B}=R\left(\frac{1}{2^2}-\frac{1}{\infty}\right)=\frac{R}{4} $
$ \Rightarrow \lambda_B=\frac{4}{R}$
Hence, from Eq. (i) and Eq. (ii), we get
$ \frac{\lambda_L}{\lambda_B}=\frac{1 / R}{4 / R}=\frac{1}{4} $
$\text { or } \lambda_L: \lambda_B=1: 4 $
$ \Rightarrow \lambda_B: \lambda_L=4: 1$