Explanation

(b) The series end of Lyman series corresponds to transition from $n_{i} = \infty$ to
$n_{f} = 1$, corresponding to the wavelength
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{1}{\left( \lambda_{\min} \right)_{L}}$ $=$ $R\left\lbrack \frac{1}{1} - \frac{1}{\infty} \right\rbrack$ $= \ R$
$\Rightarrow \ \ \ \ \ \ {(\lambda_{\min})}_{L}$ $= \frac{1}{R}$ $=$ 912 $\mathring{\mathrm{A}}$ …(i)
For last line of Balmer series
$\ \ \ \ \ \ \ \ \ \ \ \frac{1}{{(\lambda_{\min})}_{B}}$ =$R\left\lbrack \frac{1}{(2)^{2}} - \frac{1}{(\infty)^{2}} \right\rbrack$= $\frac{R}{4}$
$\Rightarrow \ \ \ \ \ {(\lambda_{\min})}_{B}$=$\frac{4}{R}$ =3636 $\mathring{\mathrm{A}}$ …(ii)
Dividing Eq.(i) by Eq. (ii) .we get
$\ \ \ \ \ \ \ \ \ \ \frac{{(\lambda_{\min})}_{L}}{{(\lambda_{\min})}_{B}}$ = 0.25