\(\frac{1}{\lambda_{B}}=R c\left(\frac{1}{2^{2}}-\frac{1}{\infty^{2}}\right)=\frac{R c}{4}\)
The wavelength of last line of Lyman series
\(\frac{1}{{{\lambda _L}}} = Rc\left( {\frac{1}{{{1^2}}} - \frac{1}{{{\infty ^2}}}} \right) = Rc\)
\(\therefore \quad \frac{{{l_B}}}{{{\lambda _L}}} = \frac{4}{1} = 4\)