Answer
$4$
Work Step by Step
$\frac{1}{\lambda}=R(\frac{1}{n^2_{low}}-\frac{1}{n^2_{high}})$
For the Lyman series $n_{low}=1$ and for the Balmer series is $n_{low}=2$.
For the shortest wavelength $n_{high}=\infty$
Thus,
For the Lyman series,
$\frac{1}{\lambda_L}=R(\frac{1}{1^2}-\frac{1}{\infty^2})=R$
$\implies \lambda_L=\frac{1}{R}$
For the Balmer series
$\frac{1}{\lambda_B}=R(\frac{1}{2^2}-\frac{1}{\infty^2})=\frac{R}{4}$
$\implies \lambda_B=\frac{4}{R}$
Therefore,
$\frac{\lambda_B}{\lambda_L}=\frac{4}{1}=4$
