The ratio of wavelengths of the last line of Balmer series and the last line of Lyman series is

For Lyman series, the last line has wavelength \( \frac{1}{\lambda_L} = R(1^2) \implies \lambda_L = \frac{1}{R} \). For Balmer series, \( \frac{1}{\lambda_B} = R(\frac{1}{2^2}) \implies \lambda_B = \frac{4}{R} \). Thus, \( \frac{\lambda_B}{\lambda_L} = 4 \).