The shortest wavelengths of Paschen, Lymen and Balmer series are in the ratio

The wavelength $\lambda$ for a transition in a hydrogen-like atom is given by the Rydberg formula: $\dfrac{1}{\lambda} = R Z^2 \left( \dfrac{1}{n_1^2} - \dfrac{1}{n_2^2} \right)$.

For the shortest wavelength of any series, the transition occurs from $n_2 = \infty$ to $n_1$. Thus, $\dfrac{1}{\lambda_{min}} = R \left( \dfrac{1}{n_1^2} \right)$, which implies $\lambda_{min} = \dfrac{n_1^2}{R}$.

For the Paschen series, $n_1 = 3$, so $\lambda_{Paschen} = \dfrac{3^2}{R} = \dfrac{9}{R}$.

For the Lyman series, $n_1 = 1$, so $\lambda_{Lyman} = \dfrac{1^2}{R} = \dfrac{1}{R}$.

For the Balmer series, $n_1 = 2$, so $\lambda_{Balmer} = \dfrac{2^2}{R} = \dfrac{4}{R}$.

The ratio of the shortest wavelengths of Paschen, Lyman, and Balmer series is $\lambda_{Paschen} : \lambda_{Lyman} : \lambda_{Balmer} = \dfrac{9}{R} : \dfrac{1}{R} : \dfrac{4}{R} = 9 : 1 : 4$.

Answer: $9: 1: 4$