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

Wavelength $\lambda$ of spectral lines in $\mathrm{H}$-atom is given as $$ \frac{1}{\lambda}=R\left(\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right) \quad \text{...(i)} $$ For shortest wavelength of Paschen series, $$ n_{1}=3 \text { and } n_{2}=\infty $$ $\therefore$ From Eq. (i), we get $$ \begin{aligned} & \frac{l}{\lambda_{P S}}=R\left(\frac{1}{3^{2}}-\frac{1}{\infty}\right)=\frac{R}{9} \\ \Rightarrow \quad & \lambda_{P S}=\frac{9}{R} \quad \text{...(ii)} \end{aligned} $$ For shortest wavelength in Balmer series, $$ n_{1}=2, n_{2}=\infty $$ $\therefore$ From Eq. (i), we get $\frac{1}{\lambda_{B S}}=R\left(\frac{1}{2^{2}}-\frac{1}{\infty}\right)=\frac{R}{4}$ $\Rightarrow \quad \lambda_{B S}=\frac{4}{R} \quad \text{...(iii)}$ For shortest wavelength of Lyman series, $$ n_{1}=1 \text { and } n_{2}=\infty $$ $\therefore$ From Eq. (i), we get $$ \begin{aligned} \frac{1}{\lambda_{L S}} &=R\left(\frac{1}{1^{2}}-\frac{1}{\infty}\right)=R \\ \Rightarrow \quad \lambda_{L S} &=\frac{1}{R} \end{aligned} $$ Hence, from Eq. (i), (ii) and (iii), we have $$ \begin{aligned} \lambda_{P S}: \lambda_{B S}: \lambda_{L S}=& \frac{9}{R}: \frac{4}{R}: \frac{1}{R} \\ &=9: 4: 1 \end{aligned} $$