The shortest wavelength of H atom in the Lyman series is λ _ 1 ....

For hydrogen atom :

For Lyman series ${n}_{1} = 1 \text{ and }$ {n}_{2} = \infty$</p><p>$\dfrac{1}{{\lambda}_{H}} = Rr \left[\dfrac{1}{1} - \dfrac{1}{\infty}\right]\text{So,}\lambda = \dfrac{1}{{R}_{H}}$</p><p>For${He}^{+}\text{ion}$</p><p>Balmer series${n}_{1} = 2 \text{and } {n}_{2} = 3</p><p>\dfrac{1}{\lambda{He}^{+}} = {R}_{H} \times {Z}^{2}\left[\dfrac{1}{4} - \dfrac{1}{9}\right]$</p><p>$\dfrac{1}{\lambda{He}^{+}} = {R}_{H} \times 4 \times \dfrac{5}{36}$</p><p>$\dfrac{1}{\lambda{He}^{+}} = \dfrac{5}{9}{R}_{H} = \left(\dfrac{5}{9}\right)\dfrac{1}{\lambda}$</p><p>\left(\lambda{He}^{+}\right) = \dfrac{9}{5}\lambda$