The shortest wavelength of H atom in the Lyman series is $\lambda$₁. The longest wavelength in the Balmar series of He⁺ is :

For Shortest wavelength energy should be maximum. <br><br>For maximum energy transition must be form n = $\infty$ to n = 1. <br><br>$${1 \over {{\lambda _1}}} = {R_H}{\left( 1 \right)^2}\left[ {{1 \over 1} - 0} \right]$$ <br><br>$\Rightarrow$ ${1 \over {{\lambda _1}}} = {R_H}$ <br><br>For longest wavelength, $\Delta$E = minimum. <br>For Blamer series n = 3 to n = 2 will have $\Delta$E minimum for He⁺, Z = 2 <br><br>So $${1 \over {{\lambda _2}}} = {R_H}{\left( 2 \right)^2}\left[ {{1 \over 4} - {1 \over 9}} \right]$$ <br><br>$\Rightarrow$ ${1 \over {{\lambda _2}}} = {R_H} \times {5 \over 9}$ <br><br>$\Rightarrow$ ${\lambda _2} = {\lambda _1} \times {9 \over 5}$