Ratio between longest wavelength of H atom in Lyman series to the shortest wavelength in Balmer series of He+  is

Longest wavelength in Lyman Series of hydrogen atom arises from transition between n = 2 to n = 1 whose number is given byv¯H2→1 =  R ×  121112 − 122=34RShortest wavelength in Balmer series of He+ arises from transition between n = ∞ to n = 2 whose wave number is given byv¯He+∞−2 =  R ×  22122 − 1∞2=RvHe+vH=λHλHe+∴       λHλHe+ = 4R3R=43