If shortest wavelength of hydrogen atom in Lyman series is x , then longest wavelength in Balmer series of He ^+ is:

Step-by-step solution

For the shortest wavelength in the Lyman series of the hydrogen atom ( Z=1 ), the transition is from n₂ = to n₁ = 1 . 1 x = R(1)^2 ( 1 1^2 - 1 ^2 ) = R This gives R = 1 x . For the longest wavelength in the Balmer series of He ^+ ( Z=2 ), the transition is from n₂ = 3 to n₁ = 2 . 1 = R(2)^2 ( 1 2^2 - 1 3^2 ) 1 = 4R ( 1 4 - 1 9 ) 1 = 4R ( 5 36 ) = 5R 9 Substituting R = 1 x : 1 = 5 9x = 9x 5 Answer: 9x 5