If the shortest wavelength of the Lyman series for a hydrogen atom is x , then the longest wavelength of the Paschen series for a He ⁺ ion is:

Step-by-step solution

The Rydberg formula for the wavelength of emitted radiation is given by: 1 = R Z^2 ( 1 n₁^2 - 1 n₂^2 ) For the shortest wavelength of the Lyman series for a hydrogen atom, Z = 1 , n₁ = 1 , and n₂ = . Substituting these values: 1 x = R (1)^2 ( 1 1^2 - 1 ) = R For the longest wavelength of the Paschen series for a He ⁺ ion, Z = 2 , n₁ = 3 , and n₂ = 4 . Let this wavelength be . Substituting these values: 1 = R (2)^2 ( 1 3^2 - 1 4^2 ) 1 = 4R ( 1 9 - 1 16 ) 1 = 4R ( 16 - 9 144 ) = 4R ( 7 144 ) = 7R 36 Substituting R =