JEE Main20204 Sep 2020Evening ShiftChemistryStructure of AtomActual
The shortest wavelength of H atom in the Lyman series is λ 1 . The longest wavelength in the Balmer series of He + is :
Options
- A36 λ 1 5
- B5 λ 1 9
- C9 λ 1 5
- D27 λ 1 5
Correct answer
C. 9 λ 1 5
Step-by-step solution
For hydrogen atom : For Lyman series n 1 = 1    &                 n 2 = ∞ 1 λ H = Rr 1 1 - 1 ∞            So,              λ = 1 R H For He + ion  Balmer series n 1 = 2           &          n 2 = 3 1 λ He + = R H × Z 2 1 4 - 1 9 1 λ He + = R H × 4 × 5 36 1 λ He + = 5 9 R H = 5 9 1 λ &