Question
If shortest wavelength of hydrogen atom in Lyman series is x, then longest wavelength in Balmer series of He ^+ is:
JEE Main
Chemistry
Structure of Atom
2026
JEE Main 2026 (06 April Shift 1)
JEE Main Chemistry Question (2026) — Solution
Options
- A. 9x 5
- B. 36x 5
- C. x 4
- D. 5x 9
Answer
A. 9x 5
Step-by-step solution
For the shortest wavelength in the Lyman series of the hydrogen atom (Z=1), the transition is from n_2 = to n_1 = 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 n_2 = 3 to n_1 = 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
Related: Chemistry — Structure of Atom · All PYQ Banks