JEE Main 2020 · 4 September · evening · Single correct · +4 / −1
The shortest wavelength of H atom in the Lyman series is λ₁. The longest wavelength in the Balmar series of He⁺ is :
- (a) (5λ ₁)/9
- (b) (36λ ₁)/5
- (c) (27λ ₁)/5
- (d) (9λ ₁)/5
Answer and solution
Answer (d) (9λ ₁)/5
Solution
Chemistry ma’am · 3 steps
Write the Rydberg formula for both lines; λ₁ is 1/R, so everything is a multiple of it.
- Find lambda one. The shortest Lyman line of H is the jump from n = ∞ to n = 1 with Z = 1. So 1/λ₁ = R(1/1² - 0) = R, which gives λ₁ = 1/R.
- Longest Balmer line. Balmer lines end on n = 2. The longest wavelength means the smallest energy gap, which is the jump from n = 3 to n = 2. For He⁺, Z = 2, so Z² = 4.
1/λ = R Z²(1/2² - 1/3²) = 4R × 5/36 = 5R/9
- So. λ = 9/5R = 9λ₁/5, which is option (d).
The idea to keep. Longest wavelength in a series means the smallest jump, into the series' lower level. Shortest means the jump from infinity. For hydrogen-like ions, always multiply by Z².
Answer from the NTA final answer key. Solution by PrepWiser.