The shortest wavelength of hydrogen atom in Lyman series is λ. The longest wavelength is Balmer series of He⁺ is

Solution

Chemistry ma’am · 4 steps

Write 1/λ for both lines with the Rydberg formula, then divide. R cancels out.

1. Rydberg formula. For a hydrogen-like species, 1/λ = RZ²(1/n₁²-1/n₂²). The shortest line of a series has n₂ = ∞. The longest line has n₂ = n₁ + 1.
2. Hydrogen, Lyman. Here Z = 1, n₁ = 1, n₂ = ∞. So 1/λ = R(1 - 0) = R, which means λ = 1/R.

   1/λ' = R(2)²(1/2²-1/3²) = 4R × 5/36 = 5R/9

3. Helium ion, Balmer. For He⁺, Z = 2, and the longest Balmer line is n = 3 → n = 2. That gives λ' = 9/5R.
4. So. Put 1/R = λ: λ' = 9/5 × 1/R = 9λ/5. The answer is (c).

The idea to keep. Shortest line means n₂ = ∞, longest line means n₂ = n₁ + 1. For ions like He⁺ or Li²⁺, never forget the Z² factor.

Answer from the NTA final answer key. Solution by PrepWiser.