Which transition in the hydrogen spectrum would have the same wavelength as the Balmer type transition from n=4 to n=2 of He⁺ spectrum

Solution

Chemistry ma’am · 3 steps

Put Z² into the Rydberg formula for He⁺, then find the hydrogen pair that gives the same number.

1. 1/λ = R Z²(1/n₁²-1/n₂²)

2. Helium ion line. For He⁺, Z = 2, so Z² = 4. The jump from n = 4 to n = 2 gives 1/λ = 4R(1/4-1/16) = 4R × 3/16 = 3R/4.
3. Match in hydrogen. For hydrogen, Z = 1, so the bracket alone must equal 3/4. Since 1/1²-1/2² = 3/4, the levels are n = 1 and n = 2.

   Hydrogen option	1/n₁²-1/n₂²	Same as 3/4?
   (a) 3 → 4	1/9-1/16 = 7/144	No
   (b) 2 → 1	1-1/4 = 3/4	Yes, and it is emission
   (c) 1 → 2	1-1/4 = 3/4	Same gap, but absorption
   (d) 1 → 3	1-1/9 = 8/9	No

4. So. The He⁺ line falls from n = 4 to n = 2, so it is an emission line. The hydrogen line that matches it is the fall from n = 2 to n = 1, option (b).

The idea to keep. For any one-electron species, 1/λ ∝ Z²(1/n₁²-1/n₂²). For He⁺, halve both n values to get the hydrogen line with the same wavelength.

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