Number of spectral lines obtained in He⁺ spectra, when an electron makes transition from fifth excited state to first excited state will be

Solution

Chemistry ma’am · 3 steps

Turn excited states into n values first. The fifth excited state is n = 6, not n = 5.

1. Find the levels. The ground state is n = 1, so the first excited state is n = 2 and the fifth excited state is n = 6. He⁺ has one electron, so its levels behave like those of hydrogen.
2. Count the lines. An electron falling from n₂ to n₁ can stop at any level in between. Every pair of levels from n₁ to n₂ gives one line, so the count depends only on Δ n = n₂ - n₁.

   N = (Δ n (Δ n + 1))/2 = (4 × 5)/2

3. So. With Δ n = 6 - 2 = 4, the number of spectral lines is 10.

The common slip: 15. Counting lines all the way down to the ground state with n(n-1)/2 for n = 6 gives 15. The electron stops at n = 2, so it gives 10.

The idea to keep. The k-th excited state is n = k + 1. Between two levels, the number of lines is (Δ n(Δ n + 1))/2, and the identity of the atom or ion does not change this count.

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