1. Identify the energy levels involved

For a hydrogen-like species such as $\mathrm{He}^+$, the spectral lines depend only on the initial and final principal quantum numbers.

• Fifth excited state means: $n = 6$ because ground state is $n=1$, first excited is $n=2$, ..., fifth excited is $n=6$.

• First excited state means: $n = 2$

So the electron can undergo transitions among the levels: $n = 6, 5, 4, 3, 2$

2. Count all possible downward transitions

Each distinct pair of levels gives one spectral line.

If there are $m$ energy levels involved, the total number of possible emission lines is: $N = \frac{m(m-1)}{2}$

Here the involved levels are $6,5,4,3,2$, so $m = 5$

Thus, $N = \frac{5\cdot 4}{2} = 10$

3. Alternative listing of transitions

Possible transitions are:

• $6 \to 5$
• $6 \to 4$
• $6 \to 3$
• $6 \to 2$
• $5 \to 4$
• $5 \to 3$
• $5 \to 2$
• $4 \to 3$
• $4 \to 2$
• $3 \to 2$

Total $= 10$ lines.

4. Final answer

The number of spectral lines obtained is: $\boxed{10}$

5. Comparison with stored answer

Stored correct answer = $10$.

This matches the derived answer.