A sample of hydrogen atoms undergoes de-excitation from the Bohr orbit to the Bohr orbit in multiple — Atomic Structure Chemistry Question

💡 Solution & Explanation

The total number of lines from $n=6$ to $n=3$ is $\frac{(6-3)(6-3+1)}{2} = \frac{3 \times 4}{2} = 6$. Since all transitions terminate at $n \ge 3$, they all fall in the infrared region (Paschen terminating at 3, Brackett at 4, Pfund at 5). Paschen lines: $6 \to 3, 5 \to 3, 4 \to 3$ ($3$ lines, not $4$). Brackett lines: $6 \to 4, 5 \to 4$ ($2$ lines). Pfund lines: $6 \to 5$ ($1$ line). Total IR lines = $3 + 2 + 1 = 6$.