In a sample of hydrogen atoms, all the atoms are in a particular excited state. If the emission spec — Atomic Structure Chemistry Question

💡 Solution & Explanation

Visible lines end at $n=2$. Having 4 visible lines means transitions from $n=6, 5, 4, 3$ to $n=2$, indicating the atoms are initially at $n=6$. IR lines are transitions ending at $n \ge 3$. Transitions to $n=3$ (Paschen) are $6 \to 3, 5 \to 3, 4 \to 3$ (3 lines). Transitions to $n=4$ (Brackett) are $6 \to 4, 5 \to 4$ (2 lines). Transition to $n=5$ (Pfund) is $6 \to 5$ (1 line). Total IR lines $= 3 + 2 + 1 = 6$. Therefore, correct answer is A.