In a sample of hydrogen atoms, electrons transit from the 6th orbit to the 2nd orbit in multi-steps. — Atomic Structure Chemistry Question

💡 Solution & Explanation

Total lines from $n=6$ to $n=2$ = $(6-2)(6-2+1)/2 = (4 \times 5)/2 = 10$. Lines ending exactly at $n=2$ (Balmer) are 4 ($6\to2, 5\to2, 4\to2, 3\to2$). Non-Balmer lines = $10 - 4 = 6$.