In a macroscopic sample of hydrogen gas, electrons undergo de-excitation from the orbit to the orbit — Atomic Structure Chemistry Question

💡 Solution & Explanation

The total number of spectral lines obtained in a multi-step transition from $n_2=6$ to $n_1=2$ is $\frac{(n_2-n_1)(n_2-n_1+1)}{2} = \frac{(6-2)(6-2+1)}{2} = \frac{4 \times 5}{2} = 10$. The lines belonging to the Balmer series are those terminating precisely at $n=2$ (i.e., $6 \to 2$, $5 \to 2$, $4 \to 2$, and $3 \to 2$), which are $4$ lines. Non-Balmer lines = $10 - 4 = 6$.