Question

Which hydrogen atom transition emits light in the Balmer series (visible)?

• A. n=2 $\rightarrow$ n=1
• B. n=3 $\rightarrow$ n=2
• C. n=4 $\rightarrow$ n=1
• D. n=3 $\rightarrow$ n=1

Hint:

Remember the order: Lyman (1), Balmer (2), Paschen (3). If it lands on 2, you can see it!

Answer 1

The Correct Option is B

Step 1: Concept
The spectral series of hydrogen are categorized based on the lower energy level ($n_{final}$) to which an electron falls.

Step 2: Meaning
For the Lyman series, $n_{final} = 1$ (Ultraviolet). For the Balmer series, $n_{final} = 2$ (Visible). For the Paschen series, $n_{final} = 3$ (Infrared).

Step 3: Analysis
Options (A), (C), and (D) all involve transitions to $n = 1$, which places them in the Lyman series. Only option (B) shows a transition from $n = 3$ to $n = 2$, which fits the criteria for the Balmer series.

Step 4: Conclusion
Transitions ending at the second energy level ($n = 2$) produce photons with wavelengths in the visible spectrum. Final Answer: (B)