Correct Option: (C) $n_f = 2$ and $n_i = 4$

Step 1 (Governing Law & Definition): [0.5 Mark]

The spectral lines of the hydrogen atom are governed by the Rydberg formula for transitions between energy levels. For the Balmer series, the final energy level is fixed at $n_f = 2$. The spectral lines are defined by the transition from an initial state $n_i$ to the final state $n_f$:

• First line: $n_i = n_f + 1 = 3$
• Second line: $n_i = n_f + 2 = 4$

Step 2 (Step-by-Step Mathematical Working): [0.25 Mark]

Given that the Balmer series corresponds to transitions ending at $n_f = 2$, we identify the sequence of transitions:

For the Balmer series ($n_f = 2$):

• 1st line (H-$\alpha$): $n_i = 3 \rightarrow n_f = 2$
• 2nd line (H-$\beta$): $n_i = 4 \rightarrow n_f = 2$

Substituting the values into the transition logic, the second line requires an electron to jump from the 4th orbit to the 2nd orbit.

Step 3 (Final Concluded Answer): [0.25 Mark]

Therefore, the transition corresponding to the second spectral line of the Balmer series is $n_i = 4$ to $n_f = 2$. This matches option (C).

💡 EL10 Concept Insight:

Think of the energy levels like steps on a ladder; the Balmer series is like landing on the 2nd step. The "first" line is the shortest jump from the 3rd step, and the "second" line is the next jump from the 4th step!