Question

The transition of electron that gives rise to the formation of the second spectral line of the Balmer series in the spectrum of hydrogen atom corresponds to:

• A. \(n_f = 2\) and \(n_i = 3\)
• B. \(n_f = 3\) and \(n_i = 4\)
• C. \(n_f = 2\) and \(n_i = 4\)
• D. \(n_f = 2\) and \(n_i = \infty\)

Hint:

In the Balmer series of hydrogen, all spectral lines are produced when electrons fall to \(n = 2\). The sequence starts from \(n = 3 \rightarrow 2\), \(4 \rightarrow 2\), \(5 \rightarrow 2\), and so on.

Answer 1

The Correct Option is C

Step 1: Understanding the Balmer series.
The Balmer series corresponds to electronic transitions in the hydrogen atom where an electron falls from a higher energy level to the second energy level \((n_f = 2)\). These transitions produce spectral lines in the visible region of the electromagnetic spectrum.
Step 2: First line of the Balmer series.
The first spectral line of the Balmer series occurs when an electron transitions from the third energy level to the second energy level, that is \[ n_i = 3 \rightarrow n_f = 2 \] This line is known as the H\(\alpha\) line.
Step 3: Second line of the Balmer series.
The second spectral line occurs when the electron falls from the fourth energy level to the second energy level: \[ n_i = 4 \rightarrow n_f = 2 \] This line is known as the H\(\beta\) line in the hydrogen spectrum.
Step 4: Analysis of options.

• (A) \(n_f = 2, n_i = 3\): Incorrect. This corresponds to the first spectral line of the Balmer series.
• (B) \(n_f = 3, n_i = 4\): Incorrect. This transition belongs to the Paschen series, not the Balmer series.
• (C) \(n_f = 2, n_i = 4\): Correct. This produces the second spectral line of the Balmer series.
• (D) \(n_f = 2, n_i = \infty\): Incorrect. This represents the series limit of the Balmer series.

Step 5: Conclusion.
Therefore, the second spectral line of the Balmer series corresponds to the transition of an electron from the fourth energy level to the second energy level.
Final Answer: \(n_f = 2\) and \(n_i = 4\).