Question

Consider the following spectral lines for atomic hydrogen :
A. First line of Paschen series
B. Second line of Balmer series
C. Third line of Paschen series
D. Fourth line of Bracket series
The correct arrangement of the above lines in ascending order of energy is :

• A. C \(<\) D \(<\) B \(<\) A
• B. A \(<\) B \(<\) C \(<\) D
• C. D \(<\) C \(<\) A \(<\) B
• D. D \(<\) A \(<\) C \(<\) B

Hint:

Generally, transitions ending at lower levels (smaller \( n_1 \)) have much higher energies. Balmer transitions are usually more energetic than Paschen, which are more energetic than Bracket.

Answer 1

The Correct Option is D

To solve the problem of determining the correct arrangement of the given spectral lines for atomic hydrogen in ascending order of energy, let's examine each of the series mentioned:

• Balmer Series: This series corresponds to transitions where the final energy level is \( n_2 = 2 \). It is visible and falls in the middle energy range.
• Paschen Series: This series corresponds to transitions to \( n_2 = 3 \). It is part of the infrared spectrum and has lower energy than the Balmer series.
• Bracket Series: This series involves transitions to \( n_2 = 4 \) and is in the infrared region with energy lower than both Paschen and Balmer series.

Let us arrange the given lines in ascending order of energy:

1. D. Fourth line of Bracket series: The transition ends at \( n_2 = 4 \). It is one of the lowest in energy among the given lines.
2. A. First line of Paschen series: Transition ends at \( n_2 = 3 \), and this is the lowest energy transition in the Paschen series.
3. C. Third line of Paschen series: Still within the Paschen series, but with a slightly higher energy transition than the first line.
4. B. Second line of Balmer series: Ends at \( n_2 = 2 \) with transitions from \( n_1 \) levels of higher energy than those of the lower series, making it higher in energy than A and C.

Therefore, the correct arrangement of the spectral lines in ascending order of energy is: D < A < C < B.