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

Step 1: Understanding the Concept:
The energy of a transition in the hydrogen spectrum is proportional to the difference of the reciprocals of the squares of the principal quantum numbers.

Step 2: Key Formula or Approach:
Energy \( E \propto \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \).

Step 3: Detailed Explanation:
Calculate the factor \( F = \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \) for each line :
A. First line of Paschen: \( n_1=3, n_2=4 \). \( F_A = \frac{1}{9} - \frac{1}{16} = \frac{7}{144} \approx 0.0486 \).
B. Second line of Balmer: \( n_1=2, n_2=4 \). \( F_B = \frac{1}{4} - \frac{1}{16} = \frac{3}{16} = 0.1875 \).
C. Third line of Paschen: \( n_1=3, n_2=6 \). \( F_C = \frac{1}{9} - \frac{1}{36} = \frac{3}{36} \approx 0.0833 \).
D. Fourth line of Bracket: \( n_1=4, n_2=8 \). \( F_D = \frac{1}{16} - \frac{1}{64} = \frac{3}{64} \approx 0.0469 \).
Comparing the values : \( 0.0469 (D)<0.0486 (A)<0.0833 (C)<0.1875 (B) \).
The ascending order is D \(<\) A \(<\) C \(<\) B.

Step 4: Final Answer:
The correct order is D \(<\) A \(<\) C \(<\) B.