Question

Match the LIST-I with LIST-II (Symbols have their usual meaning)

LIST-I Spectral series of Hydrogen	LIST-II Wave Number
A. Lyman series	I. \(\bar{v} = R\left[\dfrac{1}{1^2} - \dfrac{1}{n_2^2}\right]\)
B. Paschen series	II. \(\bar{v} = R\left[\dfrac{1}{2^2} - \dfrac{1}{n_2^2}\right]\)
C. Balmer series	III. \(\bar{v} = R\left[\dfrac{1}{3^2} - \dfrac{1}{n_2^2}\right]\)
D. Brackett series	IV. \(\bar{v} = R\left[\dfrac{1}{4^2} - \dfrac{1}{n_2^2}\right]\)

Choose the correct answer from the options given below:

• A. A-I, B-II, C-III, D-IV
• B. A-IV, B-III, C-II, D-I
• C. A-I, B-III, C-IV, D-II
• D. A-I, B-III, C-II, D-IV

Hint:

In each series, the first term 1/n1^2 uses the level where the electron ends: Lyman 1, Balmer 2, Paschen 3, Brackett 4.

Answer 1

The Correct Option is D

Step 1: Use the Region of the Spectrum:
Recall where each series lies. Lyman lines are in ultraviolet, Balmer lines are partly visible, and Paschen and Brackett lines are in infrared. The final level number rises in this order: Lyman 1, Balmer 2, Paschen 3, Brackett 4.

Step 2: Read the Formulas:
Formula I has $1^2$, formula II has $2^2$, formula III has $3^2$ and formula IV has $4^2$ in the first term. The number in that first term is the final level.

Step 3: Pair Them Up:
Lyman (final level 1) goes with I. Balmer (2) goes with II. Paschen (3) goes with III. Brackett (4) goes with IV. In the order of the question list, A-I, B-III, C-II, D-IV.

Step 4: Compare With the Choices:
Only option 4 gives exactly A-I, B-III, C-II, D-IV. Other options swap at least two pairs.

Final Answer:
\[\boxed{\text{A-I, B-III, C-II, D-IV (option 4)}}\]