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: Understanding the Question:
Each spectral series of hydrogen is a set of lines produced when the electron falls to one fixed lower level. We must match each series name to its wave number formula.

Step 2: Key Formula or Approach:
The general formula is \[ \bar{v} = R\left[\frac{1}{n_1^2} - \frac{1}{n_2^2}\right], \quad n_2 > n_1 \] The number under the first fraction, \(n_1\), is the level where the electron lands. Each series has its own \(n_1\).

Step 3: Fix n1 for Each Series:
Lyman series: \(n_1 = 1\), so formula I.
Balmer series: \(n_1 = 2\), so formula II.
Paschen series: \(n_1 = 3\), so formula III.
Brackett series: \(n_1 = 4\), so formula IV.

Step 4: Write the Matches:
A (Lyman) goes with I.
B (Paschen) goes with III.
C (Balmer) goes with II.
D (Brackett) goes with IV.
So the matching is A-I, B-III, C-II, D-IV.

Step 5: Checking Each Option:
Option 1 pairs B with II and C with III. Paschen and Balmer are swapped. Wrong.
Option 2 pairs A with IV and D with I. Lyman and Brackett are swapped. Wrong.
Option 3 pairs C with IV and D with II. Wrong.
Option 4 matches all four pairs. Correct.

Final Answer:
The correct matching is A-I, B-III, C-II, D-IV, which is option 4. \[ \boxed{4} \]