Question

The frequencies for series limit of Balmer and Paschen series are $v_{1}$ and $v_{3}$ respectively. If frequency of first line of Balmer series is $v_{2}$, then the relation is}

• A. $v_{1}-v_{3}=2v_{1}$
• B. $v_{1}+v_{2}=v_{3}$
• C. $v_{1}-v_{2}=v_{3}$
• D. $v_{1}+v_{3}=v_{2}$

Hint:

Energy differences between levels are additive. $E_{2 \to \infty} - E_{2 \to 3} = E_{3 \to \infty}$.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
We relate spectral line frequencies using Rydberg's formula.

Step 2: Key Formula or Approach:
\( \nu = Rc \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \).

Step 3: Detailed Explanation:
1. Balmer limit (\( \infty \rightarrow 2 \)): \( v_1 = Rc \left( \frac{1}{4} - 0 \right) = \frac{Rc}{4} \).
2. Paschen limit (\( \infty \rightarrow 3 \)): \( v_3 = Rc \left( \frac{1}{9} - 0 \right) = \frac{Rc}{9} \).
3. Balmer 1st line (\( 3 \rightarrow 2 \)): \( v_2 = Rc \left( \frac{1}{4} - \frac{1}{9} \right) \).
Observe that \( v_2 = \frac{Rc}{4} - \frac{Rc}{9} = v_1 - v_3 \).
Rearranging gives \( v_1 - v_2 = v_3 \).

Step 4: Final Answer:
The relation is \( v_1 - v_2 = v_3 \).