Question

The wave numbers of three spectral lines of hydrogen atom are considered. Identify the set of spectral lines belonging to the {Balmer series. (\(R\) = Rydberg constant)}

• A. \( \dfrac{5R}{36},\ \dfrac{8R}{9},\ \dfrac{15R}{16} \)
• B. \( \dfrac{7R}{144},\ \dfrac{3R}{16},\ \dfrac{16R}{255} \)
• C. \( \dfrac{3R}{4},\ \dfrac{3R}{16},\ \dfrac{7R}{144} \)
• D. \( \dfrac{5R}{36},\ \dfrac{3R}{16},\ \dfrac{21R}{100} \)

Hint:

For Balmer series, always factor out \(\dfrac{1}{4}\); if the remaining term is \(\dfrac{1}{n^2}\), it belongs to the series.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
Wave number (\( \bar{\nu} \)) of spectral lines in the Hydrogen atom is given by the Rydberg formula:
\[ \bar{\nu} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \] For the Balmer series, the electronic transition ends at the second energy level (\( n_1 = 2 \)). The possible values for \( n_2 \) are \( 3, 4, 5, \dots \).
Step 2: Detailed Explanation:
Let's calculate the wave numbers for the first few transitions of the Balmer series:
1. For \( n_2 = 3 \) (First line):
\[ \bar{\nu}_1 = R \left( \frac{1}{2^2} - \frac{1}{3^2} \right) = R \left( \frac{1}{4} - \frac{1}{9} \right) = R \left( \frac{9 - 4}{36} \right) = \frac{5R}{36} \]
2. For \( n_2 = 4 \) (Second line):
\[ \bar{\nu}_2 = R \left( \frac{1}{2^2} - \frac{1}{4^2} \right) = R \left( \frac{1}{4} - \frac{1}{16} \right) = R \left( \frac{4 - 1}{16} \right) = \frac{3R}{16} \]
3. For \( n_2 = 5 \) (Third line):
\[ \bar{\nu}_3 = R \left( \frac{1}{2^2} - \frac{1}{5^2} \right) = R \left( \frac{1}{4} - \frac{1}{25} \right) = R \left( \frac{25 - 4}{100} \right) = \frac{21R}{100} \]
Comparing these values with the options, Option (D) matches the calculated set.
Step 3: Final Answer:
The set of spectral lines for the Balmer series is \( \frac{5R}{36}, \frac{3R}{16}, \frac{21R}{100} \).