Question

The ratio of wavelengths of second line in Balmer series and the first line in Lyman series of hydrogen atom is

• A. 2:1
• B. 9:4
• C. 4:1
• D. 3:2

Hint:

To quickly identify lines in a series: For a series ending at \(n_f\), the "first line" comes from \(n_i = n_f + 1\), the "second line" from \(n_i = n_f + 2\), and so on. For example, the third Paschen line (\(n_f=3\)) would be from \(n_i = 3+3=6\).

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
This question requires the use of the Rydberg formula, which describes the wavelengths of spectral lines emitted by a hydrogen atom when an electron transitions between different energy levels. We need to identify the specific transitions corresponding to the first Lyman line and the second Balmer line.
Step 2: Key Formula or Approach:
The Rydberg formula for the inverse of the wavelength (\(\lambda\)) is:
\[ \frac{1}{\lambda} = R \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right) \] where \(R\) is the Rydberg constant, \(n_f\) is the principal quantum number of the final energy level, and \(n_i\) is the principal quantum number of the initial energy level (\(n_i>n_f\)).
- For the Lyman series, the final state is \(n_f = 1\).
- For the Balmer series, the final state is \(n_f = 2\).
Step 3: Detailed Explanation:
First line of the Lyman series (\(\lambda_L\)):
This corresponds to the transition from the next higher level to the ground state.
- Final level: \(n_f = 1\).
- Initial level for the first line: \(n_i = 2\).
\[ \frac{1}{\lambda_L} = R \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = R \left( 1 - \frac{1}{4} \right) = \frac{3R}{4} \quad \cdots(1) \] Second line of the Balmer series (\(\lambda_B\)):
This corresponds to the transition from the second higher level to the first excited state.
- Final level: \(n_f = 2\).
- The first line is from \(n_i = 3\). The second line is from \(n_i = 4\).
\[ \frac{1}{\lambda_B} = R \left( \frac{1}{2^2} - \frac{1}{4^2} \right) = R \left( \frac{1}{4} - \frac{1}{16} \right) = R \left( \frac{4-1}{16} \right) = \frac{3R}{16} \quad \cdots(2) \] Calculate the ratio \(\lambda_B : \lambda_L\):
To find the ratio \(\frac{\lambda_B}{\lambda_L}\), we can divide equation (1) by equation (2):
\[ \frac{1/\lambda_L}{1/\lambda_B} = \frac{3R/4}{3R/16} \] \[ \frac{\lambda_B}{\lambda_L} = \frac{3R}{4} \times \frac{16}{3R} \] The \(3R\) terms cancel out.
\[ \frac{\lambda_B}{\lambda_L} = \frac{16}{4} = 4 \] So, the ratio \(\lambda_B : \lambda_L\) is 4:1.
Step 4: Final Answer:
The ratio of the wavelengths is 4:1. Therefore, option (C) is correct.