Question

Shortest wavelength of Lyman series is \(x\). Find the difference of wavelengths of first Balmer and second Balmer line in terms of \(x\).

• A. \( \left(\frac{28}{15}\right)x \)
• B. \( \left(\frac{26}{15}\right)x \)
• C. \( \left(\frac{13}{15}\right)x \)
• D. \( \left(\frac{11}{15}\right)x \)

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
The wavelengths of hydrogen spectral lines are given by the Rydberg formula. We can express the Rydberg constant \(R\) in terms of the variable \(x\) using the shortest Lyman line, and then evaluate the Balmer lines.
Step 2: Key Formula or Approach:
Rydberg formula: \[ \frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
Step 3: Detailed Explanation:
1. Express \(R\) in terms of \(x\): The shortest Lyman line corresponds to the transition from \(n_2 = \infty\) to \(n_1 = 1\). \[ \frac{1}{x} = R \left( \frac{1}{1^2} - \frac{1}{\infty} \right) \implies R = \frac{1}{x} \]
2. Find wavelength of \(1^{\text{st}}\) Balmer line (\(\lambda_1\)): Transition is from \(n_2 = 3\) to \(n_1 = 2\). \[ \frac{1}{\lambda_1} = R \left( \frac{1}{2^2} - \frac{1}{3^2} \right) = R \left( \frac{1}{4} - \frac{1}{9} \right) = R \left( \frac{5}{36} \right) \] \[ \lambda_1 = \frac{36}{5R} = \frac{36x}{5} \]
3. Find wavelength of \(2^{\text{nd}}\) Balmer line (\(\lambda_2\)): Transition is from \(n_2 = 4\) to \(n_1 = 2\). \[ \frac{1}{\lambda_2} = R \left( \frac{1}{2^2} - \frac{1}{4^2} \right) = R \left( \frac{1}{4} - \frac{1}{16} \right) = R \left( \frac{3}{16} \right) \] \[ \lambda_2 = \frac{16}{3R} = \frac{16x}{3} \]
4. Calculate the difference (\(\lambda_1 - \lambda_2\)): \[ \lambda_1 - \lambda_2 = \frac{36x}{5} - \frac{16x}{3} \] Find a common denominator (15): \[ \lambda_1 - \lambda_2 = \frac{36(3)x - 16(5)x}{15} = \frac{108x - 80x}{15} = \frac{28}{15}x \]
Step 4: Final Answer:
The difference is \((28/15)x\).