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

Concept: Rydberg formula for hydrogen spectrum: \[ \frac{1}{\lambda} = R\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right) \] Shortest wavelength of Lyman series corresponds to transition \[ n_2 = \infty \rightarrow n_1 = 1 \] Thus \[ x = \frac{1}{R} \] Step 1: First Balmer line \((n_2=3 \rightarrow n_1=2)\).} \[ \frac{1}{\lambda_1} = R\left(\frac{1}{2^2}-\frac{1}{3^2}\right) \] \[ \frac{1}{\lambda_1} = R\left(\frac{1}{4}-\frac{1}{9}\right) \] \[ \frac{1}{\lambda_1} = R\left(\frac{5}{36}\right) \] \[ \lambda_1 = \frac{36}{5R} \] Since \(x = \frac{1}{R}\) \[ \lambda_1 = \frac{36}{5}x \]
Step 2: Second Balmer line \((n_2=4 \rightarrow n_1=2)\).} \[ \frac{1}{\lambda_2} = R\left(\frac{1}{4}-\frac{1}{16}\right) \] \[ \frac{1}{\lambda_2} = R\left(\frac{3}{16}\right) \] \[ \lambda_2 = \frac{16}{3R} \] \[ \lambda_2 = \frac{16}{3}x \]
Step 3: Find difference.} \[ \lambda_1 - \lambda_2 = \left(\frac{36}{5}-\frac{16}{3}\right)x \] \[ = \left(\frac{108-80}{15}\right)x \] \[ = \frac{28}{15}x \] Final Result \[ \lambda_1 - \lambda_2 = \frac{28}{15}x \]