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: Rydberg Formula. \[ \frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
Step 2: Wavelength of 2nd line in Balmer Series (\(\lambda_B\)). For Balmer series, \(n_1 = 2\). 1st line: \(3 \to 2\). 2nd line: \(4 \to 2\). So \(n_2 = 4\). \[ \frac{1}{\lambda_B} = R \left( \frac{1}{2^2} - \frac{1}{4^2} \right) = R \left( \frac{1}{4} - \frac{1}{16} \right) \] \[ \frac{1}{\lambda_B} = R \left( \frac{4-1}{16} \right) = \frac{3R}{16} \implies \lambda_B = \frac{16}{3R} \]
Step 3: Wavelength of 1st line in Lyman Series (\(\lambda_L\)). For Lyman series, \(n_1 = 1\). 1st line: \(2 \to 1\). So \(n_2 = 2\). \[ \frac{1}{\lambda_L} = R \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = R \left( 1 - \frac{1}{4} \right) \] \[ \frac{1}{\lambda_L} = \frac{3R}{4} \implies \lambda_L = \frac{4}{3R} \]
Step 4: Calculate Ratio. \[ \frac{\lambda_B}{\lambda_L} = \frac{16/3R}{4/3R} = \frac{16}{3R} \times \frac{3R}{4} = \frac{16}{4} = 4 \] Ratio is 4:1.