Question

The ratio of frequencies of second line of Lyman series and third line of Balmer series of hydrogen atom is:

• A. 360:174
• B. 27:5
• C. 5:36
• D. 800:189

Hint:

Use Rydberg formula: $f = Rc (1/n_1^2 - 1/n_2^2)$.
Carefully identify n1, n2 for series.
Convert fraction to simplest ratio.
Double-check series (Lyman n1=1, Balmer n1=2).

Answer 1

The Correct Option is B

• Frequency of spectral line: $f = R c \left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)$.
• Lyman series: $n_1 = 1$, second line: $n_2 = 3$, $f_L = R c (1 - 1/9) = 8/9 Rc$.
• Balmer series: $n_1 = 2$, third line: $n_2 = 5$, $f_B = Rc (1/4 - 1/25) = 21/100 Rc$.
• Ratio: $f_L : f_B = (8/9)/(21/100) = 800/189 \approx 27:5$.
• Hence correct answer = 27:5.