Question

If the difference in the frequencies of the first and second lines of Lyman series of hydrogen atom is f, then the difference in frequencies of the first and second lines of Balmer series of hydrogen atom is

• A. \( \frac{3f}{4} \)
• B. f
• C. \( \frac{7f}{20} \)
• D. \( \frac{9f}{16} \)

Hint:

Write out the Rydberg formula terms clearly. Factoring out the constant \( R c \) early helps in finding the ratio/relationship quickly.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:

Frequency of hydrogen spectral lines is given by \( \nu = RcZ^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \). We need to find the difference in frequencies for specified transitions in Lyman and Balmer series and relate them.
Step 2: Key Formula or Approach:

Lyman Series (\( n_1=1 \)): 1st line: \( 2 \to 1 \). 2nd line: \( 3 \to 1 \). Balmer Series (\( n_1=2 \)): 1st line: \( 3 \to 2 \). 2nd line: \( 4 \to 2 \).
Step 3: Detailed Explanation:

Let \( K = Rc \). For Lyman Series: Frequency of 1st line (\( \nu_{L1} \)): \( K(1 - \frac{1}{4}) = \frac{3K}{4} \). Frequency of 2nd line (\( \nu_{L2} \)): \( K(1 - \frac{1}{9}) = \frac{8K}{9} \). Difference \( f = \nu_{L2} - \nu_{L1} = K\left(\frac{8}{9} - \frac{3}{4}\right) = K\left(\frac{32-27}{36}\right) = \frac{5K}{36} \). So, \( K = \frac{36f}{5} \). For Balmer Series: Frequency of 1st line (\( \nu_{B1} \)): \( K(\frac{1}{4} - \frac{1}{9}) = K(\frac{5}{36}) \). Frequency of 2nd line (\( \nu_{B2} \)): \( K(\frac{1}{4} - \frac{1}{16}) = K(\frac{3}{16}) \). Difference \( \Delta \nu_B = \nu_{B2} - \nu_{B1} = K\left(\frac{3}{16} - \frac{5}{36}\right) \). LCM of 16 and 36 is 144. \( \frac{3 \times 9}{144} - \frac{5 \times 4}{144} = \frac{27 - 20}{144} = \frac{7K}{144} \). Substitute \( K = \frac{36f}{5} \): \( \Delta \nu_B = \frac{7}{144} \times \frac{36f}{5} \) \( = \frac{7f}{4 \times 5} \) (since \( 144/36 = 4 \)) \( = \frac{7f}{20} \).
Step 4: Final Answer:

The difference is \( \frac{7f}{20} \).