Question

In the hydrogen spectrum, the difference in frequencies of first and second Lyman lines is equal to the frequency of the second Balmer line. This corresponds to:

• A. Fourth Lyman line
• B. Third Lyman line
• C. Second Balmer line
• D. First Balmer line

Hint:

The Rydberg formula is crucial in spectroscopy for finding the wavelengths or frequencies of lines in the hydrogen spectrum.

Answer 1

The Correct Option is D

The Lyman and Balmer series in the hydrogen spectrum correspond to transitions in the hydrogen atom that result in the emission of light in the ultraviolet and visible regions, respectively.

Step 1: Frequencies of Lyman lines

The frequency of the \( n \to 1 \) transition in the Lyman series is given by:

\[ \nu_1 = R_H \left( \frac{1}{1^2} - \frac{1}{2^2} \right) \]

where \( R_H \) is the Rydberg constant for hydrogen.

The second Lyman line corresponds to the transition from \( n = 3 \) to \( n = 1 \), and the frequency is:

\[ \nu_2 = R_H \left( \frac{1}{1^2} - \frac{1}{3^2} \right) \]

Thus, the difference in frequencies between the first and second Lyman lines is:

\[ \Delta \nu_{\text{Lyman}} = \nu_2 - \nu_1 = R_H \left( \frac{1}{1^2} - \frac{1}{3^2} \right) - R_H \left( \frac{1}{1^2} - \frac{1}{2^2} \right) \]

Simplifying:

\[ \Delta \nu_{\text{Lyman}} = R_H \left( \frac{1}{1} - \frac{1}{9} - \frac{1}{1} + \frac{1}{4} \right) \] \[ \Delta \nu_{\text{Lyman}} = R_H \left( \frac{4}{9} - \frac{1}{4} \right) \] \[ \Delta \nu_{\text{Lyman}} = R_H \left( \frac{16}{36} - \frac{9}{36} \right) = R_H \times \frac{7}{36} \]

Step 2: Frequency of the second Balmer line

The second Balmer line corresponds to the transition from \( n = 4 \) to \( n = 2 \), and the frequency is:

\[ \nu_{\text{Balmer}} = R_H \left( \frac{1}{2^2} - \frac{1}{4^2} \right) \] \[ \nu_{\text{Balmer}} = R_H \left( \frac{1}{4} - \frac{1}{16} \right) = R_H \times \frac{3}{16} \]

Step 3: Equating the frequencies

We are given that the difference in frequencies of the first and second Lyman lines is equal to the frequency of the second Balmer line:

\[ \Delta \nu_{\text{Lyman}} = \nu_{\text{Balmer}} \]

Substituting the expressions for the frequencies:

\[ R_H \times \frac{7}{36} = R_H \times \frac{3}{16} \]

Thus, we see that this corresponds to the First Balmer line (which is the transition from \( n = 2 \) to \( n = 3 \)).

Thus, the correct answer is Option (4), First Balmer line.