Electromagnetic Spectrum

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 transition in the Lyman series is given by:

where is the Rydberg constant for hydrogen.

The second Lyman line corresponds to the transition from to , and the frequency is:

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

Simplifying:

Step 2: Frequency of the second Balmer line

The second Balmer line corresponds to the transition from to , and the frequency is:

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:

Substituting the expressions for the frequencies:

Thus, we see that this corresponds to the First Balmer line (which is the transition from to ).

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

To solve the problem of finding the ratio of the minimum wavelength of the Balmer series to the maximum wavelength in the Brackett series in the hydrogen spectrum, we employ the Rydberg formula for hydrogen:

where is the Rydberg constant, and are integers representing the final and initial energy levels, respectively, with .

For the Balmer series:

The Balmer series has . The minimum wavelength occurs at the transition from the to :

For the Brackett series:

The Brackett series has . The maximum wavelength occurs at the transition from to :

Ratio Calculation:

We need the ratio :

Thus, the ratio of the minimum wavelength of the Balmer series to the maximum wavelength in the Brackett series is .