Question

Which of the following statement(s) is(are) correct about the spectrum of hydrogen atom?

• A. The ratio of the longest wavelength to the shortest wavelength in Balmer series is $\frac{9}5$
• B. There is an overlap between the wavelength ranges of Balmer and Paschen series
• C. The wavelength of Lyman series are given by $\left(1+\frac{1}{ m ^{2}}\right) \lambda_{0}$, where $\lambda_{0}$ is the shortest wavelength of Lyman series and $m$ is an integer
• D. The wavelength ranges of Lyman and Balmer series do not overlap

Answer 1

The Correct Option is A, D

Step 1: Understanding the given spectrum of the hydrogen atom
The spectrum of the hydrogen atom is characterized by distinct series, such as the Lyman series, Balmer series, etc., that correspond to different electronic transitions within the atom.
In the case of the hydrogen atom, the wavelengths of the emitted radiation in these series can be described using the Rydberg formula:
\[ \frac{1}{\lambda} = R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \] where:
- \( R_H \) is the Rydberg constant for hydrogen,
- \( n_1 \) and \( n_2 \) are the principal quantum numbers of the initial and final states, respectively.
The Balmer series corresponds to transitions where \( n_1 = 2 \) and \( n_2 \) takes values from 3 to infinity.
The Lyman series corresponds to transitions where \( n_1 = 1 \) and \( n_2 \) takes values from 2 to infinity.
Step 2: Analyzing the statements
Statement (A): The ratio of the longest wavelength to the shortest wavelength in the Balmer series is \( \frac{9}{5} \)
In the Balmer series, the longest wavelength corresponds to the transition from \( n_2 = 3 \) to \( n_1 = 2 \), and the shortest wavelength corresponds to the transition from \( n_2 = \infty \) to \( n_1 = 2 \).
Using the Rydberg formula, we can calculate the wavelengths for these transitions and find that the ratio of the longest wavelength to the shortest wavelength is indeed \( \frac{9}{5} \). Hence, statement (A) is correct.
Statement (D): The wavelength ranges of the Lyman and Balmer series do not overlap
The Lyman series lies in the ultraviolet range, and the Balmer series lies in the visible range of the spectrum.
Since the wavelength for the Lyman series is shorter than the wavelength for the Balmer series, their wavelength ranges do not overlap. Hence, statement (D) is also correct.
Step 3: Conclusion
Therefore, the correct options are:
(A): The ratio of the longest wavelength to the shortest wavelength in the Balmer series is \( \frac{9}{5} \)
(D): The wavelength ranges of Lyman and Balmer series do not overlap