Question

In the hydrogen spectrum, the shortest wavelength of the Brackett series is produced during the transition:

• A. $n_{2}=5$ and $n_{1}=4$
• B. $n_{2}=4$ and $n_{1}=1$
• C. $n_{2}=4$ and $n_{1}=3$
• D. $n_{2}=\infty$ and $n_{1}=4$
• E. $n_{2}=4$ and $n_{1}=2$

Hint:

Shortest wavelength = Series Limit = Transition from $n = \infty$.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
This question is about the spectral series of the hydrogen atom. We need to identify the transition that corresponds to the shortest wavelength (and thus the highest energy) photon in the Brackett series.
Step 2: Key Formula or Approach:
1. Rydberg Formula: The wavelength (\( \lambda \)) of the photon emitted during an electron transition in a hydrogen atom is given by: \[ \frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \] where R is the Rydberg constant, \( n_1 \) is the principal quantum number of the final (lower energy) state, and \( n_2 \) is the principal quantum number of the initial (higher energy) state (\( n_2>n_1 \)). 2. Spectral Series: - Lyman Series: \( n_1 = 1 \) - Balmer Series: \( n_1 = 2 \) - Paschen Series: \( n_1 = 3 \) - Brackett Series: \( n_1 = 4 \) 3. Shortest Wavelength: To get the shortest wavelength \( \lambda \), the term \( 1/\lambda \) must be maximized. This corresponds to the largest possible energy difference. In the Rydberg formula, \( (\frac{1}{n_1^2} - \frac{1}{n_2^2}) \) must be maximized. For a fixed final state \( n_1 \), this is achieved when the initial state \( n_2 \) is as large as possible, i.e., \( n_2 = \infty \). This transition corresponds to the series limit. Step 3: Detailed Explanation:
The question asks for the shortest wavelength of the Brackett series. From the definitions, the Brackett series corresponds to all transitions that end in the final state \( n_1 = 4 \). To produce the shortest possible wavelength, we need the largest possible energy drop. This occurs when the electron transitions from the highest possible energy level, which is \( n_2 = \infty \) (representing an electron that is initially free from the atom), down to the final state \( n_1 = 4 \). Therefore, the transition is from \( n_2 = \infty \) to \( n_1 = 4 \).
Step 4: Final Answer:
The transition is between the states \( n_2 = \infty \) and \( n_1 = 4 \).