Question

Light of wavelength \(\frac{36}{5R}\,\text{m}\) is emitted by a hydrogen atom during the transition of electrons from the state

• A. \(n=3\) to \(n=2\)
• B. \(n=4\) to \(n=1\)
• C. \(n=4\) to \(n=2\)
• D. \(n=4\) to \(n=3\)
• E. \(n=3\) to \(n=1\)

Hint:

In hydrogen spectrum questions, first convert the given wavelength into \(1/\lambda\), then compare it directly with the Rydberg expression \(\frac{1}{n_1^2}-\frac{1}{n_2^2}\).

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
This question involves the emission spectrum of the hydrogen atom. When an electron makes a transition from a higher energy level (\(n_i\)) to a lower energy level (\(n_f\)), a photon of a specific wavelength is emitted. The wavelength of this photon is given by the Rydberg formula.
Step 2: Key Formula or Approach:
The Rydberg formula for the hydrogen atom is:
\[ \frac{1}{\lambda} = R \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right) \] where:
- \( \lambda \) is the wavelength of the emitted photon.
- R is the Rydberg constant.
- \( n_f \) is the principal quantum number of the final (lower) energy level.
- \( n_i \) is the principal quantum number of the initial (higher) energy level.
We are given \( \lambda = \frac{36}{5R} \), which means \( \frac{1}{\lambda} = \frac{5R}{36} \). We can test the transitions given in the options to see which one satisfies this equation.
Step 3: Detailed Explanation:
We need to find the transition for which \( \frac{1}{\lambda} = \frac{5R}{36} \). Let's check each option:
(A) n = 3 to n = 2: Here, \( n_i = 3 \) and \( n_f = 2 \).
\[ \frac{1}{\lambda} = R \left( \frac{1}{2^2} - \frac{1}{3^2} \right) = R \left( \frac{1}{4} - \frac{1}{9} \right) = R \left( \frac{9 - 4}{36} \right) = R \left( \frac{5}{36} \right) = \frac{5R}{36} \] This matches the given condition.
(B) n = 4 to n = 1: Here, \( n_i = 4 \) and \( n_f = 1 \).
\[ \frac{1}{\lambda} = R \left( \frac{1}{1^2} - \frac{1}{4^2} \right) = R \left( 1 - \frac{1}{16} \right) = \frac{15R}{16} \] This does not match.
(C) n = 4 to n = 2: Here, \( n_i = 4 \) and \( n_f = 2 \).
\[ \frac{1}{\lambda} = R \left( \frac{1}{2^2} - \frac{1}{4^2} \right) = R \left( \frac{1}{4} - \frac{1}{16} \right) = R \left( \frac{4 - 1}{16} \right) = \frac{3R}{16} \] This does not match.
(D) n = 4 to n = 3: Here, \( n_i = 4 \) and \( n_f = 3 \).
\[ \frac{1}{\lambda} = R \left( \frac{1}{3^2} - \frac{1}{4^2} \right) = R \left( \frac{1}{9} - \frac{1}{16} \right) = R \left( \frac{16 - 9}{144} \right) = \frac{7R}{144} \] This does not match.
(E) n = 3 to n = 1: Here, \( n_i = 3 \) and \( n_f = 1 \).
\[ \frac{1}{\lambda} = R \left( \frac{1}{1^2} - \frac{1}{3^2} \right) = R \left( 1 - \frac{1}{9} \right) = \frac{8R}{9} \] This does not match.
Step 4: Final Answer:
The transition that results in the emission of light with wavelength \( \lambda = \frac{36}{5R} \) is from n = 3 to n = 2. This corresponds to option (A). (This is the first line of the Balmer series, known as H-alpha).