Question

For below transition of e^–1 of H-atom find out shortest wavelength out of given transition 

• A. A
• B. B
• C. C
• D. D

Answer 1

The Correct Option is D

To determine the shortest wavelength corresponding to an electron transition in a hydrogen atom, we need to use the Rydberg formula for hydrogen spectral lines:

\(\frac{1}{\lambda} = R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)\)

Where:

• \(\lambda\) is the wavelength of emitted light.
• \(R_H\) is the Rydberg constant, approximately \(1.097 \times 10^7 \, \text{m}^{-1}\).
• \(n_1\) and \(n_2\) are the principal quantum numbers with \(n_2 > n_1\).

Looking at the transitions provided in the diagram:

1. Transition A: \(n = 5 \to n = 4\)
2. Transition B: \(n = 4 \to n = 3\)
3. Transition C: \(n = 3 \to n = 2\)
4. Transition D: \(n = 2 \to n = 1\)

The shortest wavelength corresponds to the transition with the largest energy difference, which occurs when an electron falls from a higher to the lowest possible energy level. This is usually from an outer level to \(n = 1\). In this case, it is Transition D:

Transition D Calculation: \(n_2 = 2\) and \(n_1 = 1\).

Using the Rydberg formula:

\(\frac{1}{\lambda} = R_H \left( \frac{1}{1^2} - \frac{1}{2^2} \right)\)

\(\frac{1}{\lambda} = 1.097 \times 10^7 \left( 1 - \frac{1}{4} \right)\)

\(\frac{1}{\lambda} = 1.097 \times 10^7 \times \frac{3}{4}\)

\(\frac{1}{\lambda} = 8.2275 \times 10^6 \, \text{m}^{-1}\)

Thus, transition D has the shortest wavelength, making option D the correct answer.