Question

In a hydrogen atom, an electron in an orbit with principal quantum number \( n \) jumps to the first excited state, emitting a photon of wavelength \( \lambda \). Then the value of \( n \) is:

• A. \( \sqrt{\frac{4\lambda R}{\lambda R + 4}} \)
• B. \( \sqrt{\frac{4\lambda R}{\lambda R - 4}} \)
• C. \( \sqrt{\frac{\lambda R - 4}{4\lambda R}} \)
• D. \( \sqrt{\frac{\lambda R + 4}{4\lambda R}} \) (R - Rydberg constant)

Hint:

For electronic transitions in hydrogen atoms, remember the Rydberg formula: \[ \frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right). \] It helps in determining photon wavelength and quantum numbers efficiently.

Answer 1

The Correct Option is B

Using the Rydberg formula for electronic transitions.
The wavelength of emitted photon during transition is given by: \[ \frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right). \] Since the electron jumps to the first excited state, \( n_1 = 2 \), so the equation becomes: \[ \frac{1}{\lambda} = R \left( \frac{1}{4} - \frac{1}{n^2} \right). \] Rearranging to solve for \( n \): \[ n^2 = \frac{4}{1 - \lambda R/4}. \] Taking square root, \[ n = \sqrt{\frac{4\lambda R}{\lambda R - 4}}. \]