Question

The ratio of the wavelengths of radiation emitted when an electron in the hydrogen atom jumps from the 4th orbit to the 2nd orbit and from the 3rd orbit to the 2nd orbit is:

• A. \( 27:25 \)
• B. \( 20:25 \)
• C. \( 20:27 \)
• D. \( 25:27 \)

Hint:

Use the Rydberg formula for wavelength calculations: \( \frac{1}{\lambda} = R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \).
- Wavelength is inversely proportional to the energy difference between levels.

Answer 1

The Correct Option is C

The wavelength of emitted radiation in hydrogen-like atoms is given by: \[ \frac{1}{\lambda} = R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \] where: - \( R_H \) is the Rydberg constant, - \( n_1 \) and \( n_2 \) are the initial and final energy levels. 1. For transition \( 4 \to 2 \): \[ \frac{1}{\lambda_{42}} = R_H \left( \frac{1}{2^2} - \frac{1}{4^2} \right) \] \[ = R_H \left( \frac{1}{4} - \frac{1}{16} \right) \] \[ = R_H \left( \frac{4}{16} - \frac{1}{16} \right) = R_H \left( \frac{3}{16} \right) \] \[ \lambda_{42} \propto \frac{16}{3} \] 2. For transition \( 3 \to 2 \): \[ \frac{1}{\lambda_{32}} = R_H \left( \frac{1}{2^2} - \frac{1}{3^2} \right) \] \[ = R_H \left( \frac{1}{4} - \frac{1}{9} \right) \] \[ = R_H \left( \frac{9}{36} - \frac{4}{36} \right) = R_H \left( \frac{5}{36} \right) \] \[ \lambda_{32} \propto \frac{36}{5} \] 3. Ratio of Wavelengths: \[ \frac{\lambda_{42}}{\lambda_{32}} = \frac{\frac{16}{3}}{\frac{36}{5}} \] \[ = \frac{16}{3} \times \frac{5}{36} = \frac{16 \times 5}{3 \times 36} = \frac{80}{108} = \frac{20}{27} \] Thus, the correct answer is \(\boxed{20:27}\).