Question

A hydrogen atom emits ultraviolet radiation of wavelength 1025 \(\text{Å}\). What are the quantum numbers of energy states involved in the transition?

Hint:

The wavelength of emitted radiation can be used to determine the quantum numbers of the states involved in the transition using the Rydberg formula.

Answer 1

Step 1: Energy of the Photon.
The energy of the emitted photon is given by: \[ E = \frac{hc}{\lambda} \] Where:
- \( h = 6.626 \times 10^{-34} \, \text{J·s} \),
- \( c = 3 \times 10^8 \, \text{m/s} \),
- \( \lambda = 1025 \times 10^{-10} \, \text{m} \).
Substitute the values: \[ E = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{1025 \times 10^{-10}} = 1.94 \times 10^{-18} \, \text{J} \]
Step 2: Transition and Quantum Numbers.
The energy of the photon corresponds to the energy difference between two energy levels of the hydrogen atom. Using the Rydberg formula: \[ E = -13.6 \left( \frac{1}{n_2^2} - \frac{1}{n_1^2} \right) \] Solve for \( n_1 \) and \( n_2 \) (the quantum numbers).
Final Answer:
The transition involves quantum numbers \( n_1 = 2 \) and \( n_2 = 3 \).