Question

In hydrogen atom spectrum, frequency of \( 2.7 \times 10^{15} \, \text{Hz} \) of EM wave is emitted when transmission takes place from 2 to 1, the frequency emitted will be

• A. \( 3.2 \times 10^{15} \, \text{Hz} \)
• B. \( 3.2 \times 10^{15} \, \text{Hz} \)
• C. \( 1.6 \times 10^{15} \, \text{Hz} \)
• D. \( 16 \times 10^{15} \, \text{Hz} \)

Hint:

In hydrogen atom transitions, the frequency of the emitted light is related to the energy difference between the levels involved in the transition.

Answer 1

The Correct Option is A

The frequency of the emitted radiation when an electron in a hydrogen atom transitions between energy levels is given by the Rydberg formula: \[ \Delta E = 13.6 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \, \text{eV} \] The frequency \( f \) is related to the energy \( E \) by: \[ E = h f \] Thus, for the transition from \( n = 3 \) to \( n = 1 \), the frequency is twice the frequency for the \( n = 2 \) to \( n = 1 \) transition.
Given the frequency for \( n = 2 \) to \( n = 1 \) is \( 2.7 \times 10^{15} \, \text{Hz} \), the frequency for \( n = 3 \) to \( n = 1 \) transition is \( 3.2 \times 10^{15} \, \text{Hz} \).
Thus, the correct answer is (a).