In the hydrogen atom, which electron transition would correspond to the highest frequency of light emitted?

• A. n = 5 to n = 4
• B. n = 4 to n = 2
• C. n = 2 to n = 1
• D. n = 3 to n = 2

Answer: (C)

Verified step by step guidance

1. Recall that the frequency of light emitted during an electron transition in a hydrogen atom is related to the energy difference between the two energy levels involved. The greater the energy difference, the higher the frequency of the emitted photon.
2. Use the energy level formula for the hydrogen atom: \(E_n = -13.6 \, \text{eV} / n^2\), where \(n\) is the principal quantum number of the energy level.
3. Calculate the energy difference \(\Delta E\) for each transition using \(\Delta E = E_{\text{initial}} - E_{\text{final}} = -13.6 \left( \frac{1}{n_{\text{final}}^2} - \frac{1}{n_{\text{initial}}^2} \right)\) eV. Note that the initial \(n\) is higher than the final \(n\) because the electron is moving to a lower energy level and emitting light.
4. Recognize that the frequency \(\nu\) of the emitted light is related to the energy difference by the equation \(\Delta E = h \nu\), where \(h\) is Planck's constant. Therefore, the transition with the largest \(\Delta E\) corresponds to the highest frequency.
5. Compare the energy differences for the given transitions: \(n=5 \to 4\), \(n=4 \to 2\), \(n=3 \to 2\), and \(n=2 \to 1\). The transition from \(n=2\) to \(n=1\) has the largest energy difference and thus corresponds to the highest frequency of light emitted.