Which of the following electronic transitions in a hydrogen atom represents the emission of a photon with the largest energy?

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

Answer: (A)

Verified step by step guidance

1. Recall that the energy of a photon emitted during an electronic transition in a hydrogen atom corresponds to the difference in energy between the two energy levels involved in the transition.
2. Use the energy level formula for the hydrogen atom: \(E_n = -\frac{13.6\ \text{eV}}{n^2}\), where \(n\) is the principal quantum number of the energy level.
3. Calculate the energy difference for each transition using \(\Delta E = E_{\text{final}} - E_{\text{initial}} = -\frac{13.6}{n_{\text{final}}^2} + \frac{13.6}{n_{\text{initial}}^2}\), noting that emission means the electron moves from a higher to a lower energy level, so \(\Delta E\) will be positive.
4. Compare the calculated energy differences for all given transitions: \(n=4\) to \(n=1\), \(n=3\) to \(n=2\), \(n=5\) to \(n=4\), and \(n=2\) to \(n=1\).
5. Identify the transition with the largest \(\Delta E\) value, which corresponds to the emission of the photon with the largest energy.