Which electron transition in a hydrogen atom would emit a photon of greatest frequency?

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

Answer: (C)

Verified step by step guidance

1. Recall that the frequency of the emitted photon 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: \(E_n = -\frac{13.6\,\text{eV}}{n^2}\), where \(n\) is the principal quantum number of the energy level.
3. Calculate the energy difference \(\Delta E\) for a transition from an initial level \(n_i\) to a final level \(n_f\) using \(\Delta E = E_{n_f} - E_{n_i} = -\frac{13.6}{n_f^2} + \frac{13.6}{n_i^2}\) (in electron volts).
4. Since the frequency \(\nu\) of the emitted photon is related to the energy difference by \(\Delta E = h \nu\), where \(h\) is Planck's constant, the transition with the largest \(\Delta E\) will emit the photon with the greatest frequency.
5. Compare the energy differences for each given transition (e.g., \(n=3\) to \(n=2\), \(n=2\) to \(n=1\), \(n=5\) to \(n=1\), and \(n=4\) to \(n=2\)) by plugging in the values into the formula and identify which has the largest \(\Delta E\).