Which electronic transition in a hydrogen atom would result in the absorption of a photon with the shortest wavelength?

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

Answer: (C)

Verified step by step guidance

1. Recall that the energy of a photon absorbed during an electronic transition in a hydrogen atom corresponds to the energy difference between the initial and final energy levels. The energy levels of hydrogen are given by the formula: \(E_n = -\frac{13.6\ \text{eV}}{n^2}\), where \(n\) is the principal quantum number.
2. Calculate the energy difference \(\Delta E\) for each transition using the formula: \(\Delta E = E_{final} - E_{initial} = -\frac{13.6}{n_{final}^2} + \frac{13.6}{n_{initial}^2}\).
3. Remember that the wavelength \(\lambda\) of the absorbed photon is related to the energy difference by the equation: \(\lambda = \frac{hc}{\Delta E}\), where \(h\) is Planck's constant and \(c\) is the speed of light.
4. Since \(\lambda\) is inversely proportional to \(\Delta E\), the shortest wavelength corresponds to the largest energy difference \(\Delta E\) between the two levels.
5. Compare the energy differences for all given transitions (\(n=1\) to \(n=3\), \(n=2\) to \(n=4\), \(n=1\) to \(n=5\), and \(n=2\) to \(n=3\)) to determine which transition has the greatest \(\Delta E\), and thus the shortest wavelength.