Which electronic transition in a hydrogen atom corresponds to the shortest wavelength emission?

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

Answer: (B)

Verified step by step guidance

1. Recall that the wavelength of light emitted during an electronic transition in a hydrogen atom is related to the energy difference between the initial and final energy levels. The larger the energy difference, the shorter the wavelength of the emitted photon.
2. Use the energy level formula for the hydrogen atom: \(E_n = -13.6 \frac{1}{n^2}\) eV, where \(n\) is the principal quantum number of the energy level.
3. Calculate the energy difference for a transition from an initial level \(n_i\) to a final level \(n_f\) using \(\Delta E = E_{n_f} - E_{n_i} = -13.6 \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right)\) eV. Note that \(n_i > n_f\) for emission.
4. Compare the energy differences for the given transitions: \(n=3\) to \(n=2\), \(n=4\) to \(n=2\), and \(n=2\) to \(n=1\). The transition with the largest \(\Delta E\) corresponds to the shortest wavelength.
5. Identify that the transition to \(n=1\) (the ground state) from the highest initial level will have the largest energy difference and thus the shortest wavelength emission.