Which electron transition in the hydrogen atom produces emitted light with the longest wavelength?

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

Answer: (D)

Verified step by step guidance

1. Recall that the wavelength of emitted light during an electron transition in a hydrogen atom is related to the energy difference between the two energy levels involved. The larger the energy difference, the shorter the wavelength of the emitted photon.
2. Use the Rydberg formula for the wavelength of emitted or absorbed light in hydrogen: \[\frac{1}{\lambda} = R \left( \frac{1}{n_{\text{final}}^2} - \frac{1}{n_{\text{initial}}^2} \right)\] where \(R\) is the Rydberg constant, \(n_{\text{initial}}\) is the starting energy level, and \(n_{\text{final}}\) is the ending energy level.
3. For emitted light, the electron moves from a higher energy level to a lower energy level (\(n_{\text{initial}} > n_{\text{final}}\)), so the energy difference is positive and light is emitted. The larger this difference, the higher the energy and the shorter the wavelength.
4. Compare the transitions given: - \(n=3\) to \(n=1\) - \(n=2\) to \(n=1\) - \(n=3\) to \(n=2\) Calculate or conceptually evaluate the energy differences for each transition using the formula \(E_n = -\frac{13.6\text{ eV}}{n^2}\) and find which has the smallest energy difference (which corresponds to the longest wavelength).
5. Note that the transition \(n=2\) to \(n=3\) is an absorption (electron moves to a higher energy level), so it does not produce emitted light. The longest wavelength of emitted light corresponds to the smallest energy drop among the emission transitions.