According to the Bohr model, which transition in the hydrogen atom would result in the emission of the photon with the greatest energy?

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

Answer: (D)

Verified step by step guidance

1. Recall that in the Bohr model of the hydrogen atom, the energy of an electron in a particular energy level \( n \) is given by the formula: \[E_n = -\frac{13.6}{n^2} \text{ eV}\] where \( E_n \) is the energy of the level and \( n \) is the principal quantum number.
2. Understand that when an electron transitions from a higher energy level \( n_i \) to a lower energy level \( n_f \), it emits a photon with energy equal to the difference between these two energy levels: \[E_{photon} = E_{n_f} - E_{n_i}\] Since \( E_n \) values are negative, the photon energy corresponds to the absolute value of this difference.
3. Calculate the energy difference for each transition by substituting the initial and final \( n \) values into the energy formula: \[\Delta E = -\frac{13.6}{n_f^2} + \frac{13.6}{n_i^2} = 13.6 \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right) \text{ eV}\]
4. Compare the energy differences for each transition: - From \( n=5 \) to \( n=4 \) - From \( n=4 \) to \( n=3 \) - From \( n=3 \) to \( n=2 \) - From \( n=2 \) to \( n=1 \) The transition with the greatest energy difference corresponds to the photon with the greatest energy.
5. Recognize that the greatest energy difference occurs when the electron falls to the lowest energy level (\( n=1 \)) from the next higher level (\( n=2 \)), because the energy levels are closer together at higher \( n \) values and farther apart near the nucleus.