Of the following transitions in the Bohr hydrogen atom, the transition results in the emission of the highest- energy photon.
A) n = 6 -n = 3
B) n = 1 -n = 6
C) n = 3 -n = 6
D) n = 6 -n = 1
E) n = 1 -n = 4

Question

Quizplus · Accepted Answer

The highest energy photon corresponds to the largest energy difference between the initial and final states. From the given transitions, we can calculate the energy difference using the formula ΔE = E_final - E_initial = (-13.6 eV) * (1/n_final^2 - 1/n_initial^2). The transition from n = 6 to n = 1 has the largest energy difference: ΔE = (-13.6 eV) * (1/1^2 - 1/6^2) = 10.2 eV. Therefore, this transition results in the emission of the highest energy photon.

Of the Following Transitions in the Bohr Hydrogen Atom, the Transition