Question

For a hydrogen atom, which electronic transition would result in emission with the highest energy?

a) n=1→n=2
b) n=3→n=1
c) n=4→n=2
d) n=2→n=3

Answer 1

The electronic transition from n=3 to n=1 in a hydrogen atom would emit the highest energy photon because it has the largest energy difference according to the Rydberg formula.

Step-by-step explanation:

The energy of photon emitted during an electronic transition in a hydrogen atom is determined by the difference in energy levels between the higher initial state and the lower final state. According to the Rydberg formula, the energy (E) associated with an electronic transition is:

E = R_H (1/n^2_1 - 1/n^2_2)

where R_H is the Rydberg constant, n_1 is the principal quantum number of the initial state, and n_2 is the principal quantum number of the final state. Among the given options, the transition from a higher energy level to the lowest one (
n=3→n=1) would result in emission with the highest energy because the energy difference between these levels is the greatest.