Final answer:
The energy of a photon emitted as an electron transitions from n = 4 to n = 2 level in a hydrogen atom is 2.55 eV.
Step-by-step explanation:
The energy of a photon emitted when an electron drops from the n = 4 energy level to the n = 2 level can be calculated using the Rydberg formula for hydrogen-like atoms:
E = RhcZ^2 (1/n1^2 - 1/n2^2)
Where E is the energy of the photon, Rhc is the Rydberg constant in joules, Z is the atomic number (Z = 1 for hydrogen), n1 is the lower energy level, and n2 is the higher energy level.
Putting in the values for the given transition:
E = 13.6 eV (1/2^2 - 1/4^2)
E = 13.6 eV (1/4 - 1/16)
E = 13.6 eV (4/16 - 1/16)
E = 13.6 eV (3/16)
E = 2.55 eV
Therefore, the photon would have an energy of 2.55 eV when an electron drops from the n = 4 level to the n = 2 level.