Final answer:
The energy of the emitted photon when a hydrogen atom transitions from the n = 2 level to the ground state is 10.2 eV. This calculation is based on difference in energy levels of the hydrogen atom according to Bohr's model.
Step-by-step explanation:
The question deals with the concept of electron transitions within a hydrogen atom and the energy associated with photons emitted during such a transition, based on Bohr's theory. When a hydrogen atom, initially in the n = 2 level, emits a photon and drops down to the ground state (n = 1), the energy of the photon can be calculated using the energy levels of the hydrogen atom.
For the hydrogen atom, each energy level En is given by:
En = -13.6 eV / n²
Thus, for the n = 2 level:
E2 = -13.6 eV / 2² = -3.4 eV
and for the n = 1 ground state:
E1 = -13.6 eV
The energy of the photon emitted (Ephoton) is the difference in energy between these two levels, given by:
Ephoton = E2 - E1 = -3.4 eV - (-13.6 eV) = 10.2 eV
Therefore, the correct answer is A. 10.2 eV.