Final answer:
The energy released when an atom transitions from a higher to a lower energy state is 6.15 MeV, which is the difference between the two energy levels of 8.20 MeV and 2.05 MeV.
Step-by-step explanation:
The amount of energy given off when an atom transitions from a higher energy state to a lower energy state is calculated as the difference between the two energy levels. Considering the example given, if an atom transitions from the first excited state to the ground state, the energy of the emitted photon would be the difference between the two energy levels, which is Ef = ΔE = E2 - E1.
Given E2 (the energy of the first excited state) is 8.20 MeV and E1 (the energy of the ground state) is 2.05 MeV, using the provided information, the energy of the emitted photon would be 6.15 MeV. This value represents quantum of energy released as the electron falls back to its lower energy level.