What is the result of an atom undergoing spontaneous emission?

When an atom undergoes spontaneous emission, it transitions from an excited state to a lower energy state, resulting in the release of a photon. The energy of this emitted photon corresponds precisely to the energy difference between the two states of the atom involved in the transition.

In spontaneous emission, the process occurs without external influence, meaning the atom does not require any external energy input to undergo this transition. Instead, the atom releases the energy it had gained when it was in the excited state.

The energy of the emitted photon can be calculated using the equation:

[ E = h \cdot f ]

where ( E ) is the energy difference between the excited state and the lower energy state, ( h ) is Planck's constant, and ( f ) is the frequency of the emitted photon. Therefore, the photon released during spontaneous emission indeed has energy that is equal to the energy difference between these two states, making this the correct interpretation of the process.

This understanding clarifies why the answer indicating that a photon with energy equal to the energy difference is released is accurate, as it directly reflects the fundamental principles of quantum mechanics at play in atomic transitions.