Final answer:
The second excited state of a beryllium electron suggests a transition beyond the first available energy level (2p) and could be interpreted as either the 3s orbital with quantum numbers n=3, l=0, m=0, or within the 2p level with quantum numbers n=2, l=1, m=-1, 0, or +1, depending on the definition of 'second' used.
Step-by-step explanation:
The ground state electron configuration of beryllium is 1s²2s². When one of the 2s electrons gets excited to the second excited state, this electron is typically promoted to a 2p orbital, which is the next higher available energy level with the same principal quantum number (n=2). However, if we consider the electron moving to the second excited state rather than just the next available state, we must understand what is meant by excited states in this context.
Usually, the term 'second excited state' would suggest that the electron has moved beyond the first available energy level above the ground state. Since the ground state for the valence electron is n=2, l=0, m=+1/2, the next energy level above it would be n=2, l=1 for the 2p orbital. Here, we have three possible magnetic quantum numbers m=-1, 0, +1. Going beyond this, we reach the n=3 energy level that contains 3s, 3p, and 3d orbitals.
If the question means that the second energy level above the current one, i.e., beyond 2p, is meant, this would be the 3s orbital with quantum numbers n=3, l=0, m=0. If instead, the 'second excited state' is interpreted as the second possible excited state within the 2p level, then the quantum numbers might be n=2, l=1, and any of m=-1, 0, or +1 for the three different 2p orbitals. However, it's essential to match the level of excitement with the proper definition.