With hydrogen as an exception, what is the correlation between the value of L and the energy of an electron?
The correct answer and explanation is:
Correct Answer:
With hydrogen as an exception, the energy of an electron increases as the value of L (the angular momentum quantum number) increases.
In hydrogen, energy levels depend only on the principal quantum number n because hydrogen has only one electron and there is no electron-electron repulsion. The energy levels are degenerate, meaning that all orbitals with the same n (regardless of L) have the same energy.
However, in atoms with more than one electron, the situation changes due to electron-electron interactions. In these multi-electron atoms, the energy of an orbital depends on both n and L. The angular momentum quantum number L determines the shape of the orbital and, consequently, how close an electron can get to the nucleus.
Orbitals with lower L values (like s-orbitals, L = 0) allow electrons to penetrate closer to the nucleus. This increases the attraction between the electron and the positively charged nucleus, leading to a lower energy. On the other hand, orbitals with higher L values (such as p, d, or f-orbitals, with L = 1, 2, 3 respectively) have more angular momentum and are more likely to be found farther from the nucleus. They experience more shielding from inner electrons and therefore are less tightly bound, resulting in higher energy.
This is why, for example, the 3d orbital (L = 2) is higher in energy than the 4s orbital (L = 0), even though 3d has a lower principal quantum number. The increased value of L reduces the electron’s ability to penetrate close to the nucleus, so its energy is raised due to less effective nuclear attraction.
In summary, for atoms with more than one electron, as the angular momentum quantum number L increases, the energy of the electron also increases, due to reduced nuclear attraction and greater electron shielding.