When last ē is on last shall such like 80Hg then what we called the energy level (n-1) or (n-2)

1 answer

In atomic physics, the principal quantum number \( n \) designates the energy level of an electron in an atom. The term "last \( e \)" likely refers to the last electron in a given configuration or the outermost electron in a particular energy level.

For the element Mercury (Hg), which has an atomic number of 80, the electron configuration is typically written as:

\[ [Xe] 4f^{14} 5d^{10} 6s^2 \]

If you need to determine whether the last filled energy level is considered \( n-1 \) or \( n-2 \), you look at the highest principal quantum number of the outermost electrons. In this case, the outermost electrons (the "last \( e \)") are in the \( 6s \) subshell, with \( n = 6 \). The \( (n-1) \) level would be \( n = 5 \), which corresponds to the \( 5d \) subshell, while \( (n-2) \) would refer to \( n = 4 \), corresponding to the \( 4f \) subshell.

Thus, in the case of Mercury (Hg):

  • The outermost energy level is \( n = 6 \) (from the \( 6s^2 \) electrons).
  • The electrons in the filled \( 5d^{10} \) subshell would correspond to \( n-1 \).
  • The filled \( 4f^{14} \) subshell corresponds to \( n-2 \).

Therefore, when referring to the "last \( e \)" in the \( 6s \) subshell of Mercury, the prior filled level (the \( 5d^{10} \) subshell) is considered to be at the \( n-1 \) level.