Question #f8928

1 Answer
Nov 5, 2017

In hydrogen like atom, we have only one electron. The gross energy is determined only by the principle quantum number #n# as,

#E_n = Z^2R/n^2# where #Z# is the number of protons in the Nucleus and #R# is the Rydberg constant.

However, when we consider multi-electron atoms such as Alkali atoms, even though they have just one outermost electron (just like hydrogen) but this time there are inner core electrons in the atom which make the case pretty complicated.

No matter what, there is a shielding effect on the outer electron due to #(Z - 1)# inner electrons so that the outermost electron experiences an effective nuclear charge less than #Ze#.

If #Z'e# be effective nuclear charge on the outermost electron, then

#Z'e = Ze - (Z - 1)e#
#implies Z' = 1# and the effective charge the outer electron experiences is the same as in hydrogen and as expected, the electronic energy should depend only on #n#.

However, there is one difference , due to elliptical orbits (I am assuming Sommerfeld's theory) the outermost electron penetrates into the core region during the course of it's motion with the nucleus at one of the foci. As it penetrates into the core, the shielding is less effective and electronic energies change.

However, the extend of penetration (which in turn determines how much electronic energies change) is determined by the eccentricity of the ellipse.
It can be shown that the eccentricity varies with #k#, Azimuthal Quantum number as,

#1 - epsilon^2 = (k/n)^2#

Thus for a given #n#, the most eccentric orbit is for #k = 1# where #epsilon# is maximum while the least eccentric orbit is for #k = n# where #epsilon = 0#.

Therefore, the Azimuthal Quantum number #k# determines the energies (due to different extent of penetration depending on #k#).

For #k = 1# i.e. #s# states have maximum eccentricity.

Now, if #Z''e# be the average effective nuclear charge on the outermost electron the electronic energies are given as,

#E_(nk) = Z''^2R/n^2# but as #Z''# depends on #k#, one can write,

#E_(nk) = Z^2R/(n + p)^2# where #p# is a correction factor depending on the value of #k#. It is known as Quantum defect. It is less than #1#.

The quantity, #(n + p)# is called the effective quantum number depending both on #n# and #k#. Obviously #(n + p)# is non integral.

It may be noted that in Quantum mechanics, #k# is replaced by #l = k - 1# as the Azimuthal Quantum number.

It may also be mentioned here that in this case we get two quantum numbers from Wilson-Sommerfeld quantum conditions, #n_r# and #k#.
The principal quantum number is then given by, #n = n_r + k# which shows direct dependence of #n# on #k#.

Therefore, #k# is indeed responsible for determining electronic sub-energies.