The result of Schrödinger's wave equation for the hydrogen atom is a wave function, Y. There is a different wave function for each set of quantum numbers. Recall that the square of the wave function tells us the probability of finding the electron at a particular location.
If we look at the wave function for the n = 1, l = 0, ml = 0 state, we find that the wave function is resembles a symmetric cloud (see Figure 28.13). That means it it equally probable to find the electron in any given direction from the proton.
If the probability doesn't depend on angle, then it can only depend on the radius from the proton. When viewed in this manner, we find that the probability peaks at a radius equal to the Bohr radius, a0 = 0.0529nm. It also happens that if we look at the state with n = 2, l = 0, the probability peaks at a radius of 4a0, and for n = 3, at 9a0.
In the Bohr model, an electron follows a circular orbit at a fixed radius from the proton. The wave equation solution indicates that the electron isn't in a circular orbit, however the radius of the Bohr orbit corresponds to the peak of the probability distribution.
The state of an electron in an atom is described by the four quantum numbers n, l, ml, and ms. According to the "exclusion principle", only one electron in an atom can have a given set of quantum numbers. The electrons in an atom tend to move to the lowest available energy levels. Without the exclusion principle, they'd all end up in the n = 1 level.
Because of the exclusion principle, electrons fill up the available energy levels (shells) in order of increasing energy. The lowest energy level is that with n = 1, l = 0, ml = 0, and ms = ±½. These are the levels occupied by the electron in hydrogen and the two electrons in helium. We note their electron configurations as 1s1 for hydrogen and 1s2 for helium. The first 1 indicates that the electrons are in the n = 1 level; the s indicates the l = 0 subshell; and the superscript 1 or 2 indicates whether there are 1 or 2 electrons in this subshell.
This notation is readily extended to describe the electronic configuration of all atoms (see table 28.4). The subshells with l = 0 are described with the letter "s", l = 1 with "p", l = 2 with "d", and l = 3 with "f". The "s" subshell can have up to 2 electrons; the "p" subshell can have up to 6 electrons; the "d" subshell can have up to 10 electrons; and the "f" subshell can have up to 14 electrons. These numbers are seen in the structure of the periodic table.
Suppose two electrons in the same system each have l = 0 and n = 3. (a) How many states would be possible if the exclusion principle were inoperative? (b) List the possible states, taking the exclusion principle into account.
(a) If the exclusion principle were inoperative, then either electron could be in any state consistent with l =0 and n = 3. The possibilities have ml = 0 (the only possible value) and ms = ±½. So we can arrange the two electrons in one of three possible configurations: both in the ms = +½ state, both in the ms = -½ state, or one in each. The electrons are still identical, so there's only one configuration with one in each, not two.
(b) Taking into account the exclusion principle, the electrons can't both be in the same state, so that leaves only one possible configuration: one electron in ml = 0, ms = +½, and the other in ml = 0, ms = -½.
As discussed in chapter 27, x-rays may be produced when an inner shell electron of a heavy atom is ejected, for example by another electron striking the atom at high energy. When an electron in a high energy level jumps down to fill the now vacant inner shell, an x-ray is emitted.
The emitted x-ray spectrum has a broad background with peaks corresponding to the characteristic x-rays of the atom. We can estimate the energies of a characteristic x-ray using the technique outlined in the following example.
Estimate the energy of the characteristic x-ray emitted from a tungsten target when an electron drops from an M shell (n=3 state) to a vacancy in the K shell (n=1 state).
First estimate the energy of the K shell (n=1)level.
This shell can hold two electrons, and since they are the innermost electrons, it's as if they share the force due to all the protons in the nucleus.
We can approximate this effect by saying that for one of the electrons, the effective charge looks like the full charge of the nucleus minus one for the other K shell electron, Zeff = Z - 1.
Tungsten has Z = 74.
Now treat this like a problem to calculate the ground state energy of a hydrogen-like atom with Zeff protons:
EK = -Zeff2(13.6eV) = -(74 - 1)2 (13.6eV) = -72500eV.
The electron in the M shell (n=3) is shielded from the full nuclear charge by the electrons in the n=1 and n=2 shells.
The n=2 shell has 8 electrons (2 in s and 6 in p subshells), and the n=1 shell has 1 electron.
Remember that one of the n=1 electrons is missing, so that shell has only 1 electron, not 2.
The effective charge seen by an M shell electron is Zeff = Z - 9.
To estimate the M shell energy, treat this like an n=3 level of a hydrogen-like atom with Zeff protons:
EM = -Zeff2(13.6eV)/32 = -(74 - 9)2(13.6eV)/32 = -6380eV.
The energy of the emitted x-ray is given by the difference between these levels:
Eph = EM - EK = -6380eV - (-72500eV) = 66100eV.
The wavelength of this x-ray is:
l = hc/Eph = (6.63×10-34Js)(3.0×108m/s)/(66100eV)(1.6×10-19J/eV) = 0.0188nm.