Chapter 41 Atomic Structure

Goals for Chapter 41 To see how to write the Schrödinger equation for a three-dimensional problem To learn how to find the wave functions and energies for a three-dimensional box To examine the full quantum-mechanical description of the hydrogen atom To study how an external magnetic field affects the orbital motion of an atom’s electrons To learn about the intrinsic angular momentum (spin) of the electron To understand how the exclusion principle affects the structure of many-electron atoms To study how the x-ray spectra of atoms indicate the structure of these atoms

The Schrödinger equation in 3-D Electrons in an atom can move in all three dimensions of space. If a particle of mass m moves in the presence of a potential energy function U(x, y, z), the Schrödinger equation for the particle’s wave function (x, y, z, t) is This is a direct extension of the one-dimensional Schrödinger equation from Chapter 40.

The Schrödinger equation in 3-D: Stationary states If a particle of mass m has a definite energy E, its wave function (x, y, z, t) is a product of a time-independent wave function (x, y, z) and a factor that depends on time but not position. Then the probability distribution function |(x, y, z, t)|2 = |(x, y, z)|2 does not depend on time (stationary states). The function (x, y, z) obeys the time-independent Schrödinger equation in three dimensions: The normalization of the three-dimensional wave function probability distribution function is

Particle in a three-dimensional box In analogy with our infinite square-well potential (U(x) = 0 inside, U(x) = ∞ outside), let us consider a three-dimensional region in space (box) of equal sides of length L, with the same potential (U(x) = 0 inside, U(x) = ∞ outside). We will consider the wave function as separable, that is, can be written as a product of the three independent dimensions x, y and z: The Schrödinger equation inside the box becomes Or, dividing by X(x)Y(y)Z(z) we have:

Particle in a three-dimensional box This says that the total energy is contributed to by three terms on the left, each depending separately on x, y and z. Let us write E = Ex + Ey + Ez. Then this equation can be separated into three equations: These obviously have the same solutions separately as our original particle in an infinite square well, and corresponding energies:

Particle in a three-dimensional box A particle’s wave-function is the product of these three solutions, We can use the three quantum numbers nX, nY, and nZ to label the stationary states (states of definite energy). Here is an example of a particle in three possible states (nX, nY, nZ) = (2, 1, 1), (1, 2, 1) or (1, 1, 2). The three states shown here are degenerate: Although they have different values of nX, nY, and nZ, they have the same total energy E.

Particle in a three-dimensional box Example 41.1: (a) Find the value of the normalization constant C. (b) Find the probability that the particle will be found somewhere in the region 0 ≤ x ≤ L/4, for the cases (nX, nY, nZ) = (1, 2, 1), (2, 1, 1) and (3, 1, 1). (a) The integral can be written as the product of three integrals Recall that so that The other terms give the same result, so |C|2 (L/2)3 = 1, and C = (2/L)3/2. (b) For the first case, (nX, nY, nZ) = (1, 2, 1), we now want to find Only the first integral differs from L/2, due to its different high limit

Energy Degeneracy For a particle in a three-dimensional box, the allowed energy levels are surprisingly complex. To find them, just count up the different possible states. Here are the first 6 for an equal-sided box: 6-fold degenerate Not degenerate 3-fold degenerate

The hydrogen atom: Quantum numbers For a hydrogen atom, the three-dimensional potential energy depends only on the electron’s distance from the proton: As before, we will seek separable wave functions, but this time, due to the spherical symmetry we will use spherical coordinates: (r, , ) rather than (x, y, z). Then Following the same procedure as before, we can separate the problem into three separate equations Energies are Same as for Bohr model! Quantum numbers are n, l and ml.

Description of solution We will not go through the mathematics of the solution, but note that we can only accept solutions for which the wave function is normalizable (does not blow up). Some radial solutions blow up at r = 0 or r = ∞, and so must be discarded. As in the harmonic oscillator problem, the radial solutions R(r) turn out to be an exponential function e-ar, multiplied by a polynomial in r. The angular solutions Q(q) are polynomials containing powers of sinq and cosq. The azimuthal solutions F(f) have to be periodic (have the same value at F(f) and F(f+2p), etc. They turn out to depend on eimlf, where ml is an integer, either positive, negative or zero.

Quantization of Angular Momentum In addition to quantized energy (specified by principle quantum number n), the solutions subject to physical boundary conditions also have quantized orbital angular momentum L. The magnitude of the vector L is required to obey where l is the orbital quantum number. Recall that the Bohr model of the hydrogen atom also had quantized angular moment L = nħ, but the lowest energy state n = 1 would have L = ħ. In contrast, the Schrödinger equation shows that the lowest state has L = 0. This lowest energy-state wave function is a perfectly symmetric sphere. For higher energy states, the vector L has in addition only certain allowed directions, such that the z-component is quantized as

The hydrogen atom: Degeneracy States with different quantum numbers l and n are often referred to with letters as follows: Hydrogen atom states with the same value of n but different values of l and ml are degenerate (have the same energy). Figure at right shows radial probability distribution for states with l = 0, 1, 2, 3 and different values of n = 1, 2, 3, 4. l value letter s 1 p 2 d 3 f 4 g 5 h n value shell 1 K 2 L 3 M 4 N

The hydrogen atom: Quantum states Table 41.1 (below) summarizes the quantum states of the hydrogen atom. For each value of the quantum number n, there are n possible values of the quantum number l. For each value of l, there are 2l + 1 values of the quantum number ml. Example 41.2: How many distinct states of the hydrogen atom (n, l, ml) are there for the n = 3 state? What are their energies? The n = 3 state has possible l values 0, 1, or 2. Each l value has ml possible values of (0), (-1, 0, 1), or (-2, -1, 0, 1, 2). The total number of states is then 1 + 3 + 5 = 9. We will see later that there is another quantum number s, for electron spin (±½), so there are actually 18 possible states for n = 3. Each of these states have the same n, so they all have the same energy.

The hydrogen atom: Probability distributions I States of the hydrogen atom with l = 0 (zero orbital angular momentum) have spherically symmetric wave functions that depend on r but not on  or . These are called s states. Figure 41.9 (below) shows the electron probability distributions for three of these states. Follow Example 41.4.

The hydrogen atom: Probability distributions II States of the hydrogen atom with nonzero orbital angular momentum, such as p states (l = 1) and d states (l = 2), have wave functions that are not spherically symmetric. Figure 41.10 (below) shows the electron probability distributions for several of these states, as well as for two spherically symmetric s states.

Magnetic moments and the Zeeman effect Electron states with nonzero orbital angular momentum (l = 1, 2, 3, …) have a magnetic dipole moment due to the electron motion. Hence these states are affected if the atom is placed in a magnetic field. The result, called the Zeeman effect, is a shift in the energy of states with nonzero ml. This is shown in Figure (below). The potential energy associated with a magnetic dipole moment m in a magnetic field of strength B is U = -mB , and the magnetic dipole moment due to the orbital angular momentum of the electron is in units of the Bohr magneton, Lower Same Higher energy energy energy Think of magnets, which like to be anti-aligned. This is a lower energy state.

Zeeman Effect (Sunspots)

Zeeman Effect (Sunspots)

The Zeeman effect and selection rules Figure 41.14 (lower left) shows the shift in energy of the five l = 2 states (each with a different value of ml) as the magnetic field strength is increased. An atom in a magnetic field can make transitions between different states by emitting or absorbing a photon A transition is allowed if l changes by 1 and ml changes by 0, 1, or –1. (This is because a photon itself carries angular momentum.) A transition is forbidden if it violates these selection rules. See Figure 41.15 (lower right).

The anomalous Zeeman effect and electron spin For certain atoms the Zeeman effect does not follow the simple pattern that we have described (see Figure 41.16 below). This is because an electron also has an intrinsic angular momentum, called spin angular momentum.

Electron spin and the Stern-Gerlach experiment The experiment of Stern and Gerlach demonstrated the existence of electron spin (see Figure 41.17 below). The z-component of the spin angular momentum has only two possible values (corresponding to ms = +1/2 and ms = –1/2).

Quantum states and the Pauli exclusion principle The allowed quantum numbers for an atomic electron (see Table 41.2 below) are n ≥ 1; 0 ≤ l ≤ n – 1; –l ≤ ml ≤ l; and ms = ±1/2. The Pauli exclusion principle states that if an atom has more than one electron, no two electrons can have the same set of quantum numbers.

A multielectron atom Figure 41.21 (at right) is a sketch of a lithium atom, which has 3 electrons. The allowed electron states are naturally arranged in shells of different size centered on the nucleus. The n = 1 states make up the K shell, the n = 2 states make up the L shell, and so on. Due to the Pauli exclusion principle, the 1s subshell of the K shell (n = 1, l = 0, ml = 0) can accommodate only two electrons (one with ms = + 1/2, one with ms = –1/2). Hence the third electron goes into the 2s subshell of the L shell (n = 2, l = 0, ml = 0).

Ground-state electron configurations

Screening in multielectron atoms An atom of atomic number Z has a nucleus of charge +Ze and Z electrons of charge –e each. Electrons in outer shells “see” a nucleus of charge +Zeffe, where Zeff < Z, because the nuclear charge is partially “screened” by electrons in the inner shells. Example 41.8: Determine Zeff experimentally. The measured energy of a 3s state of a sodium atom is -5.138 eV. Calculate the value of Zeff. From the table, Na (Z=11) has 1s22s22p63s. In analogy with hydrogen, which has a ground state of -13.6 eV, we can define an effective (reduced) Z by knowing the energy of a state, i.e. implies

X-ray spectroscopy When atoms are bombarded with high-energy electrons, x rays are emitted. There is a continuous spectrum of x rays (described in Chapter 38) as well as strong characteristic x-ray emission at certain definite wavelengths (see the peaks labeled K and K in Figure 41.23 at right). Atoms of different elements emit characteristic x rays at different frequencies and wavelengths. Hence the characteristic x-ray spectrum of a sample can be used to determine the atomic composition of the sample.

X-ray spectroscopy: Moseley’s law Moseley showed that the square root of the x-ray frequency in K emission is proportional to Z – 1, where Z is the atomic number of the atom (see Figure 41.24 below). Larger Z means a higher frequency and more energetic emitted x-ray photons. Follow Example 41.10. This is consistent with our model of multielectron atoms. Bombarding an atom with a high-energy electron can knock an atomic electron out of the innermost K shell. K x rays are produced when an electron from the L shell falls into the K-shell vacancy. The energy of an electron in each shell depends on Z, so the x-ray energy released does as well. The same principle applies to K emission (in which an electron falls from the M shell to the K shell) and K emission (in which an electron falls from the N shell to the K shell).