Chapter 6 - Hydrogen Atom PDF

Chapter 6 Hydrogen Atom 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...

Chapter 6 Hydrogen Atom 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 ℏ2 𝜕 2 Ψ 𝑥, 𝑦, 𝑧, 𝑡 𝜕 2 Ψ 𝑥, 𝑦, 𝑧, 𝑡 𝜕 2 Ψ 𝑥, 𝑦, 𝑧, 𝑡 − 2 + 2 + 2𝑚 𝜕𝑥 𝜕𝑦 𝜕𝑧 2 𝜕Ψ 𝑥, 𝑦, 𝑧, 𝑡 +𝑈 𝑥, 𝑦, 𝑧 Ψ 𝑥, 𝑦, 𝑧, 𝑡 = 𝑖ℏ 𝜕𝑡 This is a direct extension of the one-dimensional Schrödinger equation. 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). Ψ 𝑥, 𝑦, 𝑧, 𝑡 = 𝜓 𝑥, 𝑦, 𝑧 𝑒 −𝑖𝐸𝑡/ℏ The function (x, y, z) obeys the time-independent Schrödinger equation in three dimensions: ℏ2 𝜕 2 𝜓 𝑥, 𝑦, 𝑧 𝜕 2 𝜓 𝑥, 𝑦, 𝑧 𝜕 2 𝜓 𝑥, 𝑦, 𝑧 − + + 2𝑚 𝜕𝑥 2 𝜕𝑦 2 𝜕𝑧 2 +𝑈 𝑥, 𝑦, 𝑧 𝜓 𝑥, 𝑦, 𝑧 = 𝐸𝜓 𝑥, 𝑦, 𝑧 The normalization of the three-dimensional wave function probability distribution function is 2 න 𝜓 𝑥, 𝑦, 𝑧 𝑑𝑉 = 1 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: 𝜓 𝑥, 𝑦, 𝑧 = 𝑋(𝑥)𝑌(𝑦)𝑍(𝑧) The Schrödinger equation inside the box becomes ℏ2 𝜕2𝑋 𝑥 𝜕2𝑌 𝑦 𝜕2𝑍 𝑧 − 𝑌(𝑦)𝑍(𝑧) + 𝑋(𝑥)𝑍(𝑧) + 𝑋(𝑥)𝑌(𝑦) = 𝐸𝑋(𝑥)𝑌(𝑦)𝑍(𝑧) 2𝑚 𝜕𝑥 2 𝜕𝑦 2 𝜕𝑧 2 Or, dividing by X(x)Y(y)Z(z) we have: ℏ2 1 𝜕2𝑋 𝑥 1 𝜕 2𝑌 𝑦 1 𝜕2𝑍 𝑧 − + + =𝐸 2𝑚 𝑋(𝑥) 𝜕𝑥 2 𝑌(𝑦) 𝜕𝑦 2 𝑍(𝑧) 𝜕𝑧 2 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: ℏ2 𝜕 2 𝑋 𝑥 − = 𝐸𝑋 𝑥 2𝑚 𝜕𝑥 2 ℏ2 𝜕 2 𝑌 𝑦 − = 𝐸𝑌 𝑦 2𝑚 𝜕𝑦 2 ℏ2 𝜕 2 𝑍 𝑧 − = 𝐸𝑍 𝑧 2𝑚 𝜕𝑧 2 These obviously have the same solutions separately as our original particle in an infinite square well, and corresponding energies: 𝑛𝑋 𝜋𝑥 𝑛𝑋 2 𝜋 2 ℏ2 𝑋𝑛𝑋 (𝑥) = 𝐶𝑋 sin (𝑛𝑋 = 1,2,3... ) 𝐸𝑋 = (𝑛𝑋 = 1,2,3... ) 𝐿 2𝑚𝐿2 𝑛𝑌 𝜋𝑦 𝑛𝑌 2 𝜋 2 ℏ 2 𝑌𝑛𝑌 (𝑦) = 𝐶𝑌 sin (𝑛𝑌 = 1,2,3... ) 𝐸𝑌 = (𝑛𝑌 = 1,2,3... ) 𝐿 2𝑚𝐿2 𝑛𝑍 𝜋𝑧 𝑍𝑛𝑍 (𝑧) = 𝐶𝑍 sin (𝑛𝑍 = 1,2,3... ) 𝑛𝑍 2 𝜋 2 ℏ 2 𝐿 𝐸𝑍 = (𝑛𝑍 = 1,2,3... ) 2𝑚𝐿2 Particle in a three-dimensional box A particle’s wave-function is the product of these three solutions, 𝑛𝑋 𝜋𝑥 𝑛𝑌 𝜋𝑦 𝑛𝑍 𝜋𝑧 𝜓 𝑥, 𝑦, 𝑧 = 𝑋(𝑥)𝑌(𝑦)𝑍(𝑧) = 𝐶 sin sin sin 𝐿 𝐿 𝐿 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. 4𝜋 2 ℏ2 𝜋 2 ℏ2 𝜋 2 ℏ2 3𝜋 2 ℏ2 𝐸 = 𝐸𝑋 + 𝐸𝑌 + 𝐸𝑍 = + + = 2𝑚𝐿2 2𝑚𝐿2 2𝑚𝐿2 𝑚𝐿2 Particle in a three-dimensional box Example: (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). 𝐿 𝐿 𝐿 2 2 𝑛𝑋 𝜋𝑥 2 𝑛 𝜋𝑦 2 𝑌 𝑛𝑍 𝜋𝑧 න 𝜓 𝑥, 𝑦, 𝑧 𝑑𝑉 = 𝐶 න sin 𝑑𝑥 න sin 𝑑𝑦 න sin2 𝑑𝑧 = 1 0 𝐿 0 𝐿 0 𝐿 1 sin2 𝜃 = (1 − cos 2 𝜃) 2 𝐿 𝐿 𝑛𝑋 𝜋𝑥 𝐿 𝑛𝑋 𝜋𝑥 1 2𝑛𝑋 𝜋𝑥 𝐿 න sin2 𝑑𝑥 = − sin = 0 𝐿 2𝑛𝑋 𝜋 𝐿 2 𝐿 0 2 3 𝐿/4 𝐿 𝐿 2 𝜋𝑥 2𝜋𝑦 𝜋𝑧 න 𝜓 𝑥, 𝑦, 𝑧 2 𝑑𝑉 = න 2 sin 𝑑𝑥 න sin 2 𝑑𝑦 න sin2 𝑑𝑧 𝐿 0 𝐿 0 𝐿 0 𝐿 𝐿/4 𝐿/4 3 𝜋𝑥 𝐿 𝜋𝑥 1 2𝜋𝑥 𝐿 𝐿 2 𝐿 𝐿 𝐿 𝐿 න sin2 𝑑𝑥 = − sin = − so − = 0.091 0 𝐿 2𝜋 𝐿 2 𝐿 0 8 4𝜋 𝐿 8 4𝜋 2 2 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: 3𝜋 2 ℏ2 6-fold degenerate 𝐸1,1,1 = 2𝑚𝐿2 Not degenerate 3-fold degenerate 3-fold degenerate 3-fold degenerate If length of sides of box are different: 𝑛𝑋 2 𝑛𝑌 2 𝑛𝑍 2 𝜋 2 ℏ2 Not degenerate 𝐸= + + 𝐿𝑋 2 𝐿𝑌 2 𝐿𝑍 2 2𝑚 (breaks the degeneracy)  SCHRÖDINGER’S EQUATION FOR THE HYDROGEN ATOM a) Spherical polar coordinates. (b) A line of constant zenith angle  on a sphere is a circle whose plane is perpendicular to the z axis. (c) A line of constant azimuth angle  is a circle whose plane includes the z axis. In spherical polar coordinates Schrödinger’s equation is written  Solving the polar Schrödinger equation gives three quantum numbers. These Q# are required to describe the electron in a hydrogen atom, in place of the single quantum number of the Bohr theory. The third term is a function of azimuth angle  only, whereas the other terms are functions of r and  only.  Rearrange The differential equation for the function  is therefore Each of these is an ordinary differential equation for a single function of a single variable. Only the equation for R depends on the potential energy U(r).  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−r, multiplied by a polynomial in r. The angular solutions () are polynomials containing powers of sin and cos. The azimuthal solutions () have to be periodic (have the same value at () and (+2), etc. They turn out to depend on eiml, where ml is an integer, either positive, negative or zero. The solution for the radial part R(r) of the hydrogen atom wave function Ψ also requires that a certain condition be fulfilled. This condition is that E be positive or have one of the negative values En (signifying that the electron is bound to the atom) specified by The same formula for the energy levels of the hydrogen atom that Bohr obtained. PRINCIPAL QUANTUM NUMBER Quantization of energy ➔ n is quantized ORBITAL QUANTUM NUMBER If the last two terms in the square brackets of the equation cancel each other out, we have a differential equation for R(r) that involves functions of the radius vector r exclusively.  Since the orbital kinetic energy of the electron and the magnitude of its angular momentum are The orbital kinetic energy Compare above with Like total energy E, angular momentum is both conserved and quantized. Designation of Angular-Momentum States It is customary to specify electron angular-momentum states by a letter, with s corresponding to l = 0, p to l = 1, and so on, according to the following scheme: MAGNETIC QUANTUM NUMBER If the magnetic-field direction be parallel to the z axis, the component of L in z direction is ELECTRON PROBABILITY DENSITY The probability density |Ψ|2 that corresponds to the electron wave function Ψ = RΘΦ in the hydrogen atom is Probability of Finding the Electron 𝜋 2𝜋 𝑃 𝑟 𝑑𝑟 = 𝑟 2 𝑅∗ 𝑟 𝑅 𝑟 𝑑𝑟 න |Θ|2 𝑠𝑖𝑛𝜃𝑑𝜃 න |Φ|2 𝑑∅ = 𝑟 2 |𝑅|2 𝑑𝑟 0 0  𝜋 2𝜋 𝑃 𝑟 𝑑𝑟 = 𝑟 2 𝑅∗ 𝑟 𝑅 𝑟 𝑑𝑟 න |Θ|2 𝑠𝑖𝑛𝜃𝑑𝜃 න |Φ|2 𝑑∅ = 𝑟 2 |𝑅|2 𝑑𝑟 0 0  The most probable value of 𝑟 for a 1𝑠 electron is 𝑎0​, the orbital radius of a ground- state electron in the Bohr model. The average value of r for a 1s electron is 1.5a0. Angular Variation of Probability Density 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 below shows the electron probability distributions for three of these states. https://winter.group.shef.ac.uk/orbitron/atomic_orbitals/1s/index.html 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 below shows the electron probability distributions for several of these states, as well as for two spherically symmetric s states. The hydrogen atom: Degeneracy States with different quantum numbers l and n are often referred to with letters as follows: l value letter 0 s n value shell 1 p 1 K 2 d 2 L 3 f 3 M 4 g 4 N 5 h Hydrogen atom states with the same value of n but different values of l and ml are degenerate (have the same energy). Figure → shows radial probability distribution for states with l = 0, 1, 2, 3 and different values of n = 1, 2, 3, 4.  RADIATIVE TRANSITIONS Bohr postulated that the frequency of the radiation emitted by an atom dropping from an energy level Em to a lower level En is The expectation value of the position of such an electron is An electron that shifts from one energy state to another a*a = probability that the electron in state n b*b = probability that the electron in state m a*a + b*b = 1 Initially a=1 and b = 0; when the electron is in the excited state→ a=0 and b = 1; and ultimately a = 1 and b = 0 once more. While the electron is in either state, there is no radiation, but when it is in the midst of the transition from m to n (that is, when both a and b have nonvanishing values), electromagnetic waves are produced. The expectation value that corresponds to the composite wave function let a*a=a2 and b*b = b2. The first and last integrals do not vary with time, so the second and third integrals are the only ones able to contribute to a time variation in. The electron’s position therefore oscillates sinusoidally at the frequency When the electron is in state n or state m the expectation value of the electron’s position is constant. When the electron is undergoing a transition between these states, its position oscillates with the frequency. Such an electron, of course, is like an electric dipole and radiates electromagnetic waves of the same frequency. This result is the same as that postulated by Bohr and verified by experiment. SELECTION RULES The general condition necessary for an atom in an excited state to radiate is that the integral not be zero Integral is finite → allowed transitions Integral is zero → forbidden transitions In the case of the hydrogen atom, three quantum numbers are needed to specify the initial and final states involved in a radiative transition. If the principal, orbital, and magnetic quantum numbers of the initial state are n’, l’, m’l, respectively, and those of the final state are n, l, ml, and u represents either the x, y, or z coordinate, the condition for an allowed transition is Since the wave functions n,l,ml for the hydrogen atom are known, the allowed transitions can be evaluated for u = x, u = y, and u = z for all pairs of states differing in one or more quantum numbers. When this is done, it is found that the only transitions between states of different n that can occur are those in which the orbital quantum number l changes by 1 or −1 and the magnetic quantum number ml does not change or changes by 1 or −1. The change in total quantum number n is not restricted. ZEEMAN EFFECT H atom in an external magnetic field → semiclassical model the electron motion is described by a classical circular orbit, while the angular momentum is given by the quantum mechanical expression velocity The current causes a magnetic moment The angular momentum of the circulating electron is Since μ is proportional to l, the orbital magnetic moment is often labeled μl. In an external magnetic field the potential energy of a magnetic dipole with magnetic moment μ is When the magnetic field points into the z-direction (B = {0, 0, Bz = B}), because of lz = mℏ, where m (the projection of l onto the z-axis) is called the magnetic quantum number, It can take values −l ≤ m ≤+l. → Bohr magneton The additional energy caused by the magnetic field as This gives for the energies of the hydrogen atomic states in an external magnetic field The 2𝑙 + 1 m-sublevels that are degenerate without magnetic field split into 2𝑙 + 1 equidistant Zeeman components with energies between 𝐸𝑛𝑚𝑖𝑛 = 𝐸𝑐𝑜𝑢𝑙 𝑛 − 𝜇𝐵 𝑚𝐵 𝐸𝑛𝑚𝑎𝑥 = 𝐸𝑐𝑜𝑢𝑙 𝑛 + 𝜇𝐵 𝑚𝐵 With an energetic distance ∆𝐸 = 𝐸𝑛,𝑙,𝑚 − 𝐸𝑛,𝑙,𝑚−1 = 𝜇𝐵 𝐵 Zeeman splitting of a level with l=2 in a homogeneous magnetic between adjacent levels which is determined field (normal Zeeman effect) by the product of Bohr Magneton 𝜇𝐵 and magnetic field strength B. 𝑒ℏ 𝜇𝐵 = 2𝑚𝑒 The Zeeman effect and selection rules Lower left figure 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.