The Hydrogen Atom Atom is a 3D object, and the electron motion is three-dimensional. We ll start with the simplest case - The hydrogen atom. An electron and a proton (nucleus) are bound by the central-symmetric Coulomb interaction. Because mp>>me, we neglect the proton motion Thus, we can treat this problem as the motion of an electron in the 3D central-symmetric Coulomb potential. Three dimensions we expect that the motion will be characterized with three quantum numbers. U U r 2 e 4 r 0 The task: solve the time-independent Schrödinger equation for E< 0 (bound state), find the energy eigenvalues (the spectrum) and eigenfunctions (stationary states). Dr. Sabry El-Taher 1
The H-atom is the only atom that can be solved exactly. The results become the basis for understanding all other atoms and molecules. For the hydrogen atom the Schrödinger equation: has a spherically symmetric potential energy: V K V E V ( r ) 1 4 0 2 e r For Hydrogen-like atoms (He + or Li ++ ), replace e 2 with Ze 2. - distance from origin to point P - Zenith (polar) angle 2 2 2 r x y z cos The potential V(r) does not depend on either or hence, it is most convenient to work in - Azimuthal angle tan spherical polar coordinates, this will allow us to separate variables. In addition, the spherical geometry of the model suggests the use of a spherical Polar coordinate system. Dr. Sabry El-Taher 2 1 1 z r y x
Time-independent Schrödinger Equation in Polar Coordinates: In all cases, for better accuracy, replace m with the reduced mass,. 2 x, y, z x, y, z x, y, z 2 2 2 x, y, z x, y, z x, y, z 2 2 2 2 2 2 x y z,,,,,, V x y z x y z E x y z 2 2 2 2 x y z m m m e e p m p 2 r,, r,, r,, 1 2 1 1 2 r 2 2 2 2 sin r r r r sin r sin 2 2 E V r,, r,, 0 Dr. Sabry El-Taher 3
Separation of Variables The wave function is a function of r,,. This is a potentially complicated function. Assume instead that is separable, that is, a product of three functions, each of one variable only: r,, R r f g This would make the solution of the Schrödinger equation for H- atom much simpler. Dr. Sabry El-Taher 4
Solution of the Schrödinger Equation for Hydrogen Atom Start with Schrodinger s Equation: Substitute: Multiply both sides by r 2 sin 2 / R f g: Dr. Sabry El-Taher 5
r and appear only on the left side and appears only on the right side. The left side of the equation cannot change as changes. The right side cannot change with either r or. Each side needs to be equal to a constant for the equation to be true. Set the constant to be -m 2 l Azimuthal equation Dr. Sabry El-Taher 6
Now set the left side equal to -m l 2 : -m l 2 Rearrange it and divide by sin 2 (): Now, the left side depends only on r, and the right side depends only on. We can use the same trick again! Dr. Sabry El-Taher 7
Set each side equal to the constant l(l + 1): Radial equation Angular equation In addition to the azimuthal equation 2 d g 2 2 ml g 0 Azimuthal equation We ve separated the Schrödinger equation into three one-dimensional second-order differential equations, each containing only one variable. Dr. Sabry El-Taher 8
Solution of the Radial Equation The radial equation is called the associated Laguerre equation and the solutions R are called associated Laguerre functions. There are infinitely many of them, for values of n = 1, 2, 3, Assume that the ground state has n = 1 and l = 0. Let s find this solution. The radial equation becomes: The derivative of yields two terms: Dr. Sabry El-Taher 9
Try a solution A is a normalization constant. a 0 is a constant with the dimension of length. Take derivatives of R and insert them into the radial equation. To satisfy this equation for any r, both expressions in parentheses must be zero. Set the second expression equal to zero and solve for a 0 : Set the first expression equal to zero and solve for E: Bohr radius Both are equal to the Bohr results! Dr. Sabry El-Taher 10
Principal Quantum Number n There are many solutions to the radial wave equation, one for each positive integer, n. The radial quantum number n gives the quantized energy E n : n = 1, 2, 3, Lowest energy (ground state energy): E 0 = -13.6 ev. Energy levels of the hydrogen atom: E n 2 4 Z e 8 hn 2 2 2 0 A negative energy means that the electron and proton are bound together. In the course of solving the radial equation, n occurs naturally and must satisfy The condition that which is usually written as: n l 1 0 l n 1, n 1,2,3, Dr. Sabry El-Taher 11
The solutions of the radial wave equation depend on two quantum numbers n and l and are given by: ( n l 1)! 1/ 2 l 3 / 2 2 / 2l1 2r 3 nl na0 na0 l r na0 R ( r ) r e L nl 2 n n l! Where the L 2l 1 nl (2 r / na ) 0 are called associated Laguerre polynomials. The first few solutions for the radial equation are given in Table 7.1 The functions given by the above equation are just a polynomial multiplied by an exponential. Dr. Sabry El-Taher 12
Hydrogen Atom Radial Wave Functions Dr. Sabry El-Taher 13
Solution of the Angular and Azimuthal Equations The solutions to the azimuthal equation are: satisfies the azimuthal equation for any value of m l. The solution must be single-valued to be a valid solution for any : Specifically: So: m l must be an integer (positive or negative) for this to be true. Dr. Sabry El-Taher 14
e e e g = 0 and = 2 are same point. For arbitrary value of m: iml i 2 i 2 ml iml 1 for 2 but e 1 for 0 cos2 i sin 2 1 1 e 2 1 if m is a positive or negative integer or 0 iml Therefore, e 1 if 0 e iml iml 1 if 2 wavefunction is single-valued only if m l is a positive or negative integer or 0. Must be single-valued (Born conditions). Only if m 0, 1, 2, 3 l Dr. Sabry El-Taher 15
1 iml gm e where m 0, 1, 2, 3 l l 2 m l is called the magnetic (azimuthal) quantum number. The functions having the same m can be added and subtracted to obtain real functions. 1 0 m 2 0 m m 1 cos 1 sin m m m 1, 2, 3 The cos function is used for positive m s and the sin function is used for negative m s. Dr. Sabry El-Taher 16
Solution of the Angular and Azimuthal Equations Solutions to the angular and azimuthal equations are linked because both have m l. Physicists usually group these solutions together into functions called Spherical Harmonics: The radial wave function R and the spherical harmonics Y determine the probability density for the various quantum states. The total wave function depends on n, l, and m l. The wave function becomes Dr. Sabry El-Taher 17
The spherical harmonics The spherical harmonics are solutions of the angular Hamiltonian. The spherical harmonics are the product of the solutions to the angular and azimuthal equations. With normalization, these solutions are: Y l m (,) = N lm P l m (cos )e im The m quantum number corresponds to the z-component of angular momentum l. The normalization constant is: N lm = 2l + 1 2 (l m )! (l + m )! 1/2 Dr. Sabry El-Taher 18
The form of the spherical harmonics Including normalization, the spherical harmonics are: Y 0 0 = 1 4 Y 0 1 = 3 4 cos Y ±1 1 = 3 8 sine ±i Y 0 2 = 5 16 3cos 2 1 Y ±1 2 = 15 8 sincose±i Y 2 2 = 15 32 sin2 e ±2i These functions describe the angular distribution of atomic orbitals. The degeneracy (number of m values) of a given orbital is 2l+1 and the angular momentum of the electron is [l(l + 1)h] 1/2. Dr. Sabry El-Taher 19
Normalized Spherical Harmonics Dr. Sabry El-Taher 20
Dr. Sabry El-Taher 21
Have solved three one-dimensional equations to get R n ( r) f m ( ) g m ( ) The total wavefunction is: ( r,, ) g ( ) f ( ) R ( r ) nlm m lm nl n 1, 2, 3 n1, n 2, 0 m, 1 Dr. Sabry El-Taher 22
The Total Wavefunction (,, r ) g ( ) f ( ) R ( r ) n m m m n 1s function 3 1 2 Z 2 Zr / a 1s 100 0 00 10 2 2 a 0 (,, r ) g f R 2 e 0 1s 1 r / a a 3 0 2s function e 0 for Z = 1 No nodes. 1 s (,, r ) g f R (2 r / a ) e 2 200 0 00 20 3 0 4 2 a0 r /2a 0 Node at r = 2a 0. Dr. Sabry El-Taher 23
Hydrogen atom wavefunctions - orbitals For n =1: 1s orbital 1 ra / 0 Ae A s 1 a a 0 = 0.529 Å the Bohr radius The wavefunction is the probability amplitude. The probability is the absolute valued squared of the wavefunction. 1s 2 2 2/ ra Ae 0 This is the probability of finding the electron a distance r from the nucleus on a line where the nucleus is at r = 0. 3 0 1.4 2 1.2 1 0.8 0.6 2 1.5 1 note scale difference 0.4 0.5 0.2 0.5 1 1.5 2 2.5 3 r (Å) 0.2 0.4 0.6 0.8 1 1.2 1.4 r (Å) Dr. Sabry El-Taher 24 Copyright Michael D. Fayer, 2007
For n = 2: r/2a0 (2 / ) 2 0 s B r a e B 1 4 2 a 3 0 Probability amplitude When r = 2a 0, this term goes to zero. There is a node in the wave function. 2 /2 2 (2 / 0) r a s B r a e 0 2 a 0 = 0.529, the Bohr radius Absolute value of the wavefunction squared probability distribution. 0.5 0.25 0.4 0.3 0.2 node 2 0.2 0.15 0.1 node 0.1 0.05 2 4 6 8 1 2 3 4 r (Å) r (Å) Dr. Sabry El-Taher 25
Wavefunctions for Hydrogen-like Atoms Dr. Sabry El-Taher 26
Wavefunctions for Hydrogen-like Atoms Dr. Sabry El-Taher 27
Atomic orbitals Atomic orbitals is the wave functions of electrons in an atom. The Shape of orbital can be determined from the angular part of the wave function Shape of orbital Y 00 Y (, ) lm The Size of orbital and its spatial distribution can be determined from the radial part of the wave function R. () r nl 1 4 Y 00 has spherical shape, hence it is called s orbital. 1 2
Y Y 10 11 3 4 1 2 3 8 cos 1 2 sin i e p orbitals are called for Y 10, Y 11, and Y 1-1 since spectrum lines resulted from these orbitals are the most intense (principle)
Y Y Y 20 21 22 5 16 15 8 1 2 2 1 2 15 32 3cos 1 cossine 1 2 2 2i sin e i d orbitals are called for Y 20, Y 21, Y 2-1, Y 22, and Y 2-2 since spectrum lines resulted from these orbitals are rather diffuse.
The levels of m can be determined only when atom is put under magnetic field otherwise if they hold the same n and l, they all have the same energy (degenerate). Y 10 orients in z direction and is called p z however, Y 11, Y 1-1 are not p x and p y i i Y11 Y1 1 sin e e sincos Y p x i i Y11 Y1 1 sin e e sinsin Similary for d orbitals, linear combinations of Y 2+m d 2 z, d 2 x -y2, d xy, d xz, d yz. Y p y give
Probability Distribution Functions We use the wave functions to calculate the probability distributions of the electrons. The position of the electron is spread over space and is not well defined. We may use the radial wave function R(r) to calculate radial probability distributions of the electron. The probability of finding the electron in a differential volume element dt is: Dr. Sabry El-Taher 32
Probability Distribution Functions The infinitesimal volume element in spherical polar coordinates corresponding to dxdydz in Cartesian coordinates is: Therefore, At the moment, we re only interested in the radial dependence. The radial probability density is P(r) = r 2 R(r) 2 and it depends only on n and l. Dr. Sabry El-Taher 33
Size of orbitals can be determined from the probability of finding electron in the distance from nucleus or radial distribution function Probability 2 dv 2 dxdydz 2 2 r sin drd d 2 2 2 r dr sind d 2 2 2 r dr cos 0 2 2 r dr 4 0 4r 2 2 dr Probability P() r dr P(r) radial distribution function = r 2 R ne (r) 2
R(r) and P(r) for the lowest-lying states of the hydrogen atom. Dr. Sabry El-Taher 35
Probability distribution Charge distribution Dr. Sabry El-Taher 36
s orbitals - = 0 1s no nodes 2s 1 node 3s 2 nodes n - - 1 radial nodes, Angular nodes, n 1 total nodes Dr. Sabry El-Taher 37 Copyright Michael D. Fayer, 2007
p-orbitals presentation d-orbitals presentation Dr. Sabry El-Taher 38
Quantization of Orbital Angular Momentum It can be shown that the finiteness of the wave function on the z-axis requires that the orbital angular momentum be quantized: where is called the orbital angular-momentum quantum number or orbital quantum number. Also, it can be shown that because of angular periodicity of the wave function, the z-component of the orbital angular momentum must be quantized as well: where is called the orbital magnetic quantum number or magnetic quantum number. Dr. Sabry El-Taher 39
Orbital Angular Momentum Quantum Number l It s associated with the R(r) and f(θ) parts of the wave function. Energy levels are degenerate with respect to l (the energy is independent of l). l = 0 1 2 3 4 5.. Letter = s p d f g h... The letters s, p, d, and f are the first letters of sharp, principal, diffuse", and fundamental, respectively, used in the early days of spectroscopy. Atomic states are usually referred to by their values of n and l. A state with n = 2 and l = 1 is called a 2p state. Another widely used notation is n 1 2 3 4 Shell K L M N Dr. Sabry El-Taher 40
Magnetic Quantum Number m l The solution for g() specifies that m l is an integer and is related to the z component of L: Example: l = 2: (see the Figure) Only certain orientations of possible. This is called space quantization. are And (except when l = 0) we just don t know L x and L y! Dr. Sabry El-Taher 41
Dr. Sabry El-Taher 42
s orbitals The surfaces include 90% of the total density. The density of shading represents the probability density. The white circles, one in the 2s and two in the 3s show the nodes where the densities are zero. Dr. Sabry El-Taher 43
p-orbitals Note the nodal plane and the opposite signs of the wavefunction on either side, so unlike s-electron, p-electrons are never found at the nucleus. The p orbitals are labelled according to the axes along which the lobes are directed. Dr. Sabry El-Taher 44
d-orbitals Dark orange indicates +ve and light orange -ve. Note that the upper pair have lobes directed along the axes whereas in the lower three they are directed between the axes. This has fundamental importance in transition metal complexes. Dr. Sabry El-Taher 45
f-orbitals Properties of the lanthanides (rare earths) determined by the f orbitals. Dr. Sabry El-Taher 46
Magnetic Effects on Atomic Spectra: The Zeeman Effect In 1896, the Dutch physicist Pieter Zeeman showed that spectral lines emitted by atoms in a magnetic field split into multiple energy levels. It is called the Zeeman effect. Nucleus Consider the atom to behave like a small magnet. Think of an electron as an orbiting circular current loop of I = dq / dt around the nucleus. If the period is T = 2 r / v, then I = -e/t = -e/(2 r / v) = -e v /(2 r). The current loop has a magnetic moment: = IA = [-e v /(2 r)] r 2 = [-e/2m] mrv e L 2m Where A is the area of the circle and L = mvr is the magnitude of the orbital angular momentum. Dr. Sabry El-Taher 47
The Zeeman Effect e L 2m The potential energy due to the magnetic field is: If the magnetic field is in the z-direction, we only care about the z-component of : e e L ( m ) m 2m 2m z z l B l where B = eħ / 2m is called the: Bohr magneton. Dr. Sabry El-Taher 48
The Zeeman Effect A magnetic field splits the m l levels. The potential energy is quantized and now also depends on the magnetic quantum number m l. When a magnetic field is applied, the 2p level of atomic hydrogen is split into three different energy states with energy difference of ΔE = B B Δm l. m l Energy 1 E 0 + μ B B 0 E 0 1 E 0 μ B B Dr. Sabry El-Taher 49
The Zeeman Effect The transition from 2p to 1s, splits by a magnetic field. Dr. Sabry El-Taher 50
The Zeeman Effect An atomic beam of particles in the l = 1 state pass through a magnetic field along the z direction. Bm ( db / dz) The m l = +1 state will be deflected down, the m l = 1 state up, and the m l = 0 state will be undeflected. Dr. Sabry El-Taher 51
Spectral transition and selection rules For hydrogen, the energy level depends on the principal quantum number n. An electron can make a transition from a state of any n value to any other. But what about l and m l quantum numbers? The change in electronic state results in the emission or absorption of a photon. When electron changes state, there will be a change in angular momentum and in turns photon is released. Not all transition between any two electronic states can occur.
Angular momentum must be conserved for the transition to be possible called allowed transition, else called forbidden transition. :Allowed transitions: Electrons absorbing or emitting photons can change states when Dl = ±1 and Dm l = 0, ±1 n is unrestricted because it doesn t relate to angular momentum. For this conservation to occur: An electron in a d orbital (l = 2) cannot make a transition into an s orbital (l = 0) because the photon cannot carry away enough angular momentum. An s electron cannot make a transition to another s orbital because no angular momentum is produced for the photon. Forbidden transitions: Other transitions are possible but occur with smaller probabilities. This is the selection rule for the electronic transition in atom.
Electron Spin Pauli postulated in 1925 that an electron can exist in two distinct states and introduced in a rather ad hoc manner a fourth quantum number to describe the two states. Although with this he could explain the Stern-Gerlach experiment, no interpretation was given to the fourth quantum number. Before long, two Dutch graduate students, Uhlenbeck and Goudsmit, proposed that the electron might behave like a spinning sphere of charge instead of a point particle and the spinning motion would give an additional spin angular momentum S and spin magnetic moment s. According to this proposal, the spin angular momentum, like the quantized orbital angular momentum, is also quantized: where the spin quantum number s = ½, and what the Stern-Gerlach experiment measures is the z-component of the spin angular momentum where the spin magnetic quantum number m s, the fourth quantum number introduced by Pauli, has two values +½ and -½, corresponding to the two orientations, up and down, respectively, of the spin angular momentum. Dr. Sabry El-Taher 54
Intrinsic Spin In 1925, Samuel Goudsmit and George Uhlenbeck, in Holland proposed that: S the electron must have an intrinsic angular momentum and therefore a magnetic moment. In order to explain experimental data, Goudsmit and Uhlenbeck proposed that: the electron must have an intrinsic spin quantum number s = ½. m s = +½ called α spin or spin up m s = -½ called β spin or spin down Photon also has spin with m s = 1 Particles with half-integer m s are called fermion Particles with full-integer m s are called boson Dr. Sabry El-Taher 55
Intrinsic Spin The spinning electron reacts similarly to the orbiting electron in a magnetic field. S The magnetic spin quantum number m s has only two values: m s = ±½ The electron s spin will be either up or down and can never be spinning with its magnetic moment μ s exactly along the z axis. Dr. Sabry El-Taher 56