Atomic Physics - Pham Tan Thi

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1. Atomic Physics Pham Tan Thi, Ph.D. Department of Biomedical Engineering Faculty of Applied Sciences Ho Chi Minh University of Technology
2. History of Atomic Model
3. History of Atomic Model • Proposed an Atomic Theory which states that all atoms are small, hard, indivisible and indestructible particles made of a single material formed into different shapes and sizes. • Aristotle did not support his atomic theory. Democritus (460 BC - 370 BC)
4. History of Atomic Model • Known as the “Father of Modern Chemistry” • Was the first person to generate a list of 23 elements in his textbook • Devised the metric systems • Was married to a 13-year-old Marie-Anne Pierette Paulze, who assisted him much of his work • Discovered/proposed that combustion occurs when oxygen combines with other elements Antoine Lavoisier (1743 - 1794) • Discovered/proposed the Law of Conservation of Mass (or Matter) which states that, in a chemical reaction, matter is neither created or destroyed
5. History of Atomic Model • In 1803, he proposed an Atomic Theory which states that: • All substances are made of atoms; atoms are small particles that cannot be created, divided or destroyed. • Atoms of the same element are exactly alike, and atoms of different elements are different • Atoms join with other atoms to make new John Dalton substances (1766 - 1844) • He calculated the atomic weights of many various elements • He was a teacher at a very young age • He was color blind
6. History of Atomic Model J. J. Thomson (1856 - 1940) • He proved that an atom can be divided into smaller parts • While experimenting with cathode-ray tubes, discovered corpuscles, which were later called electrons • He stated that the atom is neutral
7. History of Atomic Model J. J. Thomson (1856 - 1940) • He proved that an atom can be divided into smaller parts • While experimenting with cathode-ray tubes, discovered corpuscles, which were later called electrons • He stated that the atom is neutral • In 1897, he proposed Plum Pudding Model which states that atoms mostly consist of positively charged material with negatively charged particles (electrons) located throughout the positive material • He won the Nobel Prize
8. History of Atomic Model Ernest Rutherford (1871 - 1937) • In 1909, he performed Gold Foil Experiment and suggested the characteristics of the atom: • It consists of a small core, or nucleus, that contains most of the mass of the atom • This nucleus is made up of particles called protons, which have a positive charge • The protons are surrounded by negative charged electrons, but most of atom is actually empty space • He did extensive work on radioactivity (alpha, beta particles, gamma rays/waves) and was referred as the “Father of Nuclear Physics” • He won the Nobel Prize • He was a student of J.J. Thomson
9. History of Atomic Model Niels Bohr (1885 - 1962) • In 1913, he proposed the Bohr Model, which suggests that electrons travel around the nucleus of an atom in orbits for definite paths. Additionally, the electron can jump from a path in one level to a path in another level (depending on their energy) • He won the Nobel Prize • He used to work with Ernest Rutherford
10. History of Atomic Model Erwin Schrodinger (1887 - 1961) • In 1913, he further explained the nature of electrons in an atom by stating that the exact location of an electron cannot be stated; therefore it is more accurate to view electrons in region called electron clouds. • Electron clouds are places where the electrons are likely to be found • He did extensive work on the Wave formula ➔ Schrodinger equation. • He won the Nobel Prize
11. History of Atomic Model James Chadwick (1891 - 1974) • In 1932, he realized that the atomic mass of most elements was double the number of protons —> discovery of the neutron • He used to work with Ernest Rutherford • He won the Nobel Prize
12. Quantum Mechanical Atomic Theory
13. Schrửdinger Equation in Three Dimensions • 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 Schrửdinger equation for the particle’s wave function ψ(x,y,z) is ~2 @2 (x, y, z) @2 (x, y, z) @2 (x, y, z) + + 2m @x2 @y2 @z2 ✓ ◆ +U(x, y, z) (x, y, z)=E (x, y, z) ~ 2 + U(x, y, z) (x, y, z)=E (x, y, z) 2mr  • This is a direct extension of the one-dimensional Schrửdinger equation ~2 @2 (x) + U(x) (x)=E (x) 2m @x2
14. 1366 CHAPTER 41 Atomic Structure both sides by the factor e-iEt U , leaving the time-independent Schrửdinger equa- tion in three dimensions for a> stationary state: 2 2 2 2 U 0 c x, y, z 0 c x, y, z 0 c x, y, z - + + + U x, y, z c x, y, z 2m 2 2 2 10x 2 10y 2 10z 2 a b 1 2 1 2 = Ec x, y, z (three-dimensional time-independent (41.5) Schrửdinger equation) 1 2 The probability distribution function for a stationary state is just the square iEt 2 of the absolute value of the spatial wave function: ƒc x, y, z e- U ƒ = * iEt iEt 2 c x, y, z e+ Uc x, y, z e- U = ƒc x, y, z ƒ . Note that this doesn’t depend> on time. (As we discussed> in Section> 40.1, that’s why we call these1 states 2stationary.) Hence1 for2 a stationary1 state2 the wave function1 2normalization condition, Eq. (41.3), becomes 2 (normalization condition for a ƒc x, y, z ƒ dV = 1 (41.6) L stationary state in three dimensions) 1 2 We won’t pretend that we have derived Eqs. (41.2) and (41.5). Like their one- dimensional versions, these equations have to be tested by comparison of their predictions with experimental results. Happily, Eqs. (41.2) and (41.5) both pass this test with flying colors, so we are confident that they are the correct equations. An important topic that we will address in this chapter is the solutions for Eq. (41.5) for the stationary states of the hydrogen atom. The potential-energy function for an electron in a hydrogen atom is spherically symmetric; it depends 2 2 2 1 2 only on the distance r = x + y + z from the origin of coordinates. To take advantage of this symmetry, it’s best to> use spherical coordinates rather than the Cartesian coordinates 1x, y, z to solve2 the Schrửdinger equation for the hydro- gen atom. Before introducing these new coordinates and investigating the hydro- gen atom, it’s useful to look1 at the2 three-dimensional version of the particle in a box that we considered in Section 40.2. Solving this simpler problem will give us insight into the more complicated stationary states found in atomic physics. Test Your Understanding of Section 41.1 In a certain region of space the potential-energy function for a quantum-mechanical particle is zero. In this region the 2 2 2 2 wave function c x, y, z for a certain stationary state satisfies 0 c 0x 7 0 , 0 c 0y 7 0 , 2 2 and 0 c 0z 7 0 . The particle has a definite energy E that is positive. What can you con- clude about c x,1y, z in2 this region? (i) It must be positive; (ii) it> must be negative;> (iii) it must be >zero; (iv) not enough information given to decide. ❙ 1 2 41.1 A particle is confined in a cubical 41.2 Particle in a Three-Dimensional Box box with walls at x 0 , x L , y 0 , = = = Consider a particle enclosed within a cubical box of side L. This could represent y = L, z = 0 , and zParticle= L . in a Three-Dimensional Box an electron that’s free to move anywhere within the interior of a solid metal cube z but cannot escape the cube. We’ll choose the origin to be at one corner of the box, z 5 L with• theIn xanalogy-, y-, and z-axes with along ouredges ofinfinite the box. Then square the particle potential is confined to the region(U( x0 ) =x 0 Linside, , 0 y UL , (0x ) z= ∞L (Fig.outside), 41.1). What let are theus station- ary states of this system? Asconsider for the particle a in three-dimensional a one-dimensional box that we region considered space in Section 40.2, (box)we’ll say ofthat equalthe potential sides energy isof zero length inside the L box, with but infinite the outside. same Hence the spatial wave function c x, y, z must be zero outside the box in order x 5 L that thepotential term U x, y, (zUc(xx,y,z, y, z in) =the 0time-independent inside, U(x,y,z) Schrửdinger = equation, ∞ x 2 O y 5 L Eq. (41.5),outside not be infinite.). Hence the1 probability2 distribution function ƒc x, y, z ƒ y is zero outside the1 box,2 and1 there2 is zero probability that the particle will be found 1 2 • 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: (x, y, z)=X(x)Y (y)Z(z) • The Schrửdinger equation inside the box becomes ~2 @2X(x) @2Y (y) @2Z(z) Y (y)Z(z) + Z(z)X(x) + X(x)Y (y) = EX(x)Y (y)Z(z) 2m @x2 @y2 @z2  • OR dividing by X(x)Y(y)Z(z), we have: ~2 1 @2X(x) 1 @2Y (y) 1 @2Z(z) + + = E 2m X(x) @x2 Y (y) @y2 Z(z) @z2 
15. Particle in a Three-Dimensional Box • This says that the total energy is contributed 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: 2 2 ~2 @2X(x) ~ @ Y (y) ~2 @2Z(z) = E X(x) = EyY (y) = E Z(z) 2m @x2 x 2m @y2 2m @z2 z • These obviously have the same solutions separately as our original particle in an infinite square well, and corresponding energies 2 2 2 nx⇡x nx⇡ h Xnx = Cxsin E = (nx = 1,2,3, ) L x 2mL2 ⇣ ⌘ 2 2 2 ny⇡y ny⇡ h Xny = Cysin E = (ny = 1,2,3, ) L y 2mL2 ⇣ ⌘ 2 2 2 nz⇡z nz⇡ h Xnz = Czsin E = (nz = 1,2,3, ) L z 2mL2 ⇣ ⌘
16. Particle in a Three-Dimensional Box • A particle’s wave function is the product of these three solutions: n ⇡x n ⇡y n ⇡z (x, y, z)=X(x)Y (y)Z(z)=Csin x sin y sin z L L L ⇣ ⌘ ⇣ ⌘ ⇣ ⌘ • We can use the three quantum numbers nx, ny and nz to label the stationary states (state 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, nz, they have the same total energy E: 41.2 Particle in a Three-Dimensional Box 1369 2 2 2 2 2 2 2 2 4⇡2 h ⇡ h ⇡ h 3⇡ h 41.2 Probability distribution function ƒ cn ,n ,n x, y, z ƒ for n X, n Y, n Z equal to (a) (2, 1, 1), (b) (1, 2, 1), and (c) (1, 1, 2). The value E2 = E + E + EX Y Z= + + = of ƒ c ƒ is proportionalx to the densityy of dots. Thez wave function is2 zero on the walls of2 the box and on the midplane2 of the box,2 so 2 ƒ c ƒ = 0 at these locations. 1 22 mL1 2 2mL 2mL mL 2 2 z 2 z (a) ͉c2, 1, 1͉ z (b) ͉c1, 2, 1͉ (c) ͉c1, 1, 2͉ x x x y y y spots” where there is zero probability of finding the particle. As an example, con- sider the case n X, n Y, n Z = 2, 1, 1 . From Eq. (41.15), the probability distri- bution function for this case is 1 2 1 2 2 2 2 2px 2 py 2 pz c2,1,1 x, y, z = C sin sin sin ƒ ƒ ƒ ƒ L L L As Fig. 41.2a shows, 1the probability2 distribution function is zero on the plane 2 2 x = L 2, where sin 2px L = sin p = 0 . The particle is most likely to be found near where all three of the sine-squared functions are greatest, at x, y, z> = L 4, L 2,1L 2 > or2 x, y, z = 3L 4, L 2, L 2 . Figures 41.2b and 41.2c show the similar cases n X, n Y, n Z = 1, 2, 1 and n X, n Y, n Z = 1 2 1 > > > 2 1 2 1 > > > 2 1, 1, 2 . For higher values of the quantum numbers n X , n Y , and n Z there are additional planes on which the probability1 distribution2 1 function2 equals1 zero, 2just 2 as1 the probability2 distribution function ƒc x ƒ for a one-dimensional box has more zeros for higher values of n (see Fig. 40.12). 1 2 Example 41.1 Probability in a three-dimensional box (a) Find the value of the constant C that normalizes the wave func- integral of the probability distribution function c x, y, z 2 ƒ nX, nY, nZ ƒ tion of Eq. (41.15). (b) Find the probability that the particle will be over the volume within the box equals 1. (The integral is actually found somewhere in the region 0 x L 4 (Fig. 41.3) for the over all space, but the particle-in-a-box wave functions1 are zero2 cases (i) (n X, n Y, n Z = 1, 2, 1 , (ii) (n X, n Y, n Z = 2, 1, 1 , outside the box.) > and (iii) n X, n Y, n Z) = 3, 1, 1 . The probability of finding the particle within a certain volume 2 1 2 2 1 2 within the box equals the integral of the probability distribution 1 1 2 SOLUTION function over that volume. Hence in part (b) we’ll integrate c x, y, z 2 for the given values of n , n , n over the IDENTIFY and SET UP: Equation (41.6) tells us that to normalize ƒ nX, nY, nZ ƒ X Y Z volume 0 x L 4 , 0 y L , 0 z L . the wave function, we have to choose the value of C so that the 1 2 1 2 EXECUTE: (a) From >Eq. (41.15), 2 2 2 n Xpx 2 n Ypy 2 n Zpz cn ,n ,n x, y, z = C sin sin sin 41.3 What is the probability that the particle is in the dark- ƒ X Y Z ƒ ƒ ƒ L L L colored quarter of the box? Hence the normalization1 2 condition is z z 5 L c x, y, z 2 dV ƒ nX,nY,nZ ƒ L x1=L y=2L z=L 2 2 n Xpx 2 n Ypy 2 n Zpz = ƒCƒ sin sin sin dxdydz Lx=0 Ly=0 Lz=0 L L L x=L y=L 2 2 n Xpx 2 n Ypy = ƒCƒ sin dx sin dy Lx 0 L Ly 0 L x 5 L = = x O a z=L ba b y 5 L n Zpz sin2 dz 1 y * = x 5 L/4 Lz=0 L a b Continued
17. 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. 41.2 Particle in a Three-Dimensional Box 1371 • Here are the first six states for an equal-side box: E 41.4 Energy-level diagram for a particle 2 2 6-fold degenerate in a three-dimensional cubical box. We 3⇡ h (3, 2, 1), (3, 1, 2), (1, 3, 2), 14 E label each level with the quantum numbers E = (2, 3, 1), (1, 2, 3), (2, 1, 3) 3 1, 1, 1 1,1,1 2 of the states nX, nY, nZ with that energy. 2mL not degenerate Several of the levels are degenerate (more (2, 2, 2) E 4 1, 1, 1 1 2 3-fold degenerate 11 than one state has the same energy). The (3, 1, 1), (1, 3, 1), (1, 1, 3) E 3 1, 1, 1 lowest (ground) level, nX, nY, nZ = 2 1, 1, 1 , has energy E1,1,1 = 1 + 3-fold degenerate 2 2 2 2 2 1 2 2 2 2 (2, 2, 1), (2, 1, 2), (1, 2, 2) 3E1, 1, 1 1 + 1 p U 2mL = 3p U 2mL ; we show1 the2 energies of the other 1levels as • If the length of sides of multiples2 of E>1,1,1 . > 3-fold degenerate box are different: (2, 1, 1), (1, 2, 1), (1, 1, 2) 2E1, 1, 1 2 2 2 2 2 nx ny nz ⇡ h not degenerate (1, 1, 1) E E = 2 + 2 + 2 1, 1, 1 Lx Ly Lz ! 2m (break the degeneracy) E 5 0 Since degeneracy is a consequence of symmetry, we can remove the degener- acy by making the box asymmetric. We do this by giving the three sides of the box different lengths LX , LY , and LZ . If we repeat the steps that we followed to solve the time-independent Schrửdinger equation, we find that the energy levels are given by 2 2 2 2 2 nX nY nZ p U E n 1, 2, 3, ; n 1, 2, 3, ; nX,nY,nZ = 2 + 2 + 2 X = Á Y = Á LX LY LZ 2m a b 1nZ = 1, 2, 3, Á (41.17) (energy levels, particle in a three-dimensional box with2 sides of length LX , LY , and LZ) If LX , LY , and LZ are all different, the states nX, nY, nZ = 2, 1, 1 , 1, 2, 1 , and 1, 1, 2 have different energies and hence are no longer degenerate. Note 1 2 1 2 1 2 that Eq. (41.17) reduces to Eq. (41.16) if the lengths are all the same (LX = 1 2 LY = LZ = L). Let’s summarize the key differences between the three-dimensional particle in a box and the one-dimensional case that we examined in Section 40.2: • We can write the wave function for a three-dimensional stationary state as a product of three functions, one for each spatial coordinate. Only a single function of the coordinate x is needed in one dimension. • In the three-dimensional case, three quantum numbers are needed to describe each stationary state. Only one quantum number is needed in the one-dimensional case. • Most of the energy levels for the three-dimensional case are degenerate: More than one stationary state has this energy. There is no degeneracy in the one-dimensional case. • For a stationary state of the three-dimensional case, there are surfaces on 2 which the probability distribution function ƒcƒ is zero. In the one- 2 dimensional case there are positions on the x-axis where ƒcƒ is zero. We’ll see these same features in the following section as we examine a three- dimensional situation that’s more realistic than a particle in a box: a hydrogen atom in which a negatively charged electron orbits a positively charged nucleus. Test Your Understanding of Section 41.2 Rank the following states of a particle in a cubical box of side L in order from highest to lowest energy: (i) nX, nY, nZ = 2, 3, 2 ; (ii) nX, nY, nZ = 4, 1, 1 ; (iii) nX, nY, nZ = 2, 2, 3 ; (iv) nX, nY, nZ = 1, 3, 3 . ❙ 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2
18. 1372 CHAPTER 41 Atomic Structure 41.3 The Hydrogen Atom Let’s continue the discussion of the hydrogen atom that we began in Chapter 39. In the Bohr model, electrons move in circular orbits like Newtonian particles, but with quantized values of angular momentum. While this model gave the correct energy levels of the hydrogen atom, as deduced from spectra, it had many con- ceptual difficulties. It mixed classical physics with new and seemingly contradic- tory concepts. It provided no insight into the process by which photons are emitted and absorbed. It could not be generalized to atoms with more than one electron. It predicted the wrong magnetic properties for the hydrogen atom. And perhaps most important, its picture of the electron as a localized point particle was inconsistent with the more general view we developed in Chapters 39 and 40. To go beyond the Bohr model, let’s apply the Schrửdinger equation to find the wave functions for stationary states (states of definite energy) of the hydrogen atom. As in Section 39.3, we include the motion of the nucleus by simply replac- ing the electron mass m with the reduced mass m r. The Schrửdinger Equation for the Hydrogen Atom We discussed the three-dimensional version of the Schrửdinger equation in Sec- tion 41.1. The potential-energy function is spherically symmetric: It depends 2 2 2 1 2 only on the distance r = x + y + z from the origin of coordinates: > 2 1 2 1 e U r =- (41.18) 4pP0 r 41.5 The Schrửdinger equation for the The hydrogen-atom problem 1 is 2 best formulated in spherical coordinates Hydrogen Atomhydrogen atom can be solved most readily r, u, f , shown in Fig. 41.5; the spherically symmetric potential-energy func- using spherical coordinates. tion depends only on r, not on u or f . The Schrửdinger equation with this z poten1 tial-energy2 function can be solved exactly; the solutions are combinations • For hydrogen atom, the three-dimensional potential Electron, charge 2e, energy depends only on the electron’s distance from at coordinates (r, u, f) of familiar functions. Without going into a lot of detail, we can describe the most important features of the procedure and the results. the proton: 1 e2 U(r)= First, we find the solutions using the same method of separation of variables that we employed for a particle in a cubical box in Section 41.2. We express the 4⇡"0 r wave function c r, u, f as a product of three functions, each one a function of only one of the three coordinates: • As before, we will seek separable wave functions, 1 2 but this time, due to the spherical symmetry we Nucleus, r c r, u, f = R r ™ u Ê f (41.19) will use spherical coordinates (r, θ, �) rather than charge 1e, u at the origin That is, the function R r depends1 only2 on1 r2, ™1 u2 depends1 2 only on u, and Ê f (x,y,z). Then: y depends only on f. Just as for a particle in a three-dimensional box, when we substitute Eq. (41.19)1 into2 the Schrửdinger equation,1 2 we get three separate ordi-1 2 (r, ✓, )=R(r)⇥(✓)() f nary differential equations. One equation involves only r and R r , a second involves only u and ™ u , and a third involves only f and Ê f : • Following the same procedure as before, we can 1 2 x 2 2 U d dR1 r2 U l l + 1 1 2 separate the problem into three separate 2 (41.20a) - 2 r + 2 + U r R r = ER r equations • Energies are 2m rr dr dr 2m rr 4 1 2 1 2 1 me a b a 21 2b 1 2 1 2 2 2 En = 1 d d™ u) m l ~ d 2 dR(r) ~ l(l + 1) (4⇡" )2 2n2h2 sinu + l l + 1 - ™ u = 0 (41.20b) – r + + U(r) R(r)=ER(r) 0 sin u du du sin 2u 2m dr dr 2m r2 1 ✓ ◆ ✓ ◆ a b a 1 d2 2 f b 1 2 2 13.6 eV Ê 2 1 d d⇥(✓) ml En = + m l Ê f = 0 (41.20c) sin✓ + l(l + 1) ⇥(✓)=E⇥(✓) n2 df2 sin✓ d✓ d✓ sin2✓ 1 2 ✓ ◆ ✓ ◆ In Eqs. (41.20) E is the energy of the stationary state and1 2l and m are constants d2() Quantum numbers: l 2 that we’ll discuss later. (Be careful! Don’t confuse the constant m l with the 2 + ml ()=E() n, l, ml d reduced mass m r. ) We won’t attempt to solve this set of three equations, but we can describe how it’s done. As for the particle in a cubical box, the physically acceptable solutions of these three equations are determined by boundary conditions. The radial func- tion R r in Eq. (41.20a) must approach zero at large r, because we are describing 1 2
19. Hydrogen Atom: Quantum States • Table 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. • Question: How many distinct states of the hydrogen atom (n, l, ml) are there for n = 3 state? What are their energies?
20. Hydrogen Atom: Quantum States • Table 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. • Question: How many distinct states of the hydrogen atom (n, l, ml) are there for 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. Each of these states have the same n, so they all have the same energy. We will see later there is another quantum number s, for electron spin (±1/2), so there actually 18 possible states for n = 3.
21. Spectral Lines of the Hydrogen Atom E 13.6 E = 1 = [eV] n n2 n2 13.6 13.6 " = E E = m n m2 n2 ✓ ◆ ✓ ◆ Lyman series: transitions from m = 2,3, ∞ to n = 1: Ultraviolet series Balmer series: transitions from m = 3,4, ∞ to n = 2: Visible series Paschen: transitions from m = 4,5 ∞ to n = 3: Infrared series Brackett: transitions from m = 5,6 ∞ to n = 4: Infrared series 2 Pfund: transitions from m = 6,7 ∞ to n = 5: Infrared series 3
22. Homework (1) Calculate wavelengths of Lyman, Balmer, Paschen, Brackett, Pfund series from n = 7 to lower levels? (2) Of which can we see by human eyes?
23. Spectral Lines of the Hydrogen Atom
24. Hydrogen Atom: Degeneracy • States with different quantum numbers l and n are often referred to with letters as follows: l value letter n value shell 0 s 1 K 1 p 2 L 2 d 3 M 3 f 4 N 4 g 5 h • Hydrogen atom states with the same value of n but different values of l and ml are degenerate (i.e. have the same energy). • Figures at the right show radial probability distribution for states with l = 0,1,2,3 and different values of n = 1,2,3,4.
25. Hydrogen Atom: Probability Distribution • States of the hydrogen atom with l = 0 (zero orbital angular momentum) have spherically symmetric wave functions that depend on r but not on θ or �. These are called s states. Figures below show the electron probability distributions for three of these states.
26. Hydrogen Atom: Probability Distribution • 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 not spherically symmetric. Figures below show the electron probability distributions for three of these states as well as for two spherically symmetric s states.
27. + Orbital Angular Momentum • Motion of electron around the nucleus: orbital motion • Intrinsic motion of electron: spin motion (rotation around its axis) Orbital Angular Momentum L~ 1. The vector L~ does not point in one specific direction 2. The magnitude of orbital angular momentum is quantized: L = ~ l(l + 1) p 3. The projection of vector L~ on a direction is quantized: Lz = ml~ ml =0, 1, 2, , l ± ± ±
28. 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: L = ~ l(l + 1) (l = 0, 1, 2, n-1) where l is thep orbital quantum number. • Recall that the Bohr model of the Hydrogen atom also had quantized angels moment L = nℏ, but the lowest energy state n = 1 would have L = ℏ. In contrast, the Schrodinger equation shows that the lowest state has L = 0. The wave function of this energy state is a perfectly symmetric sphere. For higher energy states, the vector L has only certain allowed directions, such that the z-component is quantized as Lz = ml~ (ml = 0, ±1, ±2, ±l)
29. Orbital Angular Momentum and Quantum Numbers
30. Magnetic Moments and Zeeman Effect • Electron states with nonzero orbital angular momentum (l = 1,2,3, ) have a magnetic dipole moment due to electron motion. Hence these states are affected if the atom is placed in a magnetic field. The result, called Zeeman effect, is a shift in the energy of states with nonzero ml. • The potential energy associated with a magnetic dipole moment ml in a magnetic field of strength B is U = mlàBB, and the magnetic dipole moment due to the orbital angular momentum of the electron is in units of the Bohr magneton eh àB = U = mlàBB 2me
31. Zeeman Effect and Selection Rules • The figure shows the shift in energy for the five l = 2 states (each with a different value of ml) as the magnetic field strength increases • An atom in a magnetic field can make transitions between different states by emitting or absorbing a photon ∆E transition is allowed if ∆l = 1 and ∆ml = 0, ±1. A transition is forbidden if it violates these selection rules.
32. Electron Spin and Stern-Gerlach Experiment • The experiment of Stern and Gerlach demonstrated the existence of electron spin. The z-component of the spin angular momentum has only two possible values, corresponding to ms = +1/2 and ms = -1/2
33. Spin Angular Momentum • Electrons have an intrinsic angular momentum called “spin” ~ p3 S =+ S = ~ 1 z 2 2 S~ = S = s(s + 1)~ with s = | | 2 p 1 1 Sz = ms~ with ms = , + 2 2 ~ S = z 2 • Spin also generates magnetic dipole moment e ~ B~ à~S = gS with g 2 2me ⇡ ge~ U = àSBcos✓ = Bms 1 2me ms =+ ~ 2 B>0 1 m = s 2
34. Anomalous Zeeman Effect and Electron Spin • For certain atoms the Zeeman effect does not follow the simple pattern that we have described in Figure below. This is because an electron also has an intrinsic angular momentum, called spin angular momentum. 1 S = ~ s(s + 1) Sz = mz~ (mz = ) ±2 p − UmB= 2.00232sBà (e spin magnetic interaction energy)
35. Summary of Quantum Numbers
36. Pauli Exclusive Principle • Pauli exclusive principle states that no two electrons can occupy the same quantum-mechanical state in a given system. That is, not two electrons in an atom can have the same values of all four quantum numbers: n, l, ml, ms • For a given principal quantum number, n, there are 2n2 quantum states n l m ms Maximum Maximum number of number of electrons in electrons in the subshell the shell n=1 l=0 m=0 ms=±1/2 2 2 n=2 l=0 m=0 ms=±1/2 2 8 l=1 m=0, ±1 ms=±1/2 6 n=3 l=0 m=0 ms=±1/2 2 l=1 m=0, ±1 ms=±1/2 6 18 l=2 m=0, ±1, ±2 ms=±1/2 10
37. Periodic Table