Physics A2 - Lecture 11 Continue: Building Atoms and Molecules - Huynh Quang Linh

Nội dung text: Physics A2 - Lecture 11 Continue: Building Atoms and Molecules - Huynh Quang Linh

1. Lecture 11_continue: Building Atoms and Molecules +e +e r d yeven Plane of hydrogen atoms yodd
2. Content Atomic Configurations Building atoms with the Pauli exclusion principle Selection rules  Molecular Wavefunctions  Example: H + H H2
3. Pauli Exclusion Principle We now want to start building more complicated atoms to study the Periodic Table. For atoms with many electrons (e.g., carbon: 6, iron: 26, etc.) - what energies do they have? From spectra of complex atoms, Wolfgang Pauli (1925) deduced a new rule: “Pauli Exclusion Principle” “In a given atom, no two electrons* can be in the same quantum state, i.e. they cannot have the same set of quantum numbers n, l, ml , ms” I.e., every “atomic orbital with n,l,ml” can hold 2 electrons: ()  Therefore, electrons do not pile up in the lowest energy state, i.e, the (1,0,0) orbital.  They are distributed among the higher energy levels according to the Pauli Principle.  Particles that obey the Pauli Principle are called “fermions” *Note: More generally, no two identical fermions (any particle with spin of ħ/2, 3ħ/2, etc.) can be in the same quantum state.
4. Filling the atomic orbitals according to the Pauli Principle s p d f g 13.6 eV n Energy E Z 2 l = 0 1 2 3 4 n 2 n is valid only for one electron in 4 the Coulomb potential of Z 3 protons. The energy levels shift as more electrons are added, due to electron-electron 2 interactions. Nevertheless, this hydrogenic diagram helps us keep track of where the added electrons go. Example: Na Z = 11 l label #orbitals (2l+1) 1s22s22p63s1 0 s 1 1 p 3 2 d 5 1 3 f 7 Z = atomic number = number of protons
5. exercise 1: Pauli Exclusion Principle 1. Which of the following states (n,l,ml,ms) is/are NOT allowed? (a). (2, 1, 1, -1/2) (b). (4, 0, 0, 1/2) (c). (3, 2, 3, -1/2) (d). (5, 2, 2, 1/2) (e). (4, 4, 2, -1/2) 2. Which of the following atomic electron configurations violates the Pauli Exclusion Principle? (a). 1s2, 2s2, 2p6, 3d10 (b). 1s2, 2s2, 2p6, 3d4 (c). 1s2, 2s2, 2p8, 3d8 (d). 1s1, 2s2, 2p6, 3d5 (e). 1s2, 2s2, 2p3, 3d11
6. exercise 1: Pauli Exclusion Principle 1. Which of the following states (n,l,ml,ms) is/are NOT allowed? (a). (2, 1, 1, -1/2) m = -l, -(l -1), (l-1), l (b). (4, 0, 0, 1/2) l (c). (3, 2, 3, -1/2) (d). (5, 2, 2, 1/2) n > l (e). (4, 4, 2, -1/2) 2. Which of the following atomic electron configurations violates the Pauli Exclusion Principle? 2 2 6 10 (a). 1s , 2s , 2p , 3d 2(2l +1) = 6 allowed electrons (b). 1s2, 2s2, 2p6, 3d4 (c). 1s2, 2s2, 2p8, 3d8 2(2l +1) = 10 allowed electrons (d). 1s1, 2s2, 2p6, 3d5 (e). 1s2, 2s2, 2p3, 3d11
7. Filling Procedure for Atomic Orbitals example: Bromine Due to electron-electron interactions, the hydrogen levels fail to give us the correct filling order as we go higher in the periodic table. The actual filling order is given in the table below. Electrons are added by proceeding along the arrows shown. Bromine is an element with Z = 35. Find its electronic configuration (e.g. 1s2 2s2 2p6 ).
8. As you learned in chemistry, the various behaviors of all the elements (and all the molecules made up from them) is all due to the way the electrons organize themselves, according to quantum mechanics.
9. Optical Transitions between Atomic Levels Consider the hydrogenic picture: r U(r) DE c f n = 2 DE h  n = 1 hc 1240 eV nm photon  DE DE In the field of a photon, the electron may be considered as being in a superposition of two stationary states. The time-dependent solution of the SEQ shows the wave function oscillating between the two eigenstates. Not all transitions are possible must conserve angular momentum (and photon has ħ!) Superpositions Stationary States: : Forbidden 1s ± 2s No electric- dipole moment transition Dl = 0 Allowed 1s 2s 2p Oscillating 1s ± 2p electric-dipole transition couples to Dl = ±1 www.falstad.com/qmatomrad photons Each photon carries 1ħ of angular momentum
10. Allowed Transitions for H (You observed some of these transitions in Lab 4.) s p d f g n Energy (eV) l = 0 1 2 3 4 0.00 4 -0.85 3 -1.51 Selection Rule for electric-dipole transitions: 2 -3.40 Dl 1 (A few representative transitions are shown.) 13.6 eV En n2 Selection Rule on m: Dm 0, 1 photon is photon is linearly circularly polarized 1 -13.6 eV polarized
11. Example Problem 1 What is the electronic structure of lithium (3 electrons)? That is, what quantum numbers do the electrons have? Solution: The guiding principle is to find the lowest energy. This involves (for atoms without too many electrons) putting the electrons into the smallest possible n state, because energy depends only on n (to a good approximation). Electron #1: As you saw in Act 1, the first two electrons have n = 1. 1 (n, l, ml, ms) = (1,0,0, /2) This forces them to have l = 0 and ml = 0. All electrons 1 1 1 Electron #2: have s = /2, so it is not listed. ms is always + /2 or - /2. 1 (n, l, ml, ms) = (1,0,0,- /2) The first two electrons can have n = 1, but the third must Electron #3: have n = 2. l = 0 has lower energy than l = 1. ms doesn’t 1 (n, l, ml, ms) = (2,0,0, ± /2) affect the energy, so either value is OK. Whenever an atom has a single electron in a higher energy state (n value) than the others, that electron is not tightly bound, and the atom can easily lose it. This kind of atom is chemically very reactive. All of the alkali metals (group IA) have this electronic configuration.
12. Example Problem 2 A hydrogen atom is in the n = 3, l = 2, ml = -2 state. To what states can the electron fall when it emits a photon? Which are the strongest (most likely) transitions? Solution: Final state quantum numbers: Remember the “selection rules”: n = 1, 2 Dn is negative here* (conservation of energy) l = 0, 1 Dl 0 (conservation of angular momentum) ml = any No restriction on Dml. Final states with strongest transition: l = 1 Dl = 1 for strongest transition (dipole allowed). n = 2 l = 1 requires n 2. ml = -1, 0, 1 All Dml are equally allowed. *In a different atom, it may be possible to have a Dn = 0 transition, and still conserve energy (e.g., go from 3s 3p).
13. Exercise 2: Pauli Exclusion Principle – Part 2 The Pauli exclusion principle applies to all fermions in all situations (not just to electrons in atoms). Consider electrons in a 2-dimensional infinite square well potential. 1. How many electrons can be in the first excited state? (a). 1 (b). 2 (c). 3 (d). 4 (e). 5 2. If there are 4 electrons in the well, what is the energy of the most energetic one (ignoring e-e interactions, and assuming the total energy is as low as possible)? (a). (h2/8mL2) x 2 (b). (h2/8mL2) x 5 (c). (h2/8mL2) x 10
14. Exercise 2: Pauli Exclusion Principle – Part 2 The Pauli exclusion principle applies to all fermions in all situations (not just to electrons in atoms). Consider electrons in a 2-dimensional infinite square well potential. 1. How many electrons can be in the first excited state? (a). 1 The ground state has quantum numbers (1, 1), and can hold (b). 2 2 electrons, i.e., (1,1,1/2) & (1,1,-1/2) (c). 3 The first excited state has two degenerate single- (d). 4 electron states: (2,1) and (1,2), each of which can have (e). 5 2 electrons: (2,1,1/2), (2,1,-1/2), (1,2,1/2), (1,2,-1/2) 2. If there are 4 electrons in the well, what is the energy of the most energetic one (ignoring e-e interactions, and assuming the total energy is as low as possible)? (a). (h2/8mL2) x 2 The ground state has 2 electrons. The next two are (b). (h2/8mL2) x 5 in the first excited state, the energy of which is: 2 2 2 2 2 2 (c). (h2/8mL2) x 10 (h /8mL ) x (1 + 2 ) = 5(h /8mL )
15. Bonding between atoms How can two neutral objects bind together? H + H  H2 Continuum of +e r free electron Let’s represent the atom states. in space by its Coulomb n = 3 potential centered on the n = 2 proton (+e): e2 n = 1 U( r ) r +e +e In this picture the r potential energy of the two protons in an e2 e2 H2 molecule look U( r ) something like this: r r1 r r2 The energy levels for this potential are more complicated, so we consider a simpler potential that we already know a lot about.
16. Particle in a box Finite square well potential (analogue to Molecular binding) Continuum of If this were the free electron ‘atomic’ potential, states. Bound states then this would be the ‘molecular’ potential: Again, we don’t know exactly what the energy levels are, but
17. Just consider the ground state: Let’s say that the lowest energy level is about equal to that of an infinite well: 1.505 eV nm2 E 0.4 eV (2L)2 L = 1 nm For convenience, we are going to plot the electronic wavefunction yA with the energy level as a baseline:
18. ‘Molecular’ Wavefunctions and Energies ‘Atomic’ Wavefunctions: L = 1 nm yA 1.505 eV nm2 E 0.4 eV (2L)2 ‘Molecular’ Wavefunctions: 2 ‘atomic’ states 2 ‘molecular’ states yeven yodd When the wells are far apart, ‘atomic’ functions don’t overlap. The single electron can be in either well with equal probability, and E = 0.4 eV.
19. ‘Molecular’ Wavefunctions and Energies d L = 1 nm Wells far apart: y 1.505 eV nm2 even E 0.4 eV (2L)2 2 yodd 1.505 eV nm E 0.4 eV (2L)2 (“Degenerate” states) Wells closer together: d ‘Atomic’ states are beginning 2 yeven to overlap and distort. y even 2 and y odd are not the same (note center point). Energies for these two states are not yodd equal. (The degeneracy is broken.)
20. Exercise 3: Symmetric vs. Antisymmetric states 1. Which state has the d lower energy? yeven (a). yeven (b). yodd yodd 2. What will happen to the energy of yeven as the two wells come together (i.e., as d is reduced)? (a). E increases (b). E decreases (c). E stays the same
21. Exercise 3: Symmetric vs. Antisymmetric states 1. Which state has the d lower energy? yeven (a). yeven (b). yodd The “curvature” argument can y be hard to apply (e.g., it odd doesn’t work inside the barrier!). Better to count the number of zero- crossings. yeven has none, while yodd has one. 2. What will happen to the energy of yeven as the two wells come together (i.e., as d is reduced)? (a). E increases As d becomes very small the curvature in yeven (b). E decreases is reduced, reducing the energy. What (c). E stays the same does this mean for the two “atoms”?
22. ‘Molecular’ Wavefunctions and Energies Wells just touching: 2L = 2 nm (becoming one well) 1.505 eV nm2 y E1 0.1 eV even (4L)2 (n = 1 state) yodd 1.505 eV nm2 E 22 0.4 eV 2 (4L)2 (n = 2 state) The even state has lower kinetic energy (smaller curvature) ! Eodd ≈ 0.4 eV f = 0.4 – 0.1 eV = 0.3 eV f = splitting between even and odd states. Eeven ≈ 0.1 eV d (mainly ground state lowering)
23. Molecular Wavefunctions and Energies back to Coulomb Potential +e r Atomic ground state: e2 (1s) U( r ) r yA r / a n = 1 y( r )  e 0 Molecular states: +e +e +e +e r r yodd yeven Bonding state Antibonding state Remember: These are single-electron states. Next lecture we will see how electrons fill these states (following the Pauli Exclusion Principle)
24. Insights on ‘Molecular’ Wavefunctions The even and odd states are stationary states -– d they don’t change in time. y even Eeven yodd Eodd If we put an electron into the left box, we are 1 y left( r ) (y even( r ) y odd( r )) forcing it into a non-stationary state: 2 What happens next?
25. Exercise 4: ‘Molecular’ Wavefunctions Yleft is a non-stationary state. The electron will oscillate (“tunnel”) back and forth between the wells: y left( r ) y right( r ) 1) If you attempt to measure the energy of the electron at some later time, what energy will you measure? Eeven? Eodd? (Eeven+ Eodd)/2 ? Something else? 2) What is the frequency of oscillation? (Assume Eodd - Eeven = 0.3 eV.)
26. Exercise 4: ‘Molecular’ Wavefunctions Yleft is a non-stationary state. The electron will oscillate (“tunnel”) back and forth between the wells: y left( r ) y right( r ) 1) If you attempt to measure the energy of the electron at some later time, what energy will you measure? Eeven? Eodd? (Eeven+ Eodd)/2 ? Something else? Answer: You will measure either Eeven or Eodd , with equal probability. 2) What is the frequency of oscillation? (Assume Eodd - Eeven = 0.3 eV.) 15 13 Solution: f = (Eodd – Eeven)/h = (0.3 eV) / (4.14 x 10 eV-s) = 7 x 10 Hz
27. Application: “Tunneling” & the NH3 Maser Astronomical N Masers: Plane of hydrogen atoms. U(x) E2 E1 0 x