Atomic Structure Ch , 9.6, 9.7

Transcription

1 Ch , 9.6, 9.7

2 Magnetic moment of an orbiting electron: An electron orbiting a nucleus creates a current loop. A current loop behaves like a magnet with a magnetic moment µ:! µ =! µ B " L Bohr magneton: µ B = e! 2m e The north and south poles of the magnet are aligned along the direction of µ. µ points in a direction opposite to that of the angular momentum vector L.

3 Magnetic moment of an orbiting electron: Since L is quantized, the magnetic moment µ is also quantized: L = l(l + 1)!, l = 0,1,2... L z = m!, m =!l,!l + 1,...l! µ =! µ B " L µ = l(l + 1)µ B, l = 0,1,2... µ z =!mµ B, m =!l,!l + 1,...l The north an south poles of the magnet are aligned along the direction of µ. µ points in a direction opposite to that of the angular momentum vector L.

4 Magnetic moment of an orbiting electron: A magnetic moment µ interacts with a magnetic field. The resulting potential energy is U =!! µ "! B = µ x B x + µ y B y + µ z B z Consider a beam of atoms passing through a spatially varying field in the z direction:! B = B(z)ez Magnetic force: F =! "U "z = # " $ % "z B(z) & ' ( µ z Since µ z is quantized (2l+1 values), the force is quantized. Hence the atomic beam will separate into 2l+1 parts depending on the quantum number m. If l=0, the atomic beam does not split into parts.

5 Stern-Gerlach experiment: A beam of silver atoms is sent through a spatially variable magnetic field in the z-direction. Silver atoms in the ground state have l=0. Hence no splitting of the beam was expected: F =! "U "z = # " $ % "z B(z) & ' ( µ z In the experiment the beam did split into two components! Conclusion: The magnetic moment must have a contribution from something other than the orbital angular momentum: spin

6 Intrinsic magnetic moment of spinning electron Electron acts like a (quantized) bar magnet that can interact with a magnetic field. The intrinsic magnetic moment due to the electron spin S is Total magnetic moment is thus! µ s =! 2µ B " S! µ =! µ o +! µ s =! µ B " L + 2S ( )

7 Electron spin: The magnitude of an electron s spin is fixed at 3! 2. It is an intrinsic property of the electron: S = s(s + 1)! = 3 2!, s = 1 2 S z = m s!, m =!s,!s + 1,...s =! 1 2!, + 1 2! µ s = s(s + 1)µ B = 3 2 µ B, s = 1 2 µ z =!mµ B, m =!s,!s + 1,...s =! 1 2 µ B, µ State specified by 4 quantum numbers: n,l, m l,m s What are the possible states for n=3?

8 Spin-orbit coupling: In the electron s frame, the nucleus is orbiting around it with orbital angular momentum L. This orbiting charge acts as a current loop which creates a magnetic field: 1 e B = 4! " 0 mc 2 r L 3 The intrinsic magnetic moment of the electron interacts with this field: U = -! µ s! B = 1 8" # 0 e 2 m 2 c 2 r 3 S! L This spin-orbit coupling energy must be included in the energy level structure of the hydrogen atom...

9 Spin-orbit coupling: Atomic Structure U = -! µ s! B = 1 8" # 0 e 2 m 2 c 2 r 3 S! L Define the total angular momentum to be Then J = L + S J 2 = ( L + S)!( L + S) = L 2 + S 2 + 2L!S L! S = 1 ( 2 J 2 " L 2 " S 2 ) Hence the spin-orbit energy can be expressed in terms of the total angular momentum and orbital angular momentum (spin is fixed at s=1/2).

10 Spin-orbit coupling: Atomic Structure Addition of angular momentum: J = L + S Using a vector model of addition we find that J is quantized in the usual manner for angular momentum: J = j( j + 1)!, j = l! s, j = l + s J z = m!, m =! j,! j + 1,...j

11 Spin-orbit coupling: Addition of angular momentum: J = L + S J = j( j + 1)!, j = l! s, j = l + s J z = m!, m =! j,! j + 1,...j Example: l = 0, s = 1 / 2 j = l! s = 1 / 2 j = l + s = 1 / 2 J = j( j + 1)! = 3 4! J z = m!, m =! j,! j + 1,...j What are the possible values of j for l =1? J z =! 1 2!, + 1 2!

12 Spin-orbit coupling: Fine structure of hydrogen The energy level structure of the hydrogen is modified by the spinorbit coupling: U = -! µ s! B = For each value of l > 0 there are two values of j (l+1/2, l-1/2). The energy of the state corresponding to each value of j is different. Energies without fine structure 1 8" # 0 e 2 m 2 c 2 r 3 S! L = 1 16" # 0 e 2 m 2 c 2 r 3 J 2 $ L 2 $ S 2 ( ) Energies with fine structure

13 Spin-orbit coupling: Fine structure of hydrogen The familiar red H-alpha line of hydrogen (or deuterium) is a single line according to the Bohr theory. In reality this line is a closely spaced doublet due to the fine structure splitting.

14 Spin-orbit coupling: Fine structure The well-known sodium doublet arising from the transitions from the 3p levels to 3s occur at wavelengths of nm and nm. What is the spin-orbit coupling energy? ( )!E = hc # hc = hc " # " 2 1 " 1 " 2 " 1 " 2 = 2.13meV

15 Indistinguishable Particles Classical particles are distinguishable by their paths. In quantum mechanics we look at probabilities (square of the wave function). Consider a two particle wave function. The two particles are indistinguishable if the probability distribution does not change when the particles are interchanged:! (r 1,r 2 2 =! (r 2,r 1 2 This implies that! (r 1,r 2 ) = ±! (r 2,r 1 ) Bosons:! (r 1,r 2 ) =! (r 2,r 1 ) Example: photons Fermions:! (r 1,r 2 ) = "! (r 2,r 1 ) Example: electrons

16 Pauli exclusion principle The two-particle wavefunction for fermions must be antisymmetric:! (r 1,r 2 ) =! a (r 1 )! b (r 2 ) "! b (r 1 )! a (r 2 ) in order to satisfy! (r 1,r 2 ) = "! (r 2,r 1 ) Note that if a and b label the same state, then the two particle function is zero. Thus no two fermions can be in the same state (described by the same wave function). This does not apply to bosons. Consequently, no two electrons in an atom can be in a state labeled by the same set of quantum numbers. This is the exclusion principle.

17 Periodic table Atomic Structure By applying the exclusion principle, we can build up the structure of atoms in their ground state: Each state is labeled by 4 quantum numbers: n, l, m l, m s. Only 1 electron is allowed per state (exclusion principle) For each n, there are n-1 values of l. For each l there are 2(2l+1) possible combinations of l, m l, m s. Thus 2(2l+1) electrons can occupy an orbital labeled by l. s orbitals (l=0) can hold 2 electrons. p orbitals (l=1) can hold 6 electrons. d orbitals (l=2) can hold 10 electrons etc We can build up the periodic table by filling in each l orbital as we increase n.

18 Periodic table Electron configuration is specified by n, the letter corresponding to l and a superscript with the number of electrons in the last orbital. Hydrogen: 1 electron: 1s 1 Helium: 2 electrons 1s 2 Write the electron configuration of sodium in its ground state. Atomic Structure Why is 4s filled before 3d? 4s orbital has a small bump near origin and penetrates shielding of core electrons better than 3d orbital, resulting in a larger effective nuclear charge and lower energy.

19 l = 2 (d) l = 3 (f) Noble Gas Halogen Group VI Group V l = 1 (p) 1 2 Group IV l = 0 (s) Group III Alkali n Periodic table

20 Chemical properties of the elements largely determined by the electrons in the outermost shell. Hence elements in a column of the periodic table behave similarly. Ionization energies increase as the electrons are more strongly attracted to the larger positive charge of the nucleus and then decrease due to the shielding effect of filled inner shells. Atomic volume shows the same behaviour for the same reason.

21 X-ray spectra X-rays are emitted when high energy (kev) transitions occur within the inner shells of heavy atoms. When an electron is excited from the K (n=1) shell then another electron from a higher shell drops into this shell, along with the emission of an X- ray photon. The spectral series are labeled by the letter of the final state in the transition.

22 Moseley s law Moseley measured X-ray spectra and deduced a formula for the K α series: E(K! ) = 3ke2 ( Z " 1) 2 8a 0 Moseley s formula accurately determined the atomic number Z of known elements. It also predicted the existence of new elements with Z=43, 61 and 75 that were not known at the time.

23 Moseley s law Atomic Structure A certain element emits x-rays with wavelength nm when it jumps from the L to K shell. What is the atomic number of the element? E(K! ) = hc " = 3ke2 ( Z # 1) 2 8a 0 $ Z = 8hca 0 3ke 2 " + 1 = 42