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Does instruction lead to learning?PowerPoint Presentation

Does instruction lead to learning?

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A mini-quiz – 5 minutes

- Write down the ground state wavefunction of the hydrogen atom?
- What is the radius of the ground state of the hydrogen atom?
- What is the radius of the n=120 state (L=119) of the hydrogen atom?
- What is the radius of the n=120 state (L=119) of the 91-times ionized uranium atom? (ignore relativity in this and the next question)
- What is the radius of the ground state of neutral uranium ?
- How does relativity change the answers to Q4 and Q5?
- Draw an energy level diagram indicating binding energies of the lowest 10 states in neutral helium – as close to scale as possible (giving their approximate binding energies)

Initial theory of hyperfine structure

(1) The nucleus also can have an intrinsic angular momentum –

The dipole form of this is called the nuclear spin.

Note – for later:

Higher order multipole distributions of the nuclear charge and “current” can also occur.

(2) The electromagnetic field caused by the electronic wavefunction at the nucleus leads to an interaction between the two:

#1 – the Coulomb interaction with the nuclear charge – done

#2 – the interaction between the nuclear spin and the electronic B-field

#3 – higher order multipole interactions – we will consider a more complete theory later

#2 – The dipole interaction energy (Hamiltonian) is

W(hfs-dipole) = - μ (nuclear) . B (elect)

looks just like spin orbit interaction, except μ is less by m/M

Hyperfine structuremagnetic dipole term

We can define the magnetic dipole moment of a nucleus in terms of its intrinsic spin (known as the nuclear spin) I.

μ = (μI/I) I

Where μ, and I are vectors, and μI is the nuclear magnetic moment usually quoted in nuclear tables.

Important note:The Wigner-Eckart Theorem:The components of any vector in a given subspace are proportional to the components of the angular momentum vector of that subspace.

You just need to find one proportionality factor – as given here.

(usually use the z-direction mJ=J)

Hence, the magnetic field due to the electron wavefunction(s)

B = J B0/J ħ

where B0 is the magnetic field at the nucleus in the (e.g.) z-direction

(Jħ=mZ)

Hyperfine structuremagnetic dipole term

The interaction energy is

ΔE(hfs) = -μ ● B = A I ● J A is a constant

= - <BZ> μ/(IJħ2)I ● J

= - <BZ> μ/(2IJħ2){F(F+1)-I(I+1)-J(J+1)}

=a/2 {F(F+1)-I(I+1)-J(J+1)}

Note: A and a are fairly standard notation in text books

What is the magnetic field at the nucleus? – s-states

S-states: The Fermi contact term

– only spin contributes to the magnetic field

We have a spherically symmetric charge distribution, with a non-zero spin-magnetization density at r=0, P -> B=8πP/3 (classical)

=-8πμ0/3 | Ψ(0) |2.

(μ0 = vacuum permeability)

Collecting terms as = (μ0/4π) ● (2μBμI/I) ● (8π|/3) |Ψ(0)|2

In hydrogenic systems, with nuclear charge Z, |Ψ(0)|2 is just the normalization constant: Z3/(πa03n3)

In other 1-electron systems (e.g. alkalis),

just put in the shielded nuclear charge ….

The directions of the B-field (labeled H) and magnetic moments in one electron S-states and P-states

What is the magnetic field at the nucleus? – non-s-states (1)

Two parts – orbital motion and electron spin:

The magnetic field at the nucleus

B = (μ0/4πr3) ● {(e v x r) – μs+ 3(μs●r )r/r2}

Where μs = 2μBS e r x v =2 μBL

Define the resulting vector in {} brackets by N. 2μB gives us an interaction energy proportional to I●N

Use Wigner-Eckart theorem to find component of N along J:

Then ΔE = (μ0/4π)● (2μBμI/I) ● <(N●J)>/J(J+1) ●<1/r3>proportional to <I●J>

= aJ/2 ● {F(F+1)-J(J+1)-I(I+1)} - definition of aJ

What is the magnetic field at the nucleus? – non-s-states (2)

- Calculate N●J …… = { L – S + 3(S·r)r/r2) } ● (L + S)
- = L2 – S2 + 3(S·r)(r·L)/r2) + 3(S·r)(r·S)/r2)
- But r·L = r·(rxp) = 0 -> 3rd term is zero
- The second term and the last term cancel!
- check by components and commutation relations for spin
- Thus < N●J > = < L2 > a nice result!
- And aJ = (μ0/4π) ● (2μBμI/I) ● <1/r3> L(L+1)/[J(J+1)]

Result for hydrogen ground state (2)

The ground-state has L=0, S=1/s -> J=1/2

Spin of the proton I = ½

HFS can be calculated exactly – to give E = 1.42 GHZ

What are these 2 ground-state levels?

Result: experiment differs from theory by about 1 part in 103!

Why? – What is the problem (too much for relativity)?

Result for hydrogen ground state (2)

The ground-state has L=0, S=1/s -> J=1/2

Spin of the proton I = ½

HFS can be calculated exactly – to give E = 1.42 GHZ

What are these 2 ground-state levels?

Result: experiment differs from theory by about 1 part in 103!

Why? –

Answer: - the anomalous magnetic moment of the electron…

gs = 2(1+ α/2π + higher order terms….)

The hydrogen maser uses this frequency

E = 1,420,405,751.7662 +- 0.0030 Hz

Example of hyperfine structure (2)

What is the spin

of this nucleus?

9/2 + I = 8, 7,…..1

->> I=7/2

Note: the regularity of the separations, gradually decreasing, as F gets smaller

ΔE{(F-1) – F)} = k. F

The HFS Electric quadrupole interaction (2)

We will first look at the next term in the electromagnetic interaction between the nucleus and the electron(s) – this will show us how to generalize the interaction to all interaction multipole moments…

The “and beyond” part will require the introduction of spherical tensors and their manipulation…

References for the first part: (these notes should be self-contained and independently understandable!)

G.K. Woodgate – Elementary atomic structure

H.B.G. Casimir - On the Interaction between Atomic Nuclei and electrons(Tweedegenootschap Prize – about 1935)

The electrostatic interaction is given by H .

For a point nucleus, this reduces to the Coulomb central field. For a finite nucleus and several electrons H can be expanded in specific multipole contributions.

HFS E2 expansion (2)

We can expand the expression for H, using the coordinate system shown:

1st term – the Coulomb central field;

2nd term – zero - the nucleus has no electric dipole moment! Hence, in general all “odd” terms should be zero.

3rd term – the electric quadrupole term.

HFS E2 (2)separations

Now we use the following identity to expand in separated spherical harmonics:

The rank k and the projection q are the same in each angular momentum space (nuclear and electronic).

Substituting in the quadrupole term, rearranging, and defining Q2 and F2gives us 5 terms

HFS E2 nuclear definitions (2)

We can now evaluate the expression for a state with given I,J,F |IJF>. Since the interaction energy is very small relative to other electronic energies, we can assume that I, J and F and their components are all good quantum numbers.

It is conventional to define the Nuclear Quadrupole Moment in the MI=I direction and have cylindrical symmetry for the nucleus – we can then define this one component, and use the Wigner-Eckart theorem to automatically define the other components

HFS E2 electronic definitions (2)

Again, define the interaction in the z-direction MJ=J,

and use the W.E. theorem to evaluate components …..

Note: we have assumed that the second derivative is proportional to 1/r3 at the origin.

HFS E2 evaluation (2)

The resulting interaction energy is

Convention-1: define the electric quadrupole interaction constant B -

- which is what gets measured in a hyperfine structure experiment…

Convention-2:

Define K

HFS - M1 plus E2 results (2)

Putting together our results for both the M1 and the E2 contributions:

The energy of any hyperfine level | nJ(LS)I F(IJ) > is

Example of measurements in 55Mn

Note that there is only a small variation from the interval rule predicted by just the M1 interaction:

Units:

1mK = 10-3 cm-1.

The transition 3d (2)54s2 6S5/2 – 3d54s4p 6P7/2 in neutral manganese

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