Today’s Outline

Today’s Outline - January 15, 2015

• Scattering from molecules and crystals

• The reciprocal lattice

• Compton (inelastic) scattering

• X-ray absorption

• Refraction and reflection of x-rays

• Magnetic interactions of x-rays

• Coherence of x-ray sources

C. Segre (IIT) PHYS 570 - Spring 2015 January 15, 2015 1 / 24

Scattering from atoms: all effects

Scattering from an atom is built up from component quantities:

Thomson scattering from a single electron

atomic form factor

anomalous scattering terms

polarization factor

−ro = − e2

4πε0mc2

f o(Q) =

∫ρ(r)e iQ·rd3r

f ′(~ω) + if ′′(~ω)

sin2 Ψ12(1 + sin2 Ψ)

− ro f (Q, ~ω) sin2 Ψ = −ro[f o(Q)

+ f ′(~ω) + if ′′(~ω)

]sin2 Ψ

atomic form factor

polarization factor

−ro = − e2

4πε0mc2

f o(Q) =

∫ρ(r)e iQ·rd3r

f ′(~ω) + if ′′(~ω)

− ro

f (Q, ~ω) sin2 Ψ

= −ro

[f o(Q)

+ f ′(~ω) + if ′′(~ω)

]sin2 Ψ

atomic form factor

polarization factor

−ro = − e2

4πε0mc2

f o(Q) =

∫ρ(r)e iQ·rd3r

f ′(~ω) + if ′′(~ω)

− ro f (Q, ~ω)

sin2 Ψ

= −ro[f o(Q)

+ f ′(~ω) + if ′′(~ω)

sin2 Ψ

atomic form factor

polarization factor

−ro = − e2

4πε0mc2

f o(Q) =

∫ρ(r)e iQ·rd3r

f ′(~ω) + if ′′(~ω)

− ro f (Q, ~ω)

sin2 Ψ

= −ro[f o(Q) + f ′(~ω) + if ′′(~ω)

sin2 Ψ

atomic form factor

polarization factor

−ro = − e2

4πε0mc2

f o(Q) =

∫ρ(r)e iQ·rd3r

f ′(~ω) + if ′′(~ω)

− ro f (Q, ~ω) sin2 Ψ = −ro[f o(Q) + f ′(~ω) + if ′′(~ω)

]sin2 Ψ

Scattering from molecules

extending to a molecule ...

Fmolecule(Q) =∑j

fj(Q)e iQ·rj

Fmolecule(Q) = f1(Q)e iQ·r1+f2(Q)e iQ·r2+f3(Q)e iQ·r3

Fmolecule(Q) =∑j

fj(Q)e iQ·rj

Fmolecule(Q) =∑j

fj(Q)e iQ·rj

Fmolecule(Q) =∑j

fj(Q)e iQ·rj

Fmolecule(Q) =

f1(Q)e iQ·r1+f2(Q)e iQ·r2+f3(Q)e iQ·r3

Fmolecule(Q) =∑j

fj(Q)e iQ·rj

Fmolecule(Q) = f1(Q)e iQ·r1+

f2(Q)e iQ·r2+f3(Q)e iQ·r3

Fmolecule(Q) =∑j

fj(Q)e iQ·rj

Fmolecule(Q) = f1(Q)e iQ·r1+f2(Q)e iQ·r2+

f3(Q)e iQ·r3

Fmolecule(Q) =∑j

fj(Q)e iQ·rj

Scattering from a crystal

and similarly, to a crystal lattice ...

... which is simply a periodic array of molecules

F crystal(Q) = FmoleculeF lattice

F crystal(Q) =∑j

fj(Q)e iQ·rj∑n

e iQ·Rn

The lattice term,∑

e iQ·Rn , is a sum over a large number so it is alwayssmall unless Q · Rn = 2πm where Rn = n1a1 + n2a2 + n3a3 is a real spacelattice vector and m is an integer. This condition is fulfilled only when Qis a reciprocal lattice vector.

F crystal(Q) =∑j

fj(Q)e iQ·rj∑n

e iQ·Rn

F crystal(Q) =∑j

fj(Q)e iQ·rj∑n

e iQ·Rn

F crystal(Q) =∑j

fj(Q)e iQ·rj∑n

e iQ·Rn

F crystal(Q) =∑j

fj(Q)e iQ·rj∑n

e iQ·Rn

e iQ·Rn , is a sum over a large number

so it is alwayssmall unless Q · Rn = 2πm where Rn = n1a1 + n2a2 + n3a3 is a real spacelattice vector and m is an integer. This condition is fulfilled only when Qis a reciprocal lattice vector.

F crystal(Q) =∑j

fj(Q)e iQ·rj∑n

e iQ·Rn

e iQ·Rn , is a sum over a large number so it is alwayssmall unless Q · Rn = 2πm where Rn = n1a1 + n2a2 + n3a3 is a real spacelattice vector and m is an integer.

This condition is fulfilled only when Qis a reciprocal lattice vector.

F crystal(Q) =∑j

fj(Q)e iQ·rj∑n

e iQ·Rn

Crystal lattices

There are 7 possible real space lattices: triclinic,

monoclinic, orthorhombic,tetragonal, hexagonal, rhombohedral, cubic

Crystal lattices

There are 7 possible real space lattices: triclinic, monoclinic,

orthorhombic,tetragonal, hexagonal, rhombohedral, cubic

Crystal lattices

There are 7 possible real space lattices: triclinic, monoclinic, orthorhombic,

tetragonal, hexagonal, rhombohedral, cubic

Crystal lattices

There are 7 possible real space lattices: triclinic, monoclinic, orthorhombic,tetragonal,

hexagonal, rhombohedral, cubic

Crystal lattices

There are 7 possible real space lattices: triclinic, monoclinic, orthorhombic,tetragonal, hexagonal,

rhombohedral, cubic

Crystal lattices

There are 7 possible real space lattices: triclinic, monoclinic, orthorhombic,tetragonal, hexagonal, rhombohedral,

Crystal lattices

There are 7 possible real space lattices: triclinic, monoclinic, orthorhombic,tetragonal, hexagonal, rhombohedral, cubic

Lattice volume

Consider the orthorhombic lattice for simplicity (the others give exactly thesame result).

a1 = ax, a2 = by, a3 = c z

a1 × a2 = abz

(a1 × a2) · a3 = abz · c z

(a1 × a2) · a3 = abc = V

A simple way of calculating the volume of the unit cell!

Lattice volume

a1 = ax, a2 = by, a3 = c z

a1 × a2 = abz

(a1 × a2) · a3 = abz · c z

(a1 × a2) · a3 = abc = V

Lattice volume

a1 = ax, a2 = by, a3 = c z

a1 × a2 = abz

(a1 × a2) · a3 = abz · c z

(a1 × a2) · a3 = abc = V

Lattice volume

a1 = ax, a2 = by, a3 = c z

a1 × a2 = abz

(a1 × a2) · a3 = abz · c z

(a1 × a2) · a3 = abc = V

Lattice volume

a1 = ax, a2 = by, a3 = c z

a1 × a2 = abz

(a1 × a2) · a3 = abz · c z

(a1 × a2) · a3 = abc = V

Lattice volume

a1 = ax, a2 = by, a3 = c z

a1 × a2 = abz

(a1 × a2) · a3 = abz · c z

(a1 × a2) · a3 = abc = V

Reciprocal lattice

Define the reciprocal lattice vectors in terms of the real space unit vectors

a∗1 = 2πa2 × a3

a1 · (a2 × a3)= 2π

a2 × a3V

a∗2 = 2πa3 × a1

a2 · (a3 × a1)= 2π

a3 × a1V

a∗3 = 2πa1 × a2

a3 · (a1 × a2)= 2π

a1 × a2V

In analogy to Rn, we can construct an arbitrary reciprocal space latticevector Ghkl

Ghkl = ha∗1 + ka∗2 + la∗3

where h, k, and l are integers called Miller indices

Reciprocal lattice

a∗1 = 2πa2 × a3

a1 · (a2 × a3)

= 2πa2 × a3

a∗2 = 2πa3 × a1

a2 · (a3 × a1)= 2π

a3 × a1V

a∗3 = 2πa1 × a2

a3 · (a1 × a2)= 2π

a1 × a2V

Reciprocal lattice

a∗1 = 2πa2 × a3

a1 · (a2 × a3)

= 2πa2 × a3

a∗2 = 2πa3 × a1

a2 · (a3 × a1)

= 2πa3 × a1

a∗3 = 2πa1 × a2

a3 · (a1 × a2)= 2π

a1 × a2V

Reciprocal lattice

a∗1 = 2πa2 × a3

a1 · (a2 × a3)

= 2πa2 × a3

a∗2 = 2πa3 × a1

a2 · (a3 × a1)

= 2πa3 × a1

a∗3 = 2πa1 × a2

a3 · (a1 × a2)

= 2πa1 × a2

Reciprocal lattice

a∗1 = 2πa2 × a3

a1 · (a2 × a3)= 2π

a2 × a3V

a∗2 = 2πa3 × a1

a2 · (a3 × a1)= 2π

a3 × a1V

a∗3 = 2πa1 × a2

a3 · (a1 × a2)= 2π

a1 × a2V

Reciprocal lattice

a∗1 = 2πa2 × a3

a1 · (a2 × a3)= 2π

a2 × a3V

a∗2 = 2πa3 × a1

a2 · (a3 × a1)= 2π

a3 × a1V

a∗3 = 2πa1 × a2

a3 · (a1 × a2)= 2π

a1 × a2V

Reciprocal lattice

a∗1 = 2πa2 × a3

a1 · (a2 × a3)= 2π

a2 × a3V

a∗2 = 2πa3 × a1

a2 · (a3 × a1)= 2π

a3 × a1V

a∗3 = 2πa1 × a2

a3 · (a1 × a2)= 2π

a1 × a2V

Reciprocal lattice

a∗1 = 2πa2 × a3

a1 · (a2 × a3)= 2π

a2 × a3V

a∗2 = 2πa3 × a1

a2 · (a3 × a1)= 2π

a3 × a1V

a∗3 = 2πa1 × a2

a3 · (a1 × a2)= 2π

a1 × a2V

Laue condition

Because of the construction of the reciprocal lattice

Ghkl · Rn = (n1a1 + n2a2 + n3a3) · (ha∗1 + ka∗2 + la∗3)

= (n1a1 + n2a2 + n3a3) · 2π(ha2 × a3

a3 × a1V

+ la1 × a2

)= 2π(hn1 + kn2 + ln3) = 2πm

and therefore, the crystal scattering factor is non-zero only when

Q = Ghkl

and a significant number of molecules scatter in phase with each other

Laue condition

Ghkl · Rn

= (n1a1 + n2a2 + n3a3) · (ha∗1 + ka∗2 + la∗3)

= (n1a1 + n2a2 + n3a3) · 2π(ha2 × a3

a3 × a1V

+ la1 × a2

)= 2π(hn1 + kn2 + ln3) = 2πm

Q = Ghkl

Laue condition

= (n1a1 + n2a2 + n3a3) · 2π(ha2 × a3

a3 × a1V

+ la1 × a2

)= 2π(hn1 + kn2 + ln3) = 2πm

Q = Ghkl

Laue condition

= (n1a1 + n2a2 + n3a3) · 2π(ha2 × a3

a3 × a1V

+ la1 × a2

= 2π(hn1 + kn2 + ln3) = 2πm

Q = Ghkl

Laue condition

= (n1a1 + n2a2 + n3a3) · 2π(ha2 × a3

a3 × a1V

+ la1 × a2

)= 2π(hn1 + kn2 + ln3) = 2πm

Q = Ghkl

Laue condition

= (n1a1 + n2a2 + n3a3) · 2π(ha2 × a3

a3 × a1V

+ la1 × a2

)= 2π(hn1 + kn2 + ln3) = 2πm

Q = Ghkl

Laue condition

= (n1a1 + n2a2 + n3a3) · 2π(ha2 × a3

a3 × a1V

+ la1 × a2

)= 2π(hn1 + kn2 + ln3) = 2πm

Q = Ghkl

Laue condition

= (n1a1 + n2a2 + n3a3) · 2π(ha2 × a3

a3 × a1V

+ la1 × a2

)= 2π(hn1 + kn2 + ln3) = 2πm

Q = Ghkl

Laue condition

A crystal is, therefore, a diffraction grating with ∼ 1020 slits!

When Q is a reciprocal lattice vector, a very strong, narrow diffractionpeak is seen at the detector.

-0.4 -0.2 0 0.2 0.4

Angle from center (radians)

nsity (

16 slits

8 slits4 slits

-0.04 -0.02 0 0.02 0.04

nsity (

1000 slits

Laue condition

-0.4 -0.2 0 0.2 0.4

nsity (

16 slits

8 slits4 slits

-0.04 -0.02 0 0.02 0.04

nsity (

1000 slits

Laue condition

-0.4 -0.2 0 0.2 0.4

nsity (

16 slits

8 slits4 slits

-0.04 -0.02 0 0.02 0.04

nsity (

1000 slits

Laue condition

-0.4 -0.2 0 0.2 0.4

nsity (

16 slits

8 slits4 slits

-0.04 -0.02 0 0.02 0.04

nsity (

1000 slits

Compton scattering

A photon-electron collision

p = ~k = 2π~/λp′ = ~k′ = 2π~/λ′

|k| 6=∣∣k′∣∣

Treat the electron relativistically and conserve energy and momentum

mc2 +hc

λ′+ γmc2 (energy)

λ′cosφ+ γmv cos θ (x-axis)

λ′sinφ+ γmv sin θ (y-axis)

Compton scattering

p = ~k = 2π~/λp′ = ~k′ = 2π~/λ′

|k| 6=∣∣k′∣∣

mc2 +hc

Compton scattering

p = ~k = 2π~/λ

p′ = ~k′ = 2π~/λ′

|k| 6=∣∣k′∣∣

mc2 +hc

Compton scattering

p = ~k = 2π~/λp′ = ~k′ = 2π~/λ′

|k| 6=∣∣k′∣∣

mc2 +hc

Compton scattering

p = ~k = 2π~/λp′ = ~k′ = 2π~/λ′

|k| 6=∣∣k′∣∣

mc2 +hc

Compton scattering

p = ~k = 2π~/λp′ = ~k′ = 2π~/λ′

|k| 6=∣∣k′∣∣

mc2 +hc

Compton scattering

p = ~k = 2π~/λp′ = ~k′ = 2π~/λ′

|k| 6=∣∣k′∣∣

mc2 +hc

Compton scattering

p = ~k = 2π~/λp′ = ~k′ = 2π~/λ′

|k| 6=∣∣k′∣∣

mc2 +hc

Compton scattering

p = ~k = 2π~/λp′ = ~k′ = 2π~/λ′

|k| 6=∣∣k′∣∣

mc2 +hc

Compton scattering derivation

squaring the momentumequations

λ− h

λ′cosφ

= γ2m2v2 cos2 θ(− h

λ′sinφ

= γ2m2v2 sin2 θ

now add them together, substitute sin2 θ + cos2 θ = 1, andsin2 φ+ cos2 φ = 1, then rearrange and substitute v = βc

γ2m2v2(sin2 θ + cos2 θ

λ− h

λ′cosφ

(− h

λ′sinφ

γ2m2v2 =h2

λ2− 2h2

λλ′cosφ+

λ′2sin2 φ+

λ′2cos2 φ

λ2− 2h2

λλ′cosφ+

λ′2=

1− β2

=m2c2β2

1− β2

λ− h

λ′cosφ

= γ2m2v2 cos2 θ

(− h

λ′sinφ

= γ2m2v2 sin2 θ

λ− h

λ′cosφ

(− h

λ′sinφ

γ2m2v2 =h2

λ2− 2h2

λλ′cosφ+

λ′2sin2 φ+

λ′2cos2 φ

λ2− 2h2

λλ′cosφ+

λ′2=

1− β2

=m2c2β2

1− β2

λ− h

λ′cosφ

λ′sinφ

= γ2m2v2 sin2 θ

λ− h

λ′cosφ

(− h

λ′sinφ

γ2m2v2 =h2

λ2− 2h2

λλ′cosφ+

λ′2sin2 φ+

λ′2cos2 φ

λ2− 2h2

λλ′cosφ+

λ′2=

1− β2

=m2c2β2

1− β2

λ− h

λ′cosφ

λ′sinφ

= γ2m2v2 sin2 θ

now add them together,

substitute sin2 θ + cos2 θ = 1, andsin2 φ+ cos2 φ = 1, then rearrange and substitute v = βc

λ− h

λ′cosφ

(− h

λ′sinφ

γ2m2v2 =h2

λ2− 2h2

λλ′cosφ+

λ′2sin2 φ+

λ′2cos2 φ

λ2− 2h2

λλ′cosφ+

λ′2=

1− β2

=m2c2β2

1− β2

λ− h

λ′cosφ

λ′sinφ

= γ2m2v2 sin2 θ

now add them together, substitute sin2 θ + cos2 θ = 1,

andsin2 φ+ cos2 φ = 1, then rearrange and substitute v = βc

λ− h

λ′cosφ

(− h

λ′sinφ

γ2m2v2 =h2

λ2− 2h2

λλ′cosφ+

λ′2sin2 φ+

λ′2cos2 φ

λ2− 2h2

λλ′cosφ+

λ′2=

1− β2

=m2c2β2

1− β2

λ− h

λ′cosφ

λ′sinφ

= γ2m2v2 sin2 θ

now add them together, substitute sin2 θ + cos2 θ = 1, andsin2 φ+ cos2 φ = 1, then rearrange

and substitute v = βc

λ− h

λ′cosφ

(− h

λ′sinφ

γ2m2v2 =h2

λ2− 2h2

λλ′cosφ+

λ′2sin2 φ+

λ′2cos2 φ

λ2− 2h2

λλ′cosφ+

λ′2=

1− β2

=m2c2β2

1− β2

λ− h

λ′cosφ

λ′sinφ

= γ2m2v2 sin2 θ

λ− h

λ′cosφ

(− h

λ′sinφ

γ2m2v2 =h2

λ2− 2h2

λλ′cosφ+

λ′2sin2 φ+

λ′2cos2 φ

λ2− 2h2

λλ′cosφ+

λ′2=

1− β2=

m2c2β2

1− β2

Now take the energy equation and square it,

then solve it for β2 which issubstituted into the equation from the momenta.

(mc2 +

λ− hc

= γ2m2c4=m2c4

1− β2

β2 = 1− m2c4(mc2 + hc

λ −hcλ′

λ2− 2h2

λλ′cosφ+

λ′2=

m2c2β2

1− β2

Now take the energy equation and square it, then solve it for β2

which issubstituted into the equation from the momenta.

(mc2 +

λ− hc

= γ2m2c4=m2c4

1− β2

β2 = 1− m2c4(mc2 + hc

λ −hcλ′

λ2− 2h2

λλ′cosφ+

λ′2=

m2c2β2

1− β2

Now take the energy equation and square it, then solve it for β2 which issubstituted into the equation from the momenta.(

mc2 +hc

λ− hc

= γ2m2c4=m2c4

1− β2

β2 = 1− m2c4(mc2 + hc

λ −hcλ′

λ2− 2h2

λλ′cosφ+

λ′2=

m2c2β2

1− β2

After substitution, expansion and cancellation, we obtain

λ′2− 2h2

λλ′cosφ = 2m

λ− hc

λ′2− 2h2

λλ′

λλ′(1− cosφ) = 2m

λ− hc

)= 2mhc

(λ′ − λλλ′

2mhc∆λ

λλ′

∆λ =h

mc(1− cosφ)

λ′2− 2h2

λλ′cosφ = 2m

λ− hc

λ′2− 2h2

λλ′

λ− hc

)= 2mhc

2mhc∆λ

λλ′

∆λ =h

mc(1− cosφ)

λ′2− 2h2

λλ′cosφ = 2m

λ− hc

λ′2− 2h2

λλ′

λ− hc

)= 2mhc

2mhc∆λ

λλ′

∆λ =h

mc(1− cosφ)

λ′2− 2h2

λλ′cosφ = 2m

λ− hc

λ′2− 2h2

λλ′

λ− hc

)= 2mhc

2mhc∆λ

λλ′

∆λ =h

mc(1− cosφ)

Compton scattering results

λc = ~/mc = 3.86× 10−3A for an electron

Comparing to the Thomson scattering length: ro/λC = 1/137

10500 11000 11500 12000 12500Energy (eV)

ay Inte

nsity (

units)

φ=160o

Compton scattering results

λc = ~/mc = 3.86× 10−3A for an electron

Comparing to the Thomson scattering length: ro/λC = 1/137

10500 11000 11500 12000 12500Energy (eV)

ay Inte

nsity (

units)

φ=160o

X-ray absorption

Absorption coefficient µ, thickness dzx-ray intensity is attenuated as

dI = −I (z)µdz

dI/I = −µdz =⇒ I = Ioe−µz

number of absorption events, W = I (z)ρaσadz = I (z)µdz

where ρa is atom density, σa is absorption cross section

µ = ρaσa =(ρmNAA

with mass density ρm , Avogadro’s number NA, atomic number A

X-ray absorption

dI = −I (z)µdz

X-ray absorption

dI = −I (z)µdz

X-ray absorption

dI = −I (z)µdz

Absorption event

• X-ray is absorbed by an atom

• Energy is transferred to a core electron

• Electron escapes atomic potential into thecontinuum

• Ion remains with a core-hole

Absorption event

Fluorescence emission

An ion with a core-hole is quite unstable (≈ 10−15s)

−→

• After a short time a higher levelelectron will drop down in energy tofill the core hole

• Energy is liberated in the form of afluorescence photon

• This leaves a second hole (not core)which is then filled from an evenhigher shell

• The result is a cascade of fluorescencephotons which are characteristic of theabsorbing atom

−→

Auger emission

While fluorescence is the most probable method of core-hole relaxationthere are other possible mechanisms

−→

k• In the Auger process, a higher level

electron will drop down in energy tofill the core hole

• The energy liberated causes thesecondary emission of an electron

• This leaves two holes which then filledfrom higher shells

• So that the secondary electron isaccompanied by fluorescence emissionsat lower energies

Auger emission

−→

• In the Auger process, a higher levelelectron will drop down in energy tofill the core hole

Auger emission

−→

Auger emission

−→

Auger emission

−→

Absorption coefficient

The absorption coefficient µ, depends strongly on the x-ray energy E , theatomic number of the absorbing atoms Z , as well as the density ρ, andatomic mass A:

µ ∼

µ ∼ ρZ 4

Isolated gas atoms show a sharp jump and a smooth curve

Atoms in a solid or liquid show fine structure after the absorption edgecalled XANES and EXAFS

−→

Isolated gas atoms show a sharp jump and a smooth curveAtoms in a solid or liquid show fine structure after the absorption edgecalled XANES and EXAFS

−→

Refraction of x-rays

X-rays can be treated like light when interaction with a medium. However,unlike visible light, the index of refraction of x-rays in matter is very closeto unity:

n = 1− δ + iβ

with δ ∼ 10−5

Snell’s Law

cosα = n cosα′

where α′ < α unlike for visible light

n = 1− δ + iβ

with δ ∼ 10−5

Snell’s Law

cosα = n cosα′

n = 1− δ + iβ

with δ ∼ 10−5

Snell’s Law

cosα = n cosα′

n = 1− δ + iβ

with δ ∼ 10−5

Snell’s Law

cosα = n cosα′

n = 1− δ + iβ

with δ ∼ 10−5

Snell’s Law

cosα = n cosα′

Reflection of x-rays

Because n < 1, at a critical angle αc , we no longer have refraction but

total external reflection

Since α′ = 0 when α = αc

n = cosαc

n ≈ 1− α2c

1− δ + iβ ≈ 1− α2c

δ =α2c

2−→ αc =

√2δ

Because n < 1, at a critical angle αc , we no longer have refraction buttotal external reflection

n = cosαc

n ≈ 1− α2c

1− δ + iβ ≈ 1− α2c

δ =α2c

2−→ αc =

√2δ

n = cosαc

n ≈ 1− α2c

1− δ + iβ ≈ 1− α2c

δ =α2c

2−→ αc =

√2δ

n = cosαc

n ≈ 1− α2c

1− δ + iβ ≈ 1− α2c

δ =α2c

2−→ αc =

√2δ

n = cosαc

n ≈ 1− α2c

1− δ + iβ ≈ 1− α2c

δ =α2c

2−→ αc =

√2δ

n = cosαc

n ≈ 1− α2c

1− δ + iβ ≈ 1− α2c

δ =α2c

2−→ αc =

√2δ

n = cosαc

n ≈ 1− α2c

1− δ + iβ ≈ 1− α2c

δ =α2c

2−→ αc =

√2δ

Uses of total external reflection

X-ray mirrors

• harmonic rejection

• focusing & collimation

Evanscent wave experiments

• studies of surfaces

• depth profiling

X-ray mirrors

• depth profiling

X-ray mirrors

• depth profiling

X-ray mirrors

• depth profiling

X-ray mirrors

• depth profiling

X-ray mirrors

• depth profiling

Magnetic interactions

We have focused on the interaction of x-rays and charged particles.However, electromagnetic radiation also consists of a traveling mag-netic field. In principle, this means it should interact with magneticmaterials as well.

Indeed, x-rays do interact with magnetic materials (and electrons whichhave magnetic moment and spin) but the strength of the interactionis comparatively weak.

Amagnetic

Acharge=

~ωmc2

=5.11× 103 eV

0.511× 106 eV= 0.01

For an x-ray of energy 5.11 keV, interacting with an electron with mass0.511 MeV. Only with the advent of synchrotron radiation sources hasmagnetic x-ray scattering become a practical experimental technique.

Amagnetic

Acharge=

~ωmc2

=5.11× 103 eV

0.511× 106 eV= 0.01

Amagnetic

Acharge=

~ωmc2

=5.11× 103 eV

0.511× 106 eV= 0.01

Amagnetic

Acharge=

~ωmc2

=5.11× 103 eV

0.511× 106 eV= 0.01

For an x-ray of energy 5.11 keV, interacting with an electron with mass0.511 MeV.

Only with the advent of synchrotron radiation sources hasmagnetic x-ray scattering become a practical experimental technique.

Amagnetic

Acharge=

~ωmc2

=5.11× 103 eV

0.511× 106 eV

= 0.01

Amagnetic

Acharge=

~ωmc2

=5.11× 103 eV

0.511× 106 eV= 0.01

Amagnetic

Acharge=

~ωmc2

=5.11× 103 eV

0.511× 106 eV= 0.01

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