Resonant photon absorption The Mossbauer effect Photon attenuation Radiation attenuation by: --...

Resonant photon absorption

The Mossbauer effect

Photon attenuation

Radiation attenuation by:-- photoelectric effect-- compton scattering (E << 1.02 MeV)

€

I Δx( ) = I 0( ) e− μ pe+μ cs( )Δx

Atomic interactions

Source Detector

Absorber

x€

Eγ

Photon attenuation

Consider nuclear resonant absorption

Source Detector

Absorber

x€

Eγ

0.0

E*

0.0

E*

Assume source and absorber are identical

€

Eγ = E*

€

E* = Eγ

Kinematics

Assume source and absorber are identical

Source Detector

Absorber

x€

Eγ

0.0

E*

0.0

E*

€

Eγ = E*

€

E* = Eγ

€

TR =pγ

2

2MR

€

Eγ = E* − TR

€

TR =pγ

2

2MR

€

rp γ

€

rp R

emission

€

rp γ

€

rp R

absorption

€

Eγ + 2TR = E*

for resonant absorption

Quantum state for source and absorber

0

0.1

0.2

0.3

0.4

0.5

0.6

5 10 15 20 25 30 35

Energy (keV)

P(E)

Source Absorber

Ignore energy scale

Estimates

€

TR =pγ

2

2MR≈

2MeV 2

2 57 ×103 MeV( )≈1.75 ×10−5 MeV

Consider an 57Fe source 57Co 57Co Fe

€

Eγ = E* − TR ≈14.413 KeV −1.75 ×10−2KeV

Eγ =14.396 KeV

0.0

E*

0.0

E*

€

Γ ≈hτ=

00.659 ×10−18KeV s

98 ×10−9 s

Γ ≈ 0.66 ×10−11KeV

Natural width of the state

Enter -- Mr. Mossbauer

Place 57Fe source bound in a metal matrix

€

TR =pγ

2

2MR≈

pγ2

2 ∞≈ 0

€

Eγ = E* − TR

Eγ = E*

Place 57Fe absorber bound in a metal matrix

€

TR =pγ

2

2MR≈

pγ2

2 ∞≈ 0

€

Eγ = E* − TR

Eγ = E*

Resonant Absorption!

Kinematics

Assume source and absorber are identical

-v+v

Source Detector

Absorber

x€

Eγ

0.0

E*

0.0

E*

€

Eγ = E*

€

E* = Eγ

move source

move source

€

Eγ' = Eγ + ED

ED ≈ Eγv

c

Doppler shift frequency:h’- h = ED

Quantum state for source and absorber

0

0.1

0.2

0.3

0.4

0.5

0.6

5 10 15 20 25 30 35

Energy (keV)

P(E)

Source Absorber

-v

no resonant

absorption

Source Absorber

Quantum state for source and absorber

0

0.1

0.2

0.3

0.4

0.5

0.6

5 10 15 20 25 30 35

Energy (keV)

P(E)

-v

no resonant

absorption

Source Absorber

Quantum state for source and absorber

0

0.1

0.2

0.3

0.4

0.5

0.6

5 10 15 20 25 30 35

Energy (keV)

P(E)

-v

small resonant

absorption

Source Absorber

Quantum state for source and absorber

0

0.1

0.2

0.3

0.4

0.5

0.6

5 10 15 20 25 30 35

Energy (keV)

P(E)

-v

more resonant

absorption

Source Absorber

Quantum state for source and absorber

0

0.1

0.2

0.3

0.4

0.5

0.6

5 10 15 20 25 30 35

Energy (keV)

P(E)

v = 0.0

maximum resonant

absorption

Source Absorber

Quantum state for source and absorber

0

0.1

0.2

0.3

0.4

0.5

0.6

5 10 15 20 25 30 35

Energy (keV)

P(E)

-v

less resonant

absorption

Source Absorber

Quantum state for source and absorber

0

0.1

0.2

0.3

0.4

0.5

0.6

5 10 15 20 25 30 35

Energy (keV)

P(E)

-v

small resonant

absorption

Source Absorber

Quantum state for source and absorber

0

0.1

0.2

0.3

0.4

0.5

0.6

5 10 15 20 25 30 35

Energy (keV)

P(E)

-v

no resonant

absorption

Source Absorber

Quantum state for source and absorber

0

0.1

0.2

0.3

0.4

0.5

0.6

5 10 15 20 25 30 35

Energy (keV)

P(E)

-v

no resonant

absorption

Transmission curve

0

0.05

0.1

0.15

0.2

0.25

0.3

0.35

0.4

0.45

5 10 15 20 25 30 35

Energy (keV)

P(E)

Resulting transmission curve

Kinematics

Assume source and absorber are NOT

identical

Source Detector

Absorber

x€

Eγ

0.0

Ea*

0.0

Es*

Doppler kinematics

Assume source and absorber are NOT

identical

Resonant absorption

€

Eγ' = Ea

*

ED = Es* − Ea

*

when -

-v+v

Source Detector

Absorber

x€

Eγ

0.0

Ea*

0.0

Es*

€

Eγ' = Eγ + ED

ED ≈ Eγv

c

move absorber!

Quantum state for source and absorber

0

0.1

0.2

0.3

0.4

0.5

0.6

5 10 15 20 25 30 35

Energy (keV)

P(E)

Source Absorber transition energy shifted

-v

no resonant

absorption

Source

Quantum state for source and absorber

0

0.1

0.2

0.3

0.4

0.5

0.6

5 10 15 20 25 30 35

Energy (keV)

P(E)

-v

small resonant

absorption

Absorber transition energy shifted

Source

Quantum state for source and absorber

0

0.1

0.2

0.3

0.4

0.5

0.6

5 10 15 20 25 30 35

Energy (keV)

P(E)

-v

more resonant

absorption

Absorber transition energy shifted

Source

Quantum state for source and absorber

0

0.1

0.2

0.3

0.4

0.5

0.6

5 10 15 20 25 30 35

Energy (keV)

P(E)

-v

more resonant

absorption

Absorber transition energy shifted

Source

Quantum state for source and absorber

0

0.1

0.2

0.3

0.4

0.5

0.6

5 10 15 20 25 30 35

Energy (keV)

P(E)

v = 0.0

less resonant

absorption

Absorber transition energy shifted

Source

Quantum state for source and absorber

0

0.1

0.2

0.3

0.4

0.5

0.6

5 10 15 20 25 30 35

Energy (keV)

P(E)

-v

smallresonant

absorption

Absorber transition energy shifted

Source

Quantum state for source and absorber

0

0.1

0.2

0.3

0.4

0.5

0.6

5 10 15 20 25 30 35

Energy (keV)

P(E)

-v

no resonant

absorption

Absorber transition energy shifted

Source

Quantum state for source and absorber

0

0.1

0.2

0.3

0.4

0.5

0.6

5 10 15 20 25 30 35

Energy (keV)

P(E)

-v

no resonant

absorption

Absorber transition energy shifted

Source

Quantum state for source and absorber

0

0.1

0.2

0.3

0.4

0.5

0.6

5 10 15 20 25 30 35

Energy (keV)

P(E)

-v

no resonant

absorption

Absorber transition energy shifted

Transmission curve

0

0.05

0.1

0.15

0.2

0.25

0.3

0.35

0.4

0.45

5 10 15 20 25 30 35

Energy (keV)

P(E)

Doppler energy shifted

€

ED = Eγv

c

“Isotope shift”

€

v = 0

Isotope shift

Resonant absorption

€

Eγ' = Ea

*

ED = Es* − Ea

*

when -

-v+v

Source Detector

Absorber

x€

Eγ

move absorber!

0.0

Ea*

0.0

Es*

Isotope shift:Level shifts due to atomic electronic

charge distribution in the

nucleus.

Constant velocity data

57FeWhat is the J for the ground state and the 14.4 Kev state?

ENSDF/NNDS

What is the multipolarity of the transition?

What is the degeneracy for the -- ground state and the -- 14.4 Kev state?

If there is a B field, then we can have a

nuclear Zeeman effect that will

remove the degeneracies

If there is a B field, then we can have a

nuclear Zeeman effect that will

remove the degeneracies

57Fe+v-v

move source with constant acceleration

Source Detector

Absorber

x€

Eγ

€

E1/2

€

E3/2

0.0

Es*

€

3

2

−

€

1

2

−

€

+3

2

€

−3

2€

+1

2

€

−1

2

m-sublevels

€

−1

2

€

+1

2

Dipole transition selection rules

€

I =1

€

m = ±1,0

Mossbauer resonant absorption with constant acceleration

-v +v0

maximum +v

maximum -vtim

e v = 0

v = 0

v = 0

Source velocity curve

Source displacement curve

data

v

t

Use MCS/MCAUse MCS/MCA

€

t

€

v= dwell time= one channel

Possible absorption transitions

Source Detector

Absorber

x€

Eγ

€

E1/2

€

E3/2

0.0

Es*

€

3

2

−

€

1

2

−

€

+3

2

€

−3

2€

+1

2

€

−1

2

m-sublevels

€

−1

2

€

+1

2

Possible absorption transitions

€

E1/2

€

E3/2

€

3

2

−

€

1

2

−

€

+3

2

€

−3

2€

+1

2

€

−1

2

m-sublevels

€

−1

2

€

+1

2

6 54 32 1

€

E1/2 > ΔE3/2

€

1,3 = ΔE3/2

€

3,5 = ΔE3/2

€

E3,6 = ΔE1/2

€

E1,4 = ΔE1/2

Possible absorption transitions

€

E1/2 > ΔE3/2

€

1,3 = ΔE3/2

€

3,5 = ΔE3/2

€

E3,6 = ΔE1/2

€

E1,4 = ΔE1/2

€

2,4 = ΔE3/2

€

4,6 = ΔE3/2

Compare these predictions with the measurements…Compare these predictions with the measurements…

…follow guidelines in Problem. 10.C. and eventually determine

€

E3/2

€

E1/2

The Pound-Rebecca Experiment

Be prepared to explain what the experiment discovered and how the Mossbauer resonant

photon absorption was essential to the measurement.

Be prepared to explain what the experiment discovered and how the Mossbauer resonant

photon absorption was essential to the measurement.

Possible absorption transitions

€

E1/2

€

E3/2

€

3

2

−

€

1

2

−

€

+3

2

€

−3

2€

+1

2

€

−1

2

m-sublevels

€

−1

2

€

+1

2

6 54 32 1

€

E1/2 > ΔE3/2

€

1,3 = ΔE3/2

€

3,5 = ΔE3/2

€

E3,6 = ΔE1/2

€

E1,4 = ΔE1/2

Case 1

Possible absorption transitions

€

E1/2

€

E3/2

€

3

2

−

€

1

2

−

€

+3

2

€

−3

2€

+1

2

€

−1

2

m-sublevels

€

−1

2

€

+1

2

5 63 41 2

€

E1/2 < ΔE3/2

€

1,3 = ΔE3/2

€

3,5 = ΔE3/2

€

E4,5 = ΔE1/2

€

E2,3 = ΔE1/2

Case 2

Possible absorption transitionsm-sublevels

€

E1/2 < ΔE3/2

€

1,2 = ΔE3/2

€

2,4 = ΔE3/2

€

E2,3 = ΔE1/2

€

E4,5 = ΔE1/2

€

E1/2

€

E3/2

€

3

2

−

€

1

2

−

€

+3

2

€

−3

2€

+1

2

€

−1

2

€

+1

2

€

−1

2

6 5 23 14

Case 3

Possible absorption transitionsm-sublevels

€

E1/2 > ΔE3/2

€

1,2 = ΔE3/2

€

E2,3 = ΔE3/2

€

E2,4 = ΔE1/2

€

E3,5 = ΔE1/2

€

E1/2

€

E3/2

€

3

2

−

€

1

2

−

€

+3

2

€

−3

2€

+1

2

€

−1

2

€

+1

2

€

−1

2

6 5 24 13

Case 4