Why Does Uranium Alpha Decay? Consider the alpha decay shown below - - PowerPoint PPT Presentation

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Why Does Uranium Alpha Decay? Consider the alpha decay shown below - - PowerPoint PPT Presentation

Jun 03, 2023 237 likes •696 views

Why Does Uranium Alpha Decay? Consider the alpha decay shown below where a uranium nucleus spontaneously breaks apart into a 4 He or alpha particle and 234 90 Th . 238 4 He + 234 E ( 4 He ) = 4 . 2 MeV 92 U 90 Th To study this reaction we

slide-1
SLIDE 1

Why Does Uranium Alpha Decay?

Consider the alpha decay shown below where a uranium nucleus spontaneously breaks apart into a 4He or alpha particle and 234

90Th. 238 92U → 4He + 234 90Th

E(4He) = 4.2 MeV To study this reaction we first map out the 4He − 234

90Th potential energy.

We reverse the decay above and use a beam of 4He nuclei striking a 234

90Th

  • target. The 4He beam comes from the radioactive decay of another

nucleus 210

84Po and E(4He) = 5.407 MeV.

1 What is the distance of closest approach of the 4He to the 234

90Th

target if the Coulomb force is the only one that matters?

2 Is the Coulomb force the only one that matters? 3 What is the lifetime of the 238

92U?

Jerry Gilfoyle Alpha Decay 1 / 32

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SLIDE 2

What Do We Know?

The

234 90Th − α Potential r (fm) V(r) (MeV)

attractive nuclear part V = Z1Z2e2

r

Jerry Gilfoyle Alpha Decay 2 / 32

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SLIDE 3

Mapping the Potential Energy Rutherford Scattering

What is the distance of closest approach of the 4He to the 234

90Th target if

  • nly the Coulomb force is active? Is the Coulomb force the only one

active? The energy of the 4He emitted by the 210

84Po to make the beam is

E(4He) = 5.407 MeV.

Alpha source

84 210

Microscope Alpha beam Collimator ZnS Scattered helium Thorium target

2 4 82

Po He + Pb

206

Demo

Jerry Gilfoyle Alpha Decay 3 / 32

slide-4
SLIDE 4

Mapping the Potential Energy Rutherford Scattering

What is the distance of closest approach of the 4He to the 234

90Th target if

  • nly the Coulomb force is active? Is the Coulomb force the only one

active? The energy of the 4He emitted by the 210

84Po to make the beam is

E(4He) = 5.407 MeV.

Alpha source

84 210

Microscope Alpha beam Collimator ZnS Scattered helium Thorium target

2 4 82

Po He + Pb

206

Demo

Jerry Gilfoyle Alpha Decay 3 / 32

slide-5
SLIDE 5

Mapping the Potential Energy Rutherford Scattering

What is the distance of closest approach of the 4He to the 234

90Th target if

  • nly the Coulomb force is active? Is the Coulomb force the only one

active? The energy of the 4He emitted by the 210

84Po to make the beam is

E(4He) = 5.407 MeV.

Alpha source

84 210

Microscope Alpha beam Collimator ZnS Scattered helium Thorium target

2 4 82

Po He + Pb

206

Demo

Jerry Gilfoyle Alpha Decay 3 / 32

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SLIDE 6

Rutherford Trajectories

Rutherford trajectories for different impact parameters x y/b Jerry Gilfoyle Alpha Decay 4 / 32

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SLIDE 7

Mapping the Potential Energy The Differential Cross Section

Th target He trajectory

4 Jerry Gilfoyle Alpha Decay 5 / 32

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SLIDE 8

Mapping the Potential Energy The Differential Cross Section

Th target He trajectory

4

dσ dΩ = Z1Z2e2 4Ecm 2 1 sin4 θ

2

  • Jerry Gilfoyle

Alpha Decay 5 / 32

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SLIDE 9

The Differential Cross Section

particle rate scattered into dA of detector = dNs dt ∝ incident beam rate × areal target density × angular detector size dNs dt ∝ dNinc dt × ntgt × dΩ dNs dt = dσ dΩ × dNinc dt × ntgt × dΩ dNinc dt = ∆Ninc ∆t = Ibeam Ze ntgt = ρtgt Atgt NAVhit 1 abeam = ρtgt Atgt NALtgt Ibeam - beam current Z - beam charge ρtgt - target density Atgt - molar mass Vhit - beam-target overlap Ltgt - target thickness

Jerry Gilfoyle Alpha Decay 6 / 32

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SLIDE 10

Areal or Surface Density of Nuclear Targets

Jerry Gilfoyle Alpha Decay 7 / 32

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SLIDE 11

The Differential Cross Section

particle rate scattered into dA of detector = dNs dt ∝ incident beam rate × areal target density × angular detector size dNs dt ∝ dNinc dt × ntgt × dΩ dNs dt = dσ dΩ × dNinc dt × ntgt × dΩ dNinc dt = ∆Ninc ∆t = Ibeam Ze ntgt = ρtgt Atgt NAVhit 1 abeam = ρtgt Atgt NALtgt dΩ = dAdet r 2

det

= ∆Adet r 2

det

= sin θdθdφ Ibeam - beam current Z - beam charge ρtgt - target density Atgt - molar mass Vhit - beam-target overlap Ltgt - target thickness dAdet - detector area rdet - target-detector distance

Jerry Gilfoyle Alpha Decay 8 / 32

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SLIDE 12

Actual Rutherford Scattering Results

20 40 60 80 100 120 140 160 180 (deg) θ 5 10 15 20 25 30 35 40 45 50 /2) θ (

4

sin × Counts H.Geiger and E.Marsden, Phil. Mag., 25, p. 604 (1913) alphas on gold

Jerry Gilfoyle Alpha Decay 9 / 32

slide-13
SLIDE 13

Actual Rutherford Scattering Results

20 40 60 80 100 120 140 160 180 (deg) θ 5 10 15 20 25 30 35 40 45 50 /2) θ (

4

sin × Counts H.Geiger and E.Marsden, Phil. Mag., 25, p. 604 (1913) alphas on gold

dσ dΩ = Z1Z2e2 4Ecm 2 1 sin4 θ

2

  • Jerry Gilfoyle

Alpha Decay 9 / 32

slide-14
SLIDE 14

Interpretation of Rutherford Scattering Results

50 100 150 0.0 0.2 0.4 0.6 0.8 1.0 1.2 Scattering Angle (deg) Measured/Predicted

4 2He +234 90 Th →4 2 He +234 90 Th dσ dΩ =

  • Z1Z2e2

4Ecm

2

1 sin4( θ

2)

DOCA = 48 fm Eα(Po) = 5.407 MeV

238 92U → 4He + 234 90Th

Eα(U) = 4.2 MeV Original decay What does this say about the 4

2He −234 90 Th potential energy?

Jerry Gilfoyle Alpha Decay 10 / 32

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SLIDE 15

Measuring the Size of the Nucleus

θcm(deg)

Jerry Gilfoyle Alpha Decay 11 / 32

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SLIDE 16

Measuring the Size of the Nucleus

θcm(deg)

Ecm = 28.3 MeV Ecm = 23.1 MeV PRL 109, 262701 (2012) Jerry Gilfoyle Alpha Decay 11 / 32

slide-17
SLIDE 17

The 4He − 234

90Th Potential

α-Th Potential Blue - known Red - a guess 10 20 30 40 50 60 70

  • 40
  • 20

20 40 r(fm) V(MeV)

Jerry Gilfoyle Alpha Decay 12 / 32

slide-18
SLIDE 18

The Paradox of Alpha Decay

1 We have probed the 248

90Th potential into an internuclear distance of

rDOCA = 48 fm using a 4He beam of E(4He) = 5.407 MeV.

2 The data are consistent with the Coulomb force and no others. 3 The radioactive decay 238

92U → 248 90Th + 4He emits an α (or 4He) with

energy Eα = 4.2 MeV.

4 For a classical ‘decay’ the emitted α should have an energy of at least

Emin = 5.407 MeV.

5 It appears the ‘decay’ α starts out at a distance remit = 62 fm. 6 How do we explain this? Jerry Gilfoyle Alpha Decay 13 / 32

slide-19
SLIDE 19

The Paradox of Alpha Decay

1 We have probed the 248

90Th potential into an internuclear distance of

rDOCA = 48 fm using a 4He beam of E(4He) = 5.407 MeV.

2 The data are consistent with the Coulomb force and no others. 3 The radioactive decay 238

92U → 248 90Th + 4He emits an α (or 4He) with

energy Eα = 4.2 MeV.

4 For a classical ‘decay’ the emitted α should have an energy of at least

Emin = 5.407 MeV.

5 It appears the ‘decay’ α starts out at a distance remit = 62 fm. 6 How do we explain this?

Quantum Tunneling!

Jerry Gilfoyle Alpha Decay 13 / 32

slide-20
SLIDE 20

The Paradox of Alpha Decay

1 We have probed the 248

90Th potential into an internuclear distance of

rDOCA = 48 fm using a 4He beam of E(4He) = 5.407 MeV.

2 The data are consistent with the Coulomb force and no others. 3 The radioactive decay 238

92U → 248 90Th + 4He emits an α (or 4He) with

energy Eα = 4.2 MeV.

4 For a classical ‘decay’ the emitted α should have an energy of at least

Emin = 5.407 MeV.

5 It appears the ‘decay’ α starts out at a distance remit = 62 fm. 6 How do we explain this?

Quantum Tunneling!

7 What do we measure? Jerry Gilfoyle Alpha Decay 13 / 32

slide-21
SLIDE 21

The Paradox of Alpha Decay

1 We have probed the 248

90Th potential into an internuclear distance of

rDOCA = 48 fm using a 4He beam of E(4He) = 5.407 MeV.

2 The data are consistent with the Coulomb force and no others. 3 The radioactive decay 238

92U → 248 90Th + 4He emits an α (or 4He) with

energy Eα = 4.2 MeV.

4 For a classical ‘decay’ the emitted α should have an energy of at least

Emin = 5.407 MeV.

5 It appears the ‘decay’ α starts out at a distance remit = 62 fm. 6 How do we explain this?

Quantum Tunneling!

7 What do we measure?

Lifetimes t1/2(238U) = 4.5 × 109 yr

Jerry Gilfoyle Alpha Decay 13 / 32

slide-22
SLIDE 22

The Plan For Calculating Nuclear Lifetimes

1 The α particle (4He) is confined by the nuclear potential and

‘bounces’ back and forth between the walls of the nucleus. Assume its energy is the same as the emitted nucleon so v =

  • 2Eα

m .

2 Each time it ‘bounces’ off the nuclear wall it has a finite probability of

tunneling through the barrier equal to the transmission coefficient T.

3 The decay rate will the product of the rate of collisions with a wall

and the probability of transmission equal to

v 2R × T.

4 The lifetime is the inverse of the decay rate 2R

vT = 2R m 2E 1 T .

5 The radius of a nucleus has been found to be described by

rnuke = 1.2A1/3 where A is the mass number of the nucleus.

6 We are liberally copying the work of Gamow, Condon, and Gurney.

Like them we will assume V = 0 inside the nucleus and V = 0 from the classical turning point to infinity.

Jerry Gilfoyle Alpha Decay 14 / 32

slide-23
SLIDE 23

The 4He − 234

90Th Potential

4.2 MeV 20 40 60 80 10 20 30 r(fm) V(MeV)

Gamow, Condon and Gurney

Jerry Gilfoyle Alpha Decay 15 / 32

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SLIDE 24

The Transfer-Matrix Solution

4.2 MeV 20 40 60 80 10 20 30 r(fm) V(MeV)

Jerry Gilfoyle Alpha Decay 16 / 32

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SLIDE 25

Recall For a Single Barrier

ζ1 = tζ3 = d12p2d21p−1

2 ζ3 =

t11 t12 t21 t22

  • ζ3

T = 1 |t11|2 d12 = 1 2

  • 1 + k2

k1

1 − k2

k1

1 − k2

k1

1 + k2

k1

  • d21 = 1

2

  • 1 + k1

k2

1 − k1

k2

1 − k1

k2

1 + k1

k2

  • p−1

1

= eik22a e−ik22a

  • p2 =

e−ik22a eik22a

  • k1 =
  • 2mE

2 k2 =

  • 2m(E − V )

2 t11 = 1 4

  • 1 + k2

k1

  • e−ik22a
  • 1 + k1

k2

  • +
  • 1 − k2

k1

  • eik22a
  • 1 − k1

k2

  • Jerry Gilfoyle

Alpha Decay 17 / 32

slide-26
SLIDE 26

The Transfer-Matrix Solution

4.2 MeV 20 40 60 80 10 20 30 r(fm) V(MeV)

Jerry Gilfoyle Alpha Decay 18 / 32

slide-27
SLIDE 27

The Transfer-Matrix Solution

4.2 MeV 1 2 3 4 5 6 7 20 40 60 80 10 20 30 r(fm) V(MeV)

Jerry Gilfoyle Alpha Decay 19 / 32

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SLIDE 28

The Transfer-Matrix Solution

4.2 MeV 1 2 3 4 5 6 7 20 40 60 80 10 20 30 r(fm) V(MeV)

dnm = 1 2

  • 1 + km

kn

1 − km

kn

1 − km

kn

1 + km

kn

  • pm =

e−ikms eikms

  • k0 =
  • 2mE

2 = k6 kn =

  • 2m(E − Vn)

2 n - left side of barrier m - right side of barrier Vn - potential of nth step. s - step size.

Jerry Gilfoyle Alpha Decay 20 / 32

slide-29
SLIDE 29

The Transfer-Matrix Solution

4.2 MeV 1 2 3 4 5 6 7 20 40 60 80 10 20 30 r(fm) V(MeV)

dnm = 1 2

  • 1 + km

kn

1 − km

kn

1 − km

kn

1 + km

kn

  • pm =

e−ikms eikms

  • k0 =
  • 2mE

2 = k6 kn =

  • 2m(E − Vn)

2 n - left side of barrier m - right side of barrier Vn - potential of nth step. s - step size.

ψ′

1 ∼

= d12p2 · d23p3 · d34p4 · d45p5

unit cell

·d56p6 · d67p7 ψ′

7 ∼

Jerry Gilfoyle Alpha Decay 20 / 32

slide-30
SLIDE 30

The Transfer-Matrix Solution

4.2 MeV 20 40 60 80 10 20 30 r(fm) V(MeV)

Jerry Gilfoyle Alpha Decay 21 / 32

slide-31
SLIDE 31

Nuclear Data

Jerry Gilfoyle Alpha Decay 22 / 32

slide-32
SLIDE 32

Difference Between Lecture and Readings

There are some differences between the formula for Rutherford scattering in the reading (go here) that are discussed below. The lecture formula is dσ dΩ = Z1Z2e2 4Ecm 2 1 sin4

θ 2

  • (1)

while the expression in the reading is the following. dσ dcos θ = π 2 z2Z 2α2 c KE 2 1 (1 − cos θ)2 (2) To go from Eq 1 to Eq 2 you need to make the following changes.

1

Change some variable names so Z1 = z, Z2 = Z, Ecm = KE.

2

Use dΩ = sin θdθdφ = dcos θdφ and integrate over all φ or φ = 0 → 2π. This gives you a factor of 2π in front of Eq 1. dσ d cos θ = 2π dσ dΩ dφ = 2π dσ dΩ (3)

3

Make the following substitutions e2 = αc and sin2 θ 2 = 1 2 (1 − cos θ) (4) and you get Eq 2.

Jerry Gilfoyle Alpha Decay 23 / 32

slide-33
SLIDE 33

Results

5 6 7 8 10-7 0.001 10.000 105 109 1013 1017 Eα(MeV) Lifetime (s) Points->Data, Line->Theory

Jerry Gilfoyle Alpha Decay 24 / 32

slide-34
SLIDE 34

Additional slides.

Jerry Gilfoyle Alpha Decay 25 / 32

slide-35
SLIDE 35

What is an Angle?

Jerry Gilfoyle Alpha Decay 26 / 32

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SLIDE 36

What is an Angle?

θ d θ ds

x y r

dθ = ds | r|

Jerry Gilfoyle Alpha Decay 26 / 32

slide-37
SLIDE 37

Solid Angle

Jerry Gilfoyle Alpha Decay 27 / 32

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SLIDE 38

Solid Angle

Jerry Gilfoyle Alpha Decay 28 / 32

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SLIDE 39

Solid Angle

Jerry Gilfoyle Alpha Decay 29 / 32

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SLIDE 40

Solid Angle

Jerry Gilfoyle Alpha Decay 30 / 32

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SLIDE 41

Solid Angle

Jerry Gilfoyle Alpha Decay 31 / 32

slide-42
SLIDE 42

Solid Angle

Jerry Gilfoyle Alpha Decay 32 / 32

slide-43
SLIDE 43

Solid Angle

dA = rdθ × r sin θdφ = r2 sin θdθdφ

Jerry Gilfoyle Alpha Decay 32 / 32

slide-44
SLIDE 44

Solid Angle

dA = rdθ × r sin θdφ = r2 sin θdθdφ

dΩ = dA

r2 = sin θdθdφ

Jerry Gilfoyle Alpha Decay 32 / 32