Why Does Uranium Alpha Decay?

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 92U 4 He + 234 90Th E( 4 He) = 4.2 MeV To study this reaction we first map out the 4 He 234 90Th potential energy. We reverse the decay above and use a beam of 4 He nuclei striking a 234 90 Th target. The 4 He beam comes from the radioactive decay of another nucleus 210 84 Po and E(4 He) = 5.407 MeV. 1 What is the distance of closest approach of the 4 He to the 234 90 Th 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 92 U? Jerry Gilfoyle Alpha Decay 1 / 26

What Do We Know? The 234 90Th α Potential V = Z 1Z 2 e 2 r V(r) (MeV) attractive nuclear part r (fm) Jerry Gilfoyle Alpha Decay 2 / 26

Mapping the Potential Energy Rutherford Scattering What is the distance of closest approach of the 4 He to the 234 90Th target if only the Coulomb force is active? Is the Coulomb force the only one active? The energy of the 4 He emitted by the 210 84Po to make the beam is E( 4 He) = 5.407 MeV. Collimator Alpha source 210 84 4 2 206 82 Po He + Pb ZnS Microscope Scattered helium Alpha beam Thorium target Jerry Gilfoyle Alpha Decay 3 / 26

Rutherford Trajectories Rutherford trajectories for different impact parameters y/b x Jerry Gilfoyle Alpha Decay 4 / 26

Mapping the Potential Energy The Differential Cross Section y z 4 He trajectory Pb target z axis Jerry Gilfoyle Alpha Decay 5 / 26

Mapping the Potential Energy The Differential Cross Section y z 4 He trajectory Pb target z axis dσ dω = ( Z1 Z 2 e 2 ) 2 1 4E cm sin 4 ( θ 2 ) Jerry Gilfoyle Alpha Decay 5 / 26

Actual Rutherford Scattering Results (θ/2) 4 Counts/sin 50 45 40 35 30 25 20 15 10 H.Geiger and E.Marsden, Phil. Mag., 25, p. 604 (1913) alphas on gold 5 0 0 20 40 60 80 100 120 140 160 180 θ (deg) 2016-12-18 10:38:14 Jerry Gilfoyle Alpha Decay 6 / 26

Actual Rutherford Scattering Results (θ/2) 4 Counts/sin 50 45 40 35 dσ dω = ( Z1 Z 2 e 2 ) 2 1 4E cm sin 4 ( θ 2 ) 30 25 20 15 10 H.Geiger and E.Marsden, Phil. Mag., 25, p. 604 (1913) alphas on gold 5 0 0 20 40 60 80 100 120 140 160 180 θ (deg) 2016-12-18 10:38:14 Jerry Gilfoyle Alpha Decay 6 / 26

Interpretation of Rutherford Scattering Results 1.2 1.0 Measured/Predicted 0.8 0.6 0.4 0.2 0.0 0 50 100 150 Scattering Angle (deg) What does this say about the 4 2He 234 90 Th potential energy? Jerry Gilfoyle Alpha Decay 7 / 26

Measuring the Size of the Nucleus θ cm (deg) Jerry Gilfoyle Alpha Decay 8 / 26

Measuring the Size of the Nucleus PRL 109, 262701 (2012) E cm = 23.1 MeV E cm = 28.3 MeV θ cm (deg) Jerry Gilfoyle Alpha Decay 8 / 26

The 4 He 234 90Th Potential 40 20 α-th Potential Blue - known Red - a guess V(MeV) 0-20 -40 0 10 20 30 40 50 60 70 r(fm) Jerry Gilfoyle Alpha Decay 9 / 26

The Paradox of Alpha Decay 1 We have probed the 248 90Th potential into an internuclear distance of r DOCA = 48 fm using a 4 He beam of E( 4 He) = 5.407 MeV. 2 The data are consistent with the Coulomb force and no others. 3 The radioactive decay 238 92 U 248 90 Th + 4 He emits an α (or 4 He) with energy E α = 4.2 MeV. 4 For a classical decay the emitted α should have an energy of at least E min = 5.407 MeV. 5 It appears the decay α starts out at a distance r emit = 62 fm. 6 How do we explain this? Jerry Gilfoyle Alpha Decay 10 / 26

The Paradox of Alpha Decay 1 We have probed the 248 90Th potential into an internuclear distance of r DOCA = 48 fm using a 4 He beam of E( 4 He) = 5.407 MeV. 2 The data are consistent with the Coulomb force and no others. 3 The radioactive decay 238 92 U 248 90 Th + 4 He emits an α (or 4 He) with energy E α = 4.2 MeV. 4 For a classical decay the emitted α should have an energy of at least E min = 5.407 MeV. 5 It appears the decay α starts out at a distance r emit = 62 fm. 6 How do we explain this? Quantum Tunneling! Jerry Gilfoyle Alpha Decay 10 / 26

The Plan For Calculating Nuclear Lifetimes 1 The α particle ( 4 He) 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. 2 Each time it bounces off the nuclear wall it has a finite probability of tunneling through the barrier. 3 The decay rate will the product of the rate of collisions with a wall and the probability of transmission. 4 The lifetime is the inverse of the decay rate. 5 The radius of a nucleus has been found to be described by r nuke = 1.2A 1/3 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 11 / 26

The 4 He 234 90Th Potential 30 Gamow, Condon and Gurney 20 V(MeV) 10 4.2 MeV 0 0 20 40 60 80 r(fm) Jerry Gilfoyle Alpha Decay 12 / 26

The Transfer-Matrix Solution 30 20 V(MeV) 10 4.2 MeV 0 0 20 40 60 80 r(fm) Jerry Gilfoyle Alpha Decay 13 / 26

Recall For a Single Barrier ζ 1 = tζ 3 = d 12 p 2 d 21 p 1 2 ζ 3 = t 11 = 1 4 d 12 = 1 2 ( 1 + k 2 k 1 1 k 2 k 1 1 k 2 k 1 1 + k 2 k 1 ) ( p 1 e ik 2 2a 0 1 = 0 e ik 22a k 1 = ( t11 t 12 t 21 t 22 ) d 21 = 1 2 p 2 = ) ζ 3 T = 1 t 11 2 ( 1 + k 1 k 2 1 k 1 k 2 1 k 1 k 2 1 + k 1 k 2 ) ( e ik 2 2a 0 0 e ik 22a 2mE 2m(E V ) 2 k 2 = 2 [( 1 + k ) ( 2 e ik 22a 1 + k ) ( 1 + 1 k ) ( 2 e ik 22a 1 k )] 1 k 1 k 2 k 1 k 2 ) Jerry Gilfoyle Alpha Decay 14 / 26

The Transfer-Matrix Solution 30 20 V(MeV) 10 4.2 MeV 0 0 20 40 60 80 r(fm) Jerry Gilfoyle Alpha Decay 15 / 26

The Transfer-Matrix Solution 30 20 V(MeV) 10 4.2 MeV 0 0 20 40 60 80 r(fm) Jerry Gilfoyle Alpha Decay 16 / 26

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 = ( Z1 Z 2 e 2 ) 2 1 4E cm while the expression in the reading is the following. dσ dcos θ = π 2 z2 Z 2 α 2 [ c KE sin 4 ( θ 2 To go from Eq 1 to Eq 2 you need to make the following changes. ) (1) ] 2 1 (1 cos θ) 2 (2) 1 Change some variable names so Z 1 = z, Z 2 = Z, E cm = 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. 3 Make the following substitutions and you get Eq 2. dσ d cos θ = 2π 0 dσ dσ dφ = 2π dω dω e 2 = α c and sin 2 θ 2 = 1 (1 cos θ) (4) 2 Jerry Gilfoyle Alpha Decay 17 / 26 (3)

Results Points->Data, Line->Theory 10 17 10 13 Lifetime (s) 10 9 10 5 10.000 0.001 10-7 5 6 7 8 E α (MeV) Jerry Gilfoyle Alpha Decay 18 / 26

Additional slides. Jerry Gilfoyle Alpha Decay 19 / 26

What is an Angle? Jerry Gilfoyle Alpha Decay 20 / 26

What is an Angle? y dθ θ r ds x dθ = ds r Jerry Gilfoyle Alpha Decay 20 / 26

Solid Angle Jerry Gilfoyle Alpha Decay 21 / 26

Solid Angle Jerry Gilfoyle Alpha Decay 22 / 26

Solid Angle Jerry Gilfoyle Alpha Decay 23 / 26

Solid Angle Jerry Gilfoyle Alpha Decay 24 / 26

Solid Angle Jerry Gilfoyle Alpha Decay 25 / 26

Solid Angle Jerry Gilfoyle Alpha Decay 26 / 26

Solid Angle da = rdθ r sin θdφ = r 2 sin θdθdφ Jerry Gilfoyle Alpha Decay 26 / 26

Solid Angle da = rdθ r sin θdφ = r 2 sin θdθdφ dω = da r 2 = sin θdθdφ Jerry Gilfoyle Alpha Decay 26 / 26