Physics Lecture Highlights: Rutherford Scattering and Elastic Collisions

PHYS 3446 Lecture #3 Wednesday, Sept 7, 2016 Dr. Jae Jae Yu 1. Rutherford Scattering 2. Rutherford Scattering with Coulomb force 3. Scattering Cross Section 4. Measurement of Cross Sections Wednesday, Sept. 7, 2016 PHYS 3446, Fall 2016 1

Announcements 1st colloquium at 4pm today UTA Physics faculty expo Physics department picnic 12 3pm, Sat. Sept. 10 1st Floor MAC Wednesday, Sept. 7, 2016 PHYS 3446, Fall 2016 2

Reminder: Homework Assignment #1 Compute the masses of electron, proton and alpha particles in MeV/c2, using E=mc2. (9 points) Need to look up and specify the masses of electrons, protons and alpha particles in kg on your paper. Compute the gravitational and the Coulomb forces between two protons separated by 10-10m and compare their strengths (15) Derive the following equations in your book: Eq. # 1.3 (5 pts) , 1.17 (8 pts), 1.32 (12 pts) Must show detailed work and accompany explanations Copying the book or your friend will result in no credit for both of you! These assignments are due coming Monday, Sept. 12. 1. 2. 3. Wednesday, Sept. 7, 2016 PHYS 3446, Fall 2016 3

Rutherford Scattering A fixed target experiment with alpha particle as projectile shot on thin gold foil Alpha particle s energy is low Speed is well below 0.1c (non-relativistic) An elastic scattering of the particles What are the conserved quantities in an elastic scattering? Momentum Kinetic Energy Wednesday, Sept. 7, 2016 PHYS 3446, Fall 2016 4

Elastic Scattering ! v a ! v 0 After the Collision m ! mt v t From momentum conservation ! ! ! ! ! a +mtv ma 0 =mav a +mtv t t = v v ma From the kinetic energy conservation 1 2mav0 2+mt 2=1 2+1 2= va 2 v0 vt 2 2mava 2mtvt ma From these two, we obtain = 2v ! ! 21-mt a v vt t ma Wednesday, Sept. 7, 2016 PHYS 3446, Fall 2016 5

Analysis Case I = 2v ! ! 21-mt a v vt t ma If mt<<m , left-hand side becomes positive v and vt must be in the same hemisphere Using the actual masses and We obtain If mt=me, then mt/m ~10-4. Thus, momentum transfer to target is pe/p 0<10-4. Change of momentum of alpha particle is negligible ma= 4 103MeV c2 me= 0.5MeV c2 ve= vt 2va va v0 ( )ve 2 10-4mava= 2 10-4mav0 meve= mamema Wednesday, Sept. 7, 2016 PHYS 3446, Fall 2016 6

= 2v ! ! 21-mt Analysis Case II If mt>>m , left-hand side of the above becomes negative v and vt are in opposite hemisphere Using the actual masses and We obtain so vt is small If mt=mAu, then mt/m ~50. Thus, pe/p 0 could be as large as 2p 0. Change of momentum of alpha particle is large particle can even recoil a v vt t ma ma= 4 103MeV c2 mt mAu 2 105MeV c2 vt 2mavamt va v0 mtvt 2mava 2mav0 Wednesday, Sept. 7, 2016 PHYS 3446, Fall 2016 7

Rutherford Scattering with EM Force 1 Let s take into account only the EM force between the and the atom Coulomb force is a central force, so a conservative force Coulomb potential between particles with Ze and Z e electrical charge separated by distance r is 2 ' ZZ e ( ) = V r r Since the total mechanical energy is conserved, 1 2 2 m E 2 0 = = = constant>0 v E mv 0 Wednesday, Sept. 7, 2016 PHYS 3446, Fall 2016 8

Rutherford Scattering with EM Force 2 The distance vector r is always the same direction as the force throughout the entire motion, so the net torque (rxF) is 0. Since there is no net torque, the angular momentum (l=rxp) is conserved. The magnitude of the angular momentum is l=mv0b. Impact parameter From the energy relation, we obtain 2 l m E mb = 2 22 = = 2 b l mE b mE From the definition of angular momentum, we obtain an equation of motion From energy conservation, we obtain another equation of motion E =1 2m dt 2mr2 dc dt dc dt =l mr2 2 dr l2 2 m dr dt= Centrifugal barrier E-V r ( )- 2mr2 2 +1 +V r ( ) Effective potential Wednesday, Sept. 7, 2016 PHYS 3446, Fall 2016 9

Rutherford Scattering with EM Force 3 Rearranging the terms for approach, we obtain r21-V r ( ) bdr r r21-V r ( ) E -b2 dr dt= - l mrb E and dc = - 1 2 -b2 Integrating this from r0 to infinity gives the angular distribution of the outgoing alpha particle Distance of closest approach Wednesday, Sept. 7, 2016 PHYS 3446, Fall 2016 10

Rutherford Scattering with EM Force 4 What happens at the DCA? Kinetic energy reduces to 0. dr dt = 0 r r = 0 The incident alpha could turn around and accelerate We can obtain r21-V r ( ) -b2= 0 E This allows us to determine DCA for a given potential and 0. Define scattering angle as the changes in the asymptotic angles of the trajectory, we obtain dr q =p -2c0=p -2b r r21-V r ( ) 1 2 -b2 r0 E Wednesday, Sept. 7, 2016 PHYS 3446, Fall 2016 11

Rutherford Scattering with EM Force 5 For a Coulomb potential 2 ' ZZ e ( ) = V r r DCA can be obtained for the given impact parameter b, r0=Z Z e2E 2 ( ) 2 1+ 1+4b2E2 Z Z e2 And the angular distribution becomes dr + q =p -2b r r21-V r ( ) 1 2 -b2 r0 E Wednesday, Sept. 7, 2016 PHYS 3446, Fall 2016 12

Rutherford Scattering with EM Force 6 Replace the variable 1/r=x, and performing the integration, we obtain 1 q =p +2cos-1 ( ) 2 1+ 4b2E2 Z Z e2 This can be rewritten 2= cosq -p 1 ( ) 2 1+4b2E2 Z Z e2 Solving this for b, we obtain b =Z cotq Z e2 2E 2 Wednesday, Sept. 7, 2016 PHYS 3446, Fall 2016 13

Rutherford Scattering with EM Force 7 b =Z 2E 2 cotq tanq 2=Z Z e2 Z e2 2bE From the solution for b, we can learn the following 1. For fixed b, E and Z The scattering angle is larger for a larger value of Z. Makes perfect sense since Coulomb potential is stronger with larger Z. Results in larger deflection. 2. For fixed b, Z and Z The scattering angle is larger when E is smaller. If particle has low energy, its velocity is smaller Spends more time in the potential, suffering greater deflection 3. For fixed Z, Z , and E The scattering angle is larger for smaller impact parameter b Makes perfect sense also, since as the incident particle is closer to the nucleus, it feels stronger Coulomb force. Wednesday, Sept. 7, 2016 PHYS 3446, Fall 2016 14

What do we learn from scattering? Scattering of a particle in a potential is completely determined when we know both The impact parameter, b, and The energy of the incident particle, E For a fixed energy, the deflection is defined by The impact parameter, b. What do we need to perform a scattering experiment? Incident flux of beam particles with known E A device that can measure the number of scattered particles at various scattering angle, . Measurements of the number of scattered particles reflect Impact parameters of the incident particles The effective size of the scattering center By measuring the scattering angle , we can learn about the potential or the force between the target and the projectile cotq b =Z Z e2 2E 2 Wednesday, Sept. 7, 2016 PHYS 3446, Fall 2016 15

Scattering Cross Section All these land here N0: The number of particles incident on the target foil per unit area per unit time. Any incident particles entering with impact parameter b and b+db will scatter to the angle and -d In other words, they scatter into the solid angle d (=2 sin d ) So the number of particles scattered into the solid angle d per unit time is 2 N0bdb. Wednesday, Sept. 7, 2016 PHYS 3446, Fall 2016 16

Scattering Cross Section For a central potential Such as Coulomb potential Which has spherical symmetry The scattering center presents an effective transverse x-sectional area of Ds = 2pbdb For the particles to scatter into and +d Wednesday, Sept. 7, 2016 PHYS 3446, Fall 2016 17

Scattering Cross Section In more generalized cases, depends on both & . )= bdbdf = -ds dWq,f )dW = -ds Ds q,f ( ( ( )sinqdqdf dWq,f Why negative? Since the deflection and the change of b are in opposite direction!! With a spherical symmetry, can be integrated out: Ds q ( )=ds dWq ( )2psinqdq = 2pbdb What is the dimension of the differential cross section? ds dWq ( )= - b db dq Differential Cross Section reorganize sinq Area!! Wednesday, Sept. 7, 2016 PHYS 3446, Fall 2016 18

Scattering Cross Section For a central potential, measuring the yield as a function of , or the differential cross section, is equivalent to measuring the entire effect of the scattering So what is the physical meaning of the differential cross section? Measurement of yield as a function of specific experimental variable This is equivalent to measuring the probability of occurrence of a physical process in a specific kinematic phase space Cross sections are measured in the unit of barns: 1barn=10-24cm2 Where does this come from? Cross sectional area of a uranium nucleus! Wednesday, Sept. 7, 2016 PHYS 3446, Fall 2016 19 couldn't hit the broad side of a barn!!