Monday, Sep. 13, 2010PHYS 3446, Fall 2010 Andrew Brandt 1 PHYS 3446 – Lecture #3 Monday, Sep. 13 2010 Dr. Brandt 1.Rutherford Scattering with Coulomb force

Monday, Sep. 13, 2010 PHYS 3446, Fall 2010 Andrew Brandt 1

PHYS 3446 – Lecture #3Monday, Sep. 13 2010

Dr. Brandt

1. Rutherford Scattering with Coulomb force2. Scattering Cross Section3. Differential Cross Section of Rutherford Scattering4. Measurement of Cross Sections


Elastic Scattering

• From momentum and kinetic energy conservation

mt

m

m

mt

After Collision0v

v

tv

20v

2 1 tt

mv

m

2 2tt

mv v

m

2 tv v

***Eq. 1.3

***Eq. 1.2


Analysis Case 1• If mt<<m,

– change of momentum of alpha particle is negligible

2 1 2ttt

mv v v

m

• If mt>>m, – alpha particle deflected backwards

Analysis Case 2


Rutherford Scattering with EM Force 1• Need to take into account the EM force between the and

the atom • The Coulomb force is a central force->conservative force• The Coulomb potential energy between particles with Ze and

Z’e electrical charge, separated by a distance r is

• Since the total energy is conserved,

2'ZZ eV r

r

20

1constant>0

2E mv 0

2 v

E

m

PHYS 3446, Fall 2010 Andrew Brandt 5

• From the energy relation, we obtain

• From the definition of angular momentum, we obtain an equation of motion

• From energy conservation, we obtain another equation of motion

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=mvb.

2 2l m E m b b mE

2

2

2

2

dr lE V r

dt m mr

2d dt l mr

Centrifugal barrier

Effective potential

Impact parameter=distance of closestapproach if there were no Coulomb force

2 2 2b l mE

2

22

1

2

1

2

drE m

dt

dmr V r

dt

Eq. 1.17

Monday, Sep. 13, 2010


Rutherford Scattering with EM Force 3• Rearranging the terms, we obtain

• and

2 21V rdr l

r bdt mrb E

1 2

2 21

bdrd

V rr r b

E

Distance of closest approach

-sign since r decreases as alpha particle approaches target

Eq. 1.18

Asymptotic scattering angle:

02

Eq. 1.19


Rutherford Scattering with EM Force 4• What happens at the DCA?

– Kinetic energy goes to 0.

– Consider the case where the alpha particle is incident on the z-axis, it would reach the DCA, stop, and reverse direction!

– From Eq. 1.18, we can obtain

– This allows us to determine DCA for a given potential and 0.• Substituting 1.19 into the scattering angle equation gives:

0

0r r

dr

dt

02 20 1 0

V rr b

E

00 1 2

2 2

2 2

1r

drb

V rr r b

E


Rutherford Scattering with EM Force 5• For a Coulomb potential

• DCA can be obtained for a given impact parameter b,

• And the angular distribution becomes

2 22 2 2

0

'1 1 4 '

2

ZZ e Er b E ZZ e

01 2

22 2

2'

1r

drb

ZZ er r b

rE

2'ZZ e

V rr


Rutherford Scattering with EM Force 6• Replace the variable 1/r=x, and performing the

integration, we obtain

• This can be rewritten

• Solving this for b, we obtain

1

22 2 2

12 cos

1 4 'b

b E ZZ e

22 2 2

1cos

21 4 'b E ZZ e

2'cot

2 2

ZZ eb

E


Rutherford Scattering with EM Force 7

• From the solution for b, we can learn the following 1. For fixed b and E

– The scattering is larger for a larger value of Z or Z’ (large charge in projectile or target)– 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.

2'cot

2 2

ZZ eb

E


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– Device that can measure number of scattered particles at various

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 forces between the target and the projectile

2'cot

2 2

ZZ eb

E


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 d2sind• So the number of particles scattered into the solid angle d per unit time

is 2N0bdb.• Note:have assumed thin foil, and large separation between nuclei—why?

Scattering Cross Section


Scattering Cross Section• For a central potential

– Such as Coulomb potential– Which has spherical symmetry

• The scattering center presents an effective transverse cross-sectional area of

• For the particles to scatter into and +d

2 bdb


Scattering Cross Section• In more generalized cases, depends on both & .

• With a spherical symmetry, can be integrated out:

,

d

d

Differential Cross Section

What is the dimension of the differential cross section?

Area!!

reorganize

bdbd ,d

dd

, sind

d dd

2 sind

dd

2 bdb

Why negative? Since the deflection and change of b are in opposite direction!!

sin

b db

d


Scattering Cross Section• For a central potential, measuring the yield as a function of 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 variables

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:

1 barn Where does this come from?

-24 210 cmCross sectional area of a typical nucleus! nanobarn

picobarn

femptobarn


Total Cross Section• Total cross section is the integration of the differential

cross section over the entire solid angle, :

• Total cross section represents the effective size of the scattering center integrated over all possible impact parameters (and consequently all possible scattering angles)

Total 4

0,

dd

d

02 sin

dd

d


Cross Section of Rutherford Scattering• The impact parameter in Rutherford scattering is

• Thus,

• Differential cross section of Rutherford scattering is

2'cot

2 2

ZZ eb

E

db

d

d

d

224'

cosec4 2

ZZ e

E

22

4

' 1

4 sin2

ZZ e

E

221 '

cosec2 2 2

ZZ e

E

sin

b db

d

what happened? can you say “trig identity?” sin(2x)=2sin(x) cos(x)plot/applets


Measuring Cross Sections

• Rutherford scattering experiment– Used a collimated beam of particles emitted from Radon– A thin Au foil target– A scintillating glass screen with ZnS phosphor deposit– Telescope to view limited area of solid angle– Telescope only needs to move along not . Why?

• Due to the spherical symmetry, scattering only depends on not .


Total X-Section of Rutherford Scattering• To obtain the total cross section of Rutherford scattering, one

integrates the differential cross section over all :

• What is the result of this integration?– Infinity!!

• Does this make sense?– Yes

• Why?– Since the Coulomb force’s range is infinite (particle with very large impact

parameter still contributes to integral through very small scattering angle)• What would be the sensible thing to do?

– Integrate to a cut-off angle since after certain distance the force is too weak to impact the scattering. (0>0); note this is sensible since alpha particles far away don’t even see charge of nucleus due to screening effects.

Total 0

2 sind

dd

22 1

0 3

' 18 sin

4 2 sin2

ZZ ed

E


Measuring Cross Sections• With the flux of N0 per unit area per second• Any particles in range b to b+db will be

scattered into to -d• The telescope aperture limits the measurable

area to

• How could they have increased the rate of measurement?– By constructing an annular telescope– By how much would it increase?

TeleA

2/d

sinRd R d 2R d


Measuring Cross Sections• Fraction of incident particles (N0) approaching the target in

the small area =bddb at impact parameter b is dn/N0. – so dn particles scatter into R2d, the aperture of the telescope

• This fraction is the same as – The sum of over all N nuclear centers throughout the foil

divided by the total area (S) of the foil.– Or, in other words, the probability for incident particles to enter

within the N areas divided by the probability of hitting the foil. This ratio can be expressed as

0

dn

N Nbd d

S

,N

S Eq. 1.39


Measuring Cross Sections• For a foil with thickness t, mass density , atomic weight A:

• Since from what we have learned previously

• The number of scattered into the detector angle () is

N A0: Avogadro’s number of atoms per mole

dn

0

tSA

A

0 0N tA

A

0

,dNN dS d

,d

dd

0

dn

N

Nbd d

S

Eq. 1.40


Measuring Cross Sections

dn

• This is a general expression for any scattering process, independent of the theory

• This gives an observed counts per second

0 0N tA

A

0

,dNN dS d

,d

dd

Number of detected particles/sec

Projectile particle flux

Density of the target particles

Scattering cross section

Detector acceptance


• Inclusive jet production cross section as a function of transverse energy

Example Cross Section: Jet +X

triumph ofQCD

what if there werean excess at highenergy?


Assignment #2 Due Mon. Sep. 201. Plot the differential cross section of the Rutherford scattering

as a function of the scattering angle for three sensible choices of the lower limit of the angle.

(use ZAu=79, Zhe=2, E=10keV).

2. Compute the total cross section of the Rutherford scattering in unit of barns for your cut-off angles.

3. Find a plot of a cross section from a current HEP experiment, and write a few sentences about what is being measured.

• 5 points extra credit if you are one of first 10 people to email an electronic version of a figure showing Rutherford 1/sin^4(x/2) angular dependence

Documents

Monday, Sep. 13, 2010PHYS 3446, Fall 2010 Andrew Brandt 1 PHYS 3446 – Lecture #3 Monday, Sep. 13 2010 Dr. Brandt 1.Rutherford Scattering with Coulomb force