PHYS 3446 – Lecture #2

Wednesday, Aug. 30, 2006 PHYS 3446, Fall 2006Jae Yu

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PHYS 3446 – Lecture #2Wednesday, Aug. 30, 2006

Dr. Jae Yu

1. Introduction2. History on Atomic Models3. Rutherford Scattering4. Rutherford Scattering with Coulomb force5. Scattering Cross Section6. Measurement of Cross Sections


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Announcements• Mail distribution list is ready. Go ahead and subscribe!

– 5 points extra credit if done by next Wednesday, Sept. 6– 3 points extra credit if done by next Friday, Sept. 8

• Extra credit opportunities– Obtain a particle data booklet and a pocket diary

• http://pdg.lbl.gov/pdgmail/– Subscribe to the Symmetry Magazine (free!)

• http://www.symmetrymagazine.org• Request for printed copies

– Subscribe to the printed copies of CERN Courier Magazine• http://www.cerncourier.com/• Send e-mail to Janice Voss at [email protected]

– 3 points each if done by Wednesday, Sept. 6.


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Why do Physics?• To understand nature through experimental

observations and measurements (Research)• Establish limited number of fundamental laws, usually

with mathematical expressions• Predict nature’s courses⇒Theory and Experiment work hand-in-hand⇒Theory works generally under restricted conditions⇒Discrepancies between experimental measurements

and theory presents opportunities to quantum leap⇒Improves our everyday lives, though some laws can

take a while till we see amongst us

Exp.Theory


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High Energy Physics

Structure of Matter

10-10m 10-14m 10-15m

<10-18m

10-9m

Matter Molecule Atom Nucleus

u

Quark

<10-19mprotons, neutrons,

mesons, etc.π,Ω,Λ...

top, bottom,charm, strange,

up, down

Condensed matter/Nano-Science/ChemistryAtomic Physics

NuclearPhysics

Baryon(Hadron)

Electron(Lepton)

10-2m


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Theory for Microscopic Scale, QM• Difficult to describe Small scale phenom. w/ just CM and EM• The study of atomic structure led to quantum mechanics

– Long range EM force is responsible for holding atoms together (why ?)– Yet sufficiently weak for QM to estimate properties of atoms reliably

• In Nucleus regime, the simple Coulomb force does not work. Why?– The force in nucleus holds positively charged particles together

• The known forces in nature– Strong ~ 1 – Electro-magnetic ~ 10-2

– Weak ~ 10-5

– Gravitational ~ 10-38


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Evolution of Atomic Models• 1803: Dalton’s billiard ball model• 1897: J.J. Thompson Discovered

electrons– Built on all the work w/ cathode

tubes– Called corpuscles– Made a bold claim that these make

up atoms– Measured m/e ratio

• 1904: J.J. Thompson Proposed a “plum pudding” model of atoms – Negatively charged electrons

embedded in a uniformly distributed positive charge

Cathode ray tube

Thompson’s tubes


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History of Atomic Models cnt’d• 1911: Geiger and

Marsden with Rutherford performed a scattering experiment with alpha particles shot on a thin gold foil


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History of Atomic Models cnt’d• 1912: Rutherford’s

planetary model, an atomic model with a positively charged heavy core surrounded by circling electrons – Deficiency of

instability. Why?• The electrons will

eventually get pulled onto the nucleus, destroying the atom


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History of Atomic Models cnt’d• 1913: Neils

Bohr proposed the Orbit Model, where electrons occupy well quantified orbits– Electrons can

only transition to pre-defined orbits


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History of Atomic Models cnt’d

• 1926: Schrödinger and de Broglie proposed the Electron Cloud Model based on quantum mechanics


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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


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Elastic Scattering

• From momentum conservation

• From kinetic energy conservation

• From these two, we obtain

mt

mα

mα θ

mt

φAfter Collisions0v

vα

tv

0v =

20v =

2 1 tt

mvmα

⎛ ⎞− =⎜ ⎟

⎝ ⎠

ttm v m vmαα

α

+= t

tmv vmα

α

+

2 2tt

mv vmαα

+

2 tv vα ⋅


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Analysis Case I• If mt<<mα,

– left-hand side becomes positive– vα and vt must be in the same direction– Using the actual masses– and– We obtain – If mt=me, then mt/mα~10-4. – Thus, pe/pα0<10-4. – Change of momentum of alpha particle is negligible

2 1 2ttt

mv v vm

αα

⎛ ⎞− = ⋅⎜ ⎟

⎝ ⎠

20.5 /em MeV c≈

0v vα ≈2e tv v vα= ≤

3 24 10 /m MeV cα ≈ ×


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Analysis Case II• If mt>>mα,

– left-hand side of the above becomes negative– vα and vt is opposite direction– Using the actual masses– and – We obtain– If mt=mAu, then mt/mα~50. – Thus, pe/pα0~2pα0. – Change of momentum of alpha particle is large

• α particle can even recoil

2 1 2ttt

mv v vm

αα

⎛ ⎞− = ⋅⎜ ⎟

⎝ ⎠

5 22 10 /t Aum m MeV c≈ ≈ ×

0v vα ≈

3 24 10 /m MeV cα ≈ ×2t tv m v mα α≤


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Rutherford Scattering with EM Force 1• Let’s take into account only the EM force between the α

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

• Since the total energy is conserved, ( )

2'ZZ eV rr

=

20

1 constant>02

E mv= = 02 v Em

⇒ =


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• 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 mb b mE= =

( )2

22

2dr lE V rdt m mr

⎛ ⎞= ± − −⎜ ⎟

⎝ ⎠

2d dt l mrχ =

Centrifugal barrier

Effective potential

Impact parameter

2 2 2b l mE⇒ =

( )

2

22

12

12

drE mdt

dmr V rdtχ

⎛ ⎞= ⎜ ⎟⎝ ⎠

⎛ ⎞+ +⎜ ⎟⎝ ⎠


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

• and

• Integrating this from r0 to infinity gives the angular distribution of the outgoing alpha particle

( )2 21V rdr l r b

dt mrb E⎛ ⎞

= − − −⎜ ⎟⎜ ⎟⎝ ⎠

( )1 2

2 21

bdrdV r

r r bE

χ = −⎡ ⎤⎛ ⎞

− −⎢ ⎥⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦

Distance of closest approach


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

– Kinetic energy reduces to 0.

– The incident alpha could turn around and accelerate– We can obtain

– This allows us to determine DCA for a given potential and χ0.• Define scattering angle θ as the changes in the asymptotic

angles of the trajectory, we obtain

0

0r r

drdt =

=

( )02 20 1 0

V rr b

E⎛ ⎞− − =⎜ ⎟⎜ ⎟

⎝ ⎠

( )00 1 2

2 2

2 2

1r

drbV r

r r bE

θ π χ π∞

= − = −⎡ ⎤⎛ ⎞

− −⎢ ⎥⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦

∫


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Rutherford Scattering with EM Force 5• For a Coulomb potential

• DCA can be obtained for the 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 22

2 2

2'1

r

drbZZ er r b

rE

θ π∞

= −⎡ ⎤⎛ ⎞

− −⎢ ⎥⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦

∫

( )2'ZZ eV r

r=


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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 cos1 4 '

bb E ZZ e

θ π −

⎛ ⎞⎜ ⎟

= + ⎜ ⎟⎜ ⎟+⎝ ⎠

( )22 2 2

1 cos21 4 'b E ZZ e

θ π−⎛ ⎞= ⎜ ⎟⎝ ⎠+

2' cot2 2

ZZ ebE

θ=


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Rutherford Scattering with EM Force 7

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

– The scattering 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.

2' cot2 2

ZZ ebE

θ=


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Assignments1. Compute the masses of electron, proton and alpha particles

in MeV/c2, using E=mc2.• Need to look up masses of electrons, protons and alpha particles

in kg.2. Compute the gravitational and the Coulomb forces between

two protons separated by 10-10m and compare their strengths

3. Derive the following equations in your book:• Eq. # 1.3, 1.17, 1.32• Must show detailed work and accompany explanations

• These assignments are due next Wednesday, Sept. 6.

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PHYS 3446 – Lecture #2