Download Wednesday, Aug. 30, 2006

PHYS 3446 – Lecture #2
Wednesday, Aug. 30, 2006
Dr. Jae Yu
1.
2.
3.
4.
5.
6.
Introduction
History on Atomic Models
Rutherford Scattering
Rutherford Scattering with Coulomb force
Scattering Cross Section
Measurement of Cross Sections
Wednesday, Aug. 30, 2006
PHYS 3446, Fall 2006
Jae Yu
1
Announcements
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– 5 points extra credit if done by next Wednesday, Sept. 6
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• Send e-mail to Janice Voss at [email protected]
– 3 points each if done by Wednesday, Sept. 6.
Wednesday, Aug. 30, 2006
PHYS 3446, Fall 2006
Jae Yu
2
Why do Physics?
{
• To understand nature through experimental
Exp. observations and measurements (Research)
• Establish limited number of fundamental laws, usually
Theory 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
{
Wednesday, Aug. 30, 2006
PHYS 3446, Fall 2006
Jae Yu
3
Structure of Matter
Matter
Molecule
Atom
Nucleus
Baryon
Quark
(Hadron)
u
10-14m
10-9m
10-10m
10-2m
Condensed matter/Nano-Science/Chemistry
Atomic Physics
Nuclear
Physics
10-15m
protons, neutrons,
mesons, etc.
p,W,L...
<10-19m
top, bottom,
charm, strange,
up, down
Electron
(Lepton)
<10-18m
High Energy Physics
Wednesday, Aug. 30, 2006
PHYS 3446, Fall 2006
Jae Yu
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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
Wednesday, Aug. 30, 2006
PHYS 3446, Fall 2006
Jae Yu
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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
Cathode ray tube
Thompson’s tubes
• 1904: J.J. Thompson Proposed a
“plum pudding” model of atoms
– Negatively charged electrons
embedded in a uniformly distributed
positive charge
Wednesday, Aug. 30, 2006
PHYS 3446, Fall 2006
Jae Yu
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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
Wednesday, Aug. 30, 2006
PHYS 3446, Fall 2006
Jae Yu
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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
Wednesday, Aug. 30, 2006
PHYS 3446, Fall 2006
Jae Yu
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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
Wednesday, Aug. 30, 2006
PHYS 3446, Fall 2006
Jae Yu
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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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PHYS 3446, Fall 2006
Jae Yu
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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
Wednesday, Aug. 30, 2006
PHYS 3446, Fall 2006
Jae Yu
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Elastic Scattering
ma
v0
q
After Collisions
mt
f
• From momentum conservation
ma v a  mt v t
mt
 va 
vt
v0 
ma
ma
• From kinetic energy conservation
v02
mt 2
vt
 va 
ma
vt
2
• From these two, we obtain
2
vt 1 

Wednesday, Aug. 30, 2006
va
PHYS 3446, Fall 2006
Jae Yu
mt
ma

  2va  v t

12
Analysis Case I
• If mt<<ma,
–
–
–
–
–
–
–
–
vt2

mt
1 
 ma

  2va  vt

left-hand side becomes positive
va and vt must be in the same direction
Using the actual masses
me  0.5MeV / c 2 and ma  4 103 MeV / c 2
We obtain ve  vt  2va
If mt=me, then mt/ma~10-4.  va  v0
Thus, pe/pa0<10-4.
Change of momentum of alpha particle is negligible
Wednesday, Aug. 30, 2006
PHYS 3446, Fall 2006
Jae Yu
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Analysis Case II
vt2
• If mt>>ma,
–
–
–
–
–
–
–
–

mt
1 
 ma

  2va  vt

left-hand side of the above becomes negative
va and vt is opposite direction
Using the actual masses
mt  mAu  2 105 MeV / c2 and ma  4 103 MeV / c 2
We obtain vt  2ma va mt
If mt=mAu, then mt/ma~50.  va  v0
Thus, pe/pa0~2pa0.
Change of momentum of alpha particle is large
• a particle can even recoil
Wednesday, Aug. 30, 2006
PHYS 3446, Fall 2006
Jae Yu
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Rutherford Scattering with EM Force 1
• Let’s take into account only the EM force between the a
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
• Since the total energy is conserved,
ZZ ' e
V r  
r
1 2
2E
E  mv0  constant>0  v0 
2
m
Wednesday, Aug. 30, 2006
PHYS 3446, Fall 2006
Jae Yu
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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
Impact parameter
the angular momentum is l=mvb.
• From the energy relation, we obtain
l  m 2 E mb  b 2mE  b2  l 2 2mE
• From the definition of angular momentum, we obtain an equation of
motion
d  dt  l mr 2
• From energy conservation, we obtain another equation of motion
1  dr 
E  m 
2  dt 
2
1
 d 
 mr 2 
V r 
 Aug.
Wednesday,
30, 2006
2
 dt 
2
dr
2
l2 

 E V r  
2 
dt
m
2mr 
PHYS 3446, Fall 2006
Jae Yu
Centrifugal barrier
Effective potential
16
Rutherford Scattering with EM Force 3
• Rearranging the terms for approach, we obtain
V r   2
dr
l
2

r 1 
  b
dt
mrb
E 

• and
d  
bdr
 2  V r  

2
r  r 1 
b 
E 
 

12
• Integrating this from r0 to infinity gives the angular
distribution of the outgoing alpha particle
Distance of
closest
approach
Wednesday, Aug. 30, 2006
PHYS 3446, Fall 2006
Jae Yu
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Rutherford Scattering with EM Force 4
• What happens at the DCA?
– Kinetic energy reduces to 0.
dr
dt
0
r  r0
– The incident alpha could turn around and accelerate
– We can obtain
V  r0   2
2
r0 1 

E
  b  0

– This allows us to determine DCA for a given potential and 0.
• Define scattering angle q as the changes in the asymptotic
angles of the trajectory, we obtain
q  p  2  0  p  2b 

r0
Wednesday, Aug. 30, 2006
dr
 2  V r   2 
r  r 1 
b 
E 
 

PHYS 3446, Fall 2006
Jae Yu
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18
Rutherford Scattering with EM Force 5
• For a Coulomb potential
ZZ ' e 2
V r  
r
• DCA can be obtained for the given impact parameter b,
ZZ ' e 2 E 
2 2
r0 
1

1

4
b
E

2

 ZZ ' e 
2
2



• And the angular distribution becomes
q  p  2b 

r0
Wednesday, Aug. 30, 2006
dr
12
 2  ZZ ' e  2 
r  r 1 
b 
rE 
 

2
PHYS 3446, Fall 2006
Jae Yu
19
Rutherford Scattering with EM Force 6
• Replace the variable 1/r=x, and performing the
integration, we obtain

1
1 
q  p  2b cos 
 1  4b2 E 2 ZZ ' e2


• This can be rewritten
1
1  4b E
2
2
 ZZ ' e 
2
2

2





q p 
 cos 

2


• Solving this for b, we obtain
ZZ ' e2
q
b
cot
2E
2
Wednesday, Aug. 30, 2006
PHYS 3446, Fall 2006
Jae Yu
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Rutherford Scattering with EM Force 7
ZZ ' e
q
b
cot
2E
2
From the solution for b, we can learn the following
2
•
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
–
Wednesday, Aug. 30, 2006
Makes perfect sense also, since as the incident particle is closer to the
nucleus, it feels stronger Coulomb force.
PHYS 3446, Fall 2006
Jae Yu
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Assignments
1.
Compute the masses of electron, proton and alpha particles
in MeV/c2, using E=mc2.
•
2.
Compute the gravitational and the Coulomb forces between
two protons separated by 10-10m and compare their
strengths
Derive the following equations in your book:
3.
•
•
•
Need to look up masses of electrons, protons and alpha particles
in kg.
Eq. # 1.3, 1.17, 1.32
Must show detailed work and accompany explanations
These assignments are due next Wednesday, Sept. 6.
Wednesday, Aug. 30, 2006
PHYS 3446, Fall 2006
Jae Yu
22