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Presentation on Rutherford Scattering

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Alpha Particles strike gold foil . Why did they choose gold ? Observation ... Most of the particles passed through the golden foil with very little scattering ... – PowerPoint PPT presentation

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Title: Presentation on Rutherford Scattering


1
Presentation on Rutherford Scattering
  • by
  • Demetis Dionisis
  • dionisis_at_demetis.com

2
Initial Approach
3
Thompsons Model
  • The incident alpha particle would mostly see a
    neutral atom.
  • Observations were expected to show deflections of
    less than four degrees.

4
Java Applet showing the Thompsons model
  • Click here to view the Applet

5
The new Model (E.Rutherford)
  • Geiger and Marsden (Cavendish Laboratory in
    Cambridge)
  • Alpha Particles strike gold foil .
  • Why did they choose gold ?

6
Observation from the experiment
  • Most of the particles passed through the golden
    foil with very little scattering
  • Some were observed with scattering angles far in
    excess of a few degrees
  • Some were backscattered !!!

7
Rutherfords assumptions
  • The Positive charge and almost all of the mass is
    located in a very tiny nucleus.
  • The Nucleus is very dense
  • The Radius of the nucleus is about 0.510(-15)
  • The Nuclear radius is only 1/10000 of the atomic
    radius !

8
Rutherford Scattering Java Applet
  • Click here to view the applet

9
I. Classical Scattering by a central potential
  • We assume that the path of the particle is
    confined to the XZ plane.
  • The Kinetic Energy of the particle which is being
    scattered is

10
The Scattering of a particle in the field of a
repulsive central potential . The Origin of the
Potential is the Targeted nucleus.
11
Important Parameters
  • b Impact Parameter
  • T Deflection function
  • ? Angle of scattering
  • a Angle between the asymptotes to the orbit

12
The total Energy E Constant
13
..
14
The equation for the orbit ..
15
The distance of the closest approach
16
Having ro what is the orbit?
  • Once we have obtained the distance of the closest
    approach , then we use the following integral to
    find the orbit.

17
Assumptions for the acting force on the particle
  • Zero at large distances
  • At close distances , the particle experiences a
    repulsive or attractive force.
  • ____________________________________
  • As a result , the particle moves in a
    straight line trajectory (when far from the
    force) and generally gets deflected from its
    original path.

18
Calculation of the angle a
  • a is the angle between the two asymptotes to the
    orbit.
  • or

19
Deflection Function T
  • T p a
  • For central potentials ( Z-axis symmetry )
    , it is more convenient to use the angle of
    scattering ? ,
  • ?F (if )
  • 2p-F (if )

20
Cross Sections
  • We assume a uniform beam of particles
  • Particles have the same energy and they dont
    interact with each other
  • The target consists of n scattering centres
  • The target particles are away from each other so
    that each scattering , implicates only one
    target.
  • The target is thin enough to ignore multiple
    scattering

21
dN
  • n scattering centres.
  • N the number of incident particles
  • dN the number of incident particles that are
    scattered into the solid angle dO .
  • ?hen

22
What is the s(?,f) ?
  • A proportionality factor in the dN equation.
  • Another way to define s is
  • Differential scattering cross-section
  • ( ??af????? ??e???? ??at?µ? S?ed?s??)

23
Total Scattering Cross-Section
24
Calculation of s Step 1( out of 6 )
  • We assume that the force is central and vanishes
    for large r.
  • Since b is different for each particle , the
    scattering will differ too.
  • Each different value of L ( or b ) defines a
    single deflection function and a scattering angle
    ?.

25
Calculation of s Step 2
  • The particles that are scattered between angles ?
    , ?d? are those with angular momentum between L
    , LdL or impact parameter b , bdb.
  • The incident particles fall within the ring
    area 2pb db.
  • The number of the particles that fall within
    the ring is N2pb db

26
Calculation of s Step 3
  • The number of particles that fall into the ring
    with angular momentum between L , LdL are
  • And that number is equal to the number
    of the particles scattered in the dO angle.

27
Calculation of s Step 4
  • So we have ,

28
Calculation of s Step 5
29
Calculation of s Step 6
  • Finally, we have two equations for s ,

30
Additional Equations
31
Rutherford Scattering
  • Exact solution can be obtained if we use a
    Coulomb Potential .

32
Repulsive Case
The closest approach distance can be
found via ,
33
ro
  • We pose
  • and
  • ..

34
The Deflection function T
..........
35
T
or
36
The orbit of the particle
We can use the following equation ,
..
37
s ?
  • Since we know the connection between b and ? , we
    can calculate s via the equation ,

.
..
38
Final formula for the differential cross-section
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