1. Laboratory system and center of mass system
Laboratory reference frame:
Before collision After collision m1 m2 v1 u1 ϕ
u2=0 v2 α
Center of mass reference frame:
Before collision After collision m1 m2 V1 U1 U2 q q V2

2. Kinematics of the Transformation
Assume m2 is initially at rest in the lab frame.
Clearly, after m1 scatters from it, in general it will not be at rest! It will recoil due to the scattering!
Momentum IS ALWAYS conserved in a collision!
 Cannot get the lab scattering angle ϕ directly from solving the 1 body CM frame problem for θ.
Need to take the result from the1 body CM frame scattering & transform it back to the lab frame.

3. In the lab frame, the situation looks like:
m2 is initially at rest
In the CM frame,
the situation looks like:
 Looks like this
to an observer
moving with the
Center of Mass.

4. Relationship between velocities of particles in center of mass and laboratory frames of reference

5. Also note:

6. Also from conservation of kinetic energy:

7. Relationship between scattering angle in center of mass and laboratory frames of reference
In CM system, the scattering angle is given as:
In lab system, the scattering angle is given as: q q ϕ α

8. The relation between velocities is
Therefore from equation (2) q q

9. From diagram, the direction of and is same.q q

10. q q

11. For elastic collision

12. We know
Discuss the cases:
m1=m2
m1< m2
m1> m2

13. Relationship between recoil angle in lab frame and scattering angle in center of mass frame
Consider elastic collision between two particles in lab frame;
According to law of conservation of linear momentum Y ϕ X α

18. Rutherford Scattering

19. Rutherford classified radioactivity
Awarded Nobel prize in chemistry 1908 “for investigations into the disintegration of the elements and the chemistry of radioactive substances”
Together with Geiger and Marsden scattered alpha particles from atomic nuclei and produced the theory of “Rutherford Scattering”.
Postulated the existence of the neutron
Ernest Rutherford ( )

20. The Geiger – Marsden experiment

21. Scattering
Scattering is a general physical process where some forms of radiation, such as light, sound, or moving particles, are forced to deviate from a straight trajectory by one or more paths due to localized non- uniformities in the medium through which they pass
Scattering may also refer to particle-particle collisions between molecules, atoms, electrons, photons and other particles.
The types of non-uniformities which can cause scattering, sometimes known as scatterers or scattering centers

22. Scattering
If a particle approaches a heavier target at rest, they will have an interaction and the trajectory of the incident particle will change. If the incident particle comes from infinity, it will again go to infinity and the change in path will continue. The continuous change in the trajectory of the incident particle during its interaction with the target is called scattering.
During the collision of atomic particles, one cannot follow the individual particles in the scattering process since these particles are indistinguishable. To solve this problem, concept of probability is introduced.

23. Impact parameter
The impact parameter  is defined as the perpendicular distance between the path of a projectile and the center of the field  created by an object that the projectile is approaching

24. Impact parameter
If the line of motion of incident particle directly passes through the centre of force then there will be Head on Collision and impact parameter will be zero.

25. Cross-section (σ)
Probability of a nuclear reaction is measured in terms of cross- section.
It gives the quantitative measurement of the probability of a nuclear reaction.
It is defined as the target area presented by a nucleus to the incident particles.
The units of cross-section are Barn. 1 Barn=10-24m2
Cross-section =(number of scattered particles)/ (total number of incident particles)
σ=Nsc/I
where I is intensity of beam and Nsc is the number of particles scattered

26. Differential Cross-section (σ(θ))
If we are interested in the probability of scattering in a certain specified direction then this probability is expressed as differential cross-section.
Consider a beam of particles of intensity I (number crossing per unit area per second) to be incident on a single scatterer fixed in the lab system.
Let Nsc be the number of particles scattered per second per unit solid angle at an angle θ. Then the number of particles scattered per second into an annular solid angle element dΩ will be Nsc(θ)dΩ.
σ(θ) dΩ=
(number of particles scattered into solid angle)/
(intensity of incident particles)
σ(θ) dΩ=(Nsc(θ)dΩ)/I

27. Rutherford Scattering

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39. Results Rutherford’s experiment found that:
Most of the alpha particles passed through the gold foil
undeviated.
A few alpha particles were deflected from their path
but continued through the gold foil.
A small number of alpha particles rebounded.

40. Conclusion From the results of his experiment, Rutherford explained:
As most alpha particles passed through the gold foil
atoms undeviated, Rutherford concluded that most of
the atom was actually empty space.
The deviation of some alpha particles from their
original path were due to positive charges within the
foil.

41. Conclusion From the results of his experiment, Rutherford explained:
A small number of alpha particles had rebounded
because they collided with something much larger and
heavier and which contains a concentrated region of
positive charge.

42. Conclusion
As a result of his observations, Rutherford suggested that the atom had a positively charged centre which contained most of the mass.
He called the heavy positively charged centre the nucleus.
He went on to suggest that the nucleus was surrounded by orbiting electrons required for electrical neutrality.

43. Rutherford Scattering formula
Rutherford derived the scattering formula by assuming the nucleus to be sufficiently heavy, so that it remains at rest during scattering process.
Consider the nucleus of charge Ze to be at the origin of a system of polar coordinated r, θ.
The scattering of alpha-particles is due to repulsive electrostatic force between the nucleus and the alpha particle. If Ze is the charge on the nucleus and 2e charge on alpha particle then the magnitude of the repulsive force is
F=2Ze2/r2 where r is the distance between the alpha particle and the nucleus.
Due to repulsive interaction, alpha particle will be deflected along a trajectory which will be one branch of hyperbola, with nucleus as outer focus.

44. Rutherford Scattering formula
Alpha particle incident along asymptote AO makes an angle ψ at O and gets deflected along asymptote OC making an angle ϕ which is angle of scattering. Let p be the perpendicular distance of AO produced from nucleus. The distance is called impact parameter (MN). Y
Alpha particle
(r,θ)
Asymptotes r
Target nucleus ψ O X’ θ X ψ N ϕ
Angle of scattering
Impact parameter p M Y’

45. Relation between angle of scattering and impact parameter
Since the force is the central force i.e. inverse square force.
The scattered α-particle of reduced mass μ under the influence of inverse square force will move along a conic given by equation

46. The total energy of the system remains constant at all points on the trajectory. Therefore, if u be the initial velocity of the α-particle at infinity, where potential energy is zero, then the total energy is given by
The initial angular momentum of the α-particle is given by

47. Angle of scattering

48. This is the relation between angle of scattering and impact parameter p. This shows that angle of scattering is inversely proportional to impact parameter. The smaller the impact parameter, greater will be the scattering.

49. Differential Cross-section (σ(θ))
Consider a beam of particles of intensity I (number crossing per unit area per second) to be incident on a single scatterer fixed in the lab system. Let the collision occur for values of the impact parameter between p and p+dp, then the particles will be scattered within the angular range between
The probability of such collisions is proportional to the area of an annular disc of radius p and width dp.
Let Nsc be the number of particles scattered per second per unit solid angle at an angle θ. Then the number of particles scattered per second into an annular solid angle element dΩ will be Nsc(θ)dΩ.
σ(θ) dΩ= (number of particles scattered into solid angle)/(intensity of incident particles)
σ(θ) dΩ=(Nsc(θ)dΩ)/I

50. The Rutherford Cross – Section Formula∆p p p vo m s
Secret of deriving the formula quickly is to express the momentum transfer in two ways. The first way you can see in the top parallelogram diagram.
(1)

51. The Rutherford Cross – Section Formulaφ
The second way is by integrating the force on the alpha across the trajectory: since we know by Newton’s law and Coulomb’s law:

52. Eqn 3.5
At first sight this integral looks impossible because both  and r are both functions of t.
However the conservation of angular momentum helps:
Eqn 3.1
From which one sees that:
So that (3.5) becomes
Now we can equate this with the first method of finding Δp

53. This allows us to get the impact parameter b as a function of θ
Where S0 is the distance of closest approach for head on collision
Coulomb potential so
* All particles scattered by more than some value of  must have impact parameters less than b. So that cross-section for scattering into any angle greater than  must be:
Eqn(3.9)

54. b
Particles fly off into solid angle given by:
The differential scattering cross-section is defined as:

55. Distance of closest approacho o

56. The distance of closest approach “d” will be determined by:
Using: we get -
Solution of this quadratic gives:

57. Failure of the Rutherford Formula
Increasing energy and constant angle
Increasing angle and constant energy
Failure of the formula occurs because the distance of closest approach is less than the diameter of the nucleus. This can happen if (a) the angle of scatter is large or (b) the energy of the particle is large enough. With Alpha particles from radioactive sources this is difficult. But with those from accelerators it becomes possible to touch the nucleus and find out its size because the distance of closest approach is given by:

58. Relationship between center of mass and laboratory frames of reference – continued (elastic scattering) v1 V1
VCM y q