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Rutherford Scattering: Discovering Atomic Nuclei through Particle Collisions, Study notes of Particle Physics

An introduction to Rutherford scattering, a groundbreaking experiment that led to the discovery of atomic nuclei. a theory section explaining the principles behind the scattering of alpha particles from atomic nuclei and an experiment section detailing how to measure the masses of target nuclei using conservation of momentum and energy. In this lab, students will use a 1-MV Pelletron accelerator to scatter alpha particles from thin foils and infer the masses of the target nuclei.

What you will learn

  • What was the significance of Rutherford's scattering experiments for our understanding of atomic structure?
  • How does the theory of Rutherford scattering explain the scattering of alpha particles from atomic nuclei?
  • How can the masses of target nuclei be inferred from Rutherford scattering experiments?

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Rutherford Scattering
1 Introduction
In 1911 Ernest Rutherford discovered the atomic nucleus by scattering α-particles (nuclei of He-
lium atoms) from a gold foil and observing that some of the α-particles were scattered at backward
angles. These results were in startling contradiction with J. J. Thomson’s popular "plum pudding"
model in which the atom is a spherically shaped mass of positive charge in which negative elec-
trons are embedded. Rutherford concluded that these observations, however, were consistent with
a new model in which all of the atom’s positive charge and nearly all its mass are concentrated in
a small region.
This was the first demonstration of how scattering experiments can be used to study atomic struc-
ture, and was the birth of nuclear and particle physics. Today’s high energy physics experiments
employ these techniques to learn about the structure of known subatomic particles and to discover
new particles.
In this lab we will use the 1-MV Pelletron accelerator in room N008 of the Science and Engineering
Center to scatter α-particles from thin foils. By measuring the kinetic energies and angles of the
scattered α-particles and applying conservation of momentum and energy, we will infer the masses
of the target nuclei.
2 Theory
Consider a particle of mass mwith initial momentum
pithat collides with another particle of mass
Mthat is initially at rest, as shown in Figure 1. The incident particle scatters at an angle θwith a
final momentum
pfwhile the target particle recoils at an angle φwith a momentum
P.
From conservation of momentum we have
pi=
pf+
P.(1)
Using the momentum diagram in Figure 2, we can write an expression for the square of the mo-
mentum of the recoiling target particle
P2=
pi
pf2=p2
i+p2
f2pipfcosθ.(2)
Assuming the collision is elastic, we can apply conservation of kinetic energy and write
Ki=Kf+K(3)
where Ki,Kf, and Kare the kinetic energies of the incident, scattered, and recoil particles, re-
spectively. Inserting the expression for the kinetic energy of a particle in terms of its momentum,
1
pf3
pf4
pf5

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Rutherford Scattering

1 Introduction

In 1911 Ernest Rutherford discovered the atomic nucleus by scattering α-particles (nuclei of He- lium atoms) from a gold foil and observing that some of the α-particles were scattered at backward angles. These results were in startling contradiction with J. J. Thomson’s popular "plum pudding" model in which the atom is a spherically shaped mass of positive charge in which negative elec- trons are embedded. Rutherford concluded that these observations, however, were consistent with a new model in which all of the atom’s positive charge and nearly all its mass are concentrated in a small region.

This was the first demonstration of how scattering experiments can be used to study atomic struc- ture, and was the birth of nuclear and particle physics. Today’s high energy physics experiments employ these techniques to learn about the structure of known subatomic particles and to discover new particles.

In this lab we will use the 1-MV Pelletron accelerator in room N008 of the Science and Engineering Center to scatter α-particles from thin foils. By measuring the kinetic energies and angles of the scattered α-particles and applying conservation of momentum and energy, we will infer the masses of the target nuclei.

2 Theory

Consider a particle of mass m with initial momentum −→pi that collides with another particle of mass M that is initially at rest, as shown in Figure 1. The incident particle scatters at an angle θ with a final momentum −→p (^) f while the target particle recoils at an angle φ with a momentum

P.

From conservation of momentum we have

−→pi = −→p (^) f + −→ P. (1)

Using the momentum diagram in Figure 2, we can write an expression for the square of the mo- mentum of the recoiling target particle

P^2 =

pi − −→p (^) f

= p^2 i + p^2 f − 2 pi p (^) f cos θ. (2)

Assuming the collision is elastic, we can apply conservation of kinetic energy and write

Ki = Kf + K (3)

where Ki, Kf , and K are the kinetic energies of the incident, scattered, and recoil particles, re- spectively. Inserting the expression for the kinetic energy of a particle in terms of its momentum,



p

i



p

f



P

 



   





  

 



    

  

    

  

   



 

  

m 

m

M

M

Figure 1: Scattering diagram.

p

i

P p

f





    

    

  



    

    

  

Figure 2: Momentum vector diagram.

K = p^2 / 2 m, into Eq. (3) and solving for the square of the momentum of the recoiling target particle we have

P^2 = 2 M

p^2 i 2 m

p^2 f 2 m

Setting Eqs. (2) and (4) equal to each other and performing a little algebra (alright, a lot of algebra), we get an equation for the mass of the target nucleus M as a function of m, Ki, Kf , and θ

M = m

Kf Ki −^2

Kf Ki cos^ θ^ +^1 1 − Kf Ki

Table 1: Data from energy spectrum at θ = 160 ◦. Peak Kf (keV) M (u)

Narrow peak 1

Narrow peak 2

Narrow peak 3

Narrow peak 4

Narrow peak 5

Broad peak

Table 2: Data from energy spectrum at θ = 140 ◦. Peak Kf (keV) M (u)

Narrow peak 1

Narrow peak 2

Narrow peak 3

Narrow peak 4

Narrow peak 5

Broad peak

Table 3: Average masses and determination of elements. Peak Mθ = 160 ◦^ (u) Mθ = 140 ◦^ (u) MAve. (u) Element

Narrow peak 1

Narrow peak 2

Narrow peak 3

Narrow peak 4

Narrow peak 5

Broad peak