Rutherford Scattering Simulation and
Analysis with Monte Carlo Method
Rui Hou
10114117
Department of physics, Southeast University, Nanjing211189, China
2016.12.03
Abstract
As we all know, it’s difficult to verify that count rate of the α-particle scattering is
in direct proportion to the square of target atom mass number, because we can’t get
two target sheet with the same thickness. In this paper, I simulate the Rutherford
scattering experiment in the ideal situation to verify Rutherford formula and the some
of other conclusions.
Key Words: Monte Carlo; Least Squares; Metropolis Algorithm; Euler Algorithm
1 Model
1.1 Energy Distribution
We should consider that radioactive source emits α-particles’ energy follows
Gaussian distributions.
E∝
1
2&'
(
)
(+),- )/
01 /
E0 is the energy corresponding to the full-energy peak, σ is standard deviation.
1.2 Motion Equations
Rutherford scattering was first referred to as Coulomb scattering because it relies
only upon static electric(Coulomb) forces, and the minimal distance between particles
1
is set only by this potential. Asume that an 3-particle is approaching an atom, the
3-paticle mass of m and the charge carried by it is 2e, it coordinates to (x, y). The
target atom carries charge Ze and it coordinates to (x0,y0), when 3-particle is at a
distance of r from the target atom, it’s acceleration equations are
6( 0 (: − :8 )
2&78 9 < =
6( 0 (? − ?8 )
4> =
2&78 9 < =
4+ =
the velocity equations are
@0+ = @A+ + 4+ ∆D
@0> = @A> + 4> ∆D
the displacement equation
:0 = :A + @A+ ∆D
?0 = ?A + @A> ∆D
1.3 Rutherford Formula
The Rutherford formula is that:
⎛ 1 ⎞
dσ (θ )
dn
⎟
'E F =
=
= ⎜⎜
dΩ
nN 0tdΩ ⎝ 4πε 0 ⎟⎠
2
2
⎛ 2Ze 2 ⎞
1
⎜⎜
⎟⎟
⎝ 4 E ⎠ sin 4 θ
2
'E F is scattering differential cross-section, θ is scattering angle.
2 Methods
2.1 Metropolis Monte Carlo Algorithm and Euler Method
The Metropolis algorithm produces a “random walk” of points :I } whose
asymptotic probability approaches P(x) after a large number of steps. The random
walk is defined by a “transition probability” w(:I → :L ) for one value :I to another xj in
order that the distribution of points x0 ,x1 ,x2 , ... converges to P(x). In can be shown
that it is sufficient (but not necessary) to satisfy the “detailed balance” condition
P(:I )w(:I → :L ) = P(:L )w(:L → :I )
P(x)=(
)
2
(MNO- )/
/P/
This relation dos not specify w(:I → :L ) uniquely. A simple choice is
w(:I → :L )=min 1,
U(+V )
U(+W )
This choice can be described by the following steps. Suppose that the “random
walker” is a position xn. To generate xn+1 we:
1. choose a trial position xt=xn+XY , where the XY is a random number in
the interval [-X, X].
2. Calculate w=P(xt)/P(xn)
3. If w≥ 1,we accept the change and let .xn+1=xt.
4. If w≤ 1,generate a random number r.
5. If r≤ w, accept the change and let .xn+1=xt.
6. If the trial change is not accepted, the let xn+1=xn.
So that we can get the α-particles’ energy who follows Gaussian distributions.
Then I calculate their velocity, initialize the position of α-particles randomly
and make them move to the target atom and calculate their trajectory using Euler
method.
2.2 Least Square Algorithm
Least Square algorithm could be used to fit the trajectory of 3-particles so that
we can get the scattering angle.
3 Result and Analysis
3.1 Energy Distribution
Fig1 Energy distribution(E0=0.662Mev)
As we can see from the figure,10000 α-particles’ energy follow Gaussian
distributions.
3
3.2 ]-particles Scattering Trajectories
^
3.2 Relationship in N and
_`
Sample Num
1
2
3
4
1/E (Mev)
0.662
1.223
1.512
1.828
N
37
16
13
12
2
Table 2:100000 α-particles’ scattering statistical table
Fig2 Relationship in N and
A
a/
The coefficient of correlation of counting rate and square of energy is 0.9986,so
we can get the conclusion that N ∝
A
,/
3.3 Relationship in N and Z2
4
Z
Z
2
N
60
65
70
75
80
3600
4225
4900
5625
6400
23
34
38
47
55
Table 2:100000 α-particles’ scattering statistical table
Fig3: Relationship in N and Z2
The coefficient of correlation of counting rate and square of energy is 0.9761,
so N is in direct proportion to Z 0 is authentic
3.4 Relationship in N and
^
defg (h)
Fig4:30000 α-particles’ scattering frequency count
As we can see, there are very few α-particles join the small-angle scattering.
The result is in line with actual experiment.
θ
40
45
50
55
60
i
1/sin4( ) 73.091 46.650 31.343 22.000 16.005
0
N
85
47
25
5
12
3
Table 3:100000 α-particles’ scattering statistical table
Fig5: Relationship in N and
A
n
/
jklm ( )
The coefficient of correlation of counting rate and square of energy is 0.9995,
so we can get the conclusion that N ∝
1
4
F
sin ( )
2
4 Summary
This paper also researched that the relationship of counting rate and
Z2,
A
jklm (q)
A
a/
,
in Rutherford Scattering, after analyzing we conclude that there are
linear relationships between counting rate and those parameters. There is a little
point gap between the result and theoretical values, but I think, as long as we
increase the number of α-particles, errors will get smaller.
5 References
[1]
Tao Pang. An Introduction to Computational Physics. Cambrige University
Press,2006
[2]
.
.
, 2008.
[3]
.
.
, 2010.
6