Nuclear Physics
UoD Faculty of Science, Department of Physics
2013 - 2014
Dr. Tom Schulman
Lesson 5
The strong nuclear force and scattering experiments
(Lilley p. 23 – 25, Krane p. 80 – 108)
Theory
We have learnt that the atomic nucleus consists of positively charge protons and neutral
neutrons. Because there are no negatively charged particles in the nucleus the Coulomb force
cannot explain the stability of the nucleus. A force stronger than the Coulomb force must exist in
order to keep the protons together.
Scattering experiments have been and still are used to understand the nature of the strong
nuclear force. The first scattering experiments were carried out by Ernest Rutherford. In
Rutherford’s experiment helium nuclei, i.e., alpha particles, were accelerated and targeted through
thin aluminum and gold foils. Proton-nuclei, proton-proton, neutron-proton and electron-nuclei
scattering experiments have also been carried out. In scattering experiments the tracks of the
accelerated particles are measured and conclusions about the nuclear force are made from the
deviations of the particles from the straight path.
Alpha-nucleus Coulomb scattering – the Rutherford experiment
Figure 5.1 shows an alpha particle passing a heavy nucleus (e.g., an Au nucleus). The
distance b is known and we want to calculate the deviation angle θ.
detector

Fy

vy

F

v

v
y
alpha particle

v0

r

alpha source
b

z
nucleus
Figure 5.1
Let us assume that the only force effective between the alpha particle and the nucleus is the
Coulomb force. In this case we may write the vertical component of the force felt by the alpha
particle as
Fy  F sin  
Qq
2Ze 2
sin


sin 
4 0 r 2
4 0 r 2
(5.1)
In equation 5.1 Q = Ze is the charge of the atomic nucleus and q = 2e is the charge of the alpha
particle. Then the equation of motion is
Fy  ma  m
dv y
dt
dv y

dt

2Ze 2
sin 
4 0 r 2 m
(5.2)
The angular momentum of the alpha particle must remain equal to the initial angular momentum:
mv0 b  mv r  m
d 2
1
1 d
r  2 
dt
v0 b dt
r
(5.3)
Applying equation 5.3 to equation 5.2 gives
dv y

dt
2Ze 2
d
2Ze 2
sin 
 dv y 
sin d
4 0 mv0 b
dt
4 0 mv0 b
(5.4)
Integrating both sides (with vy = 0 and  = 0 at the beginning and vy = v0sin and  =  -  at the
end) gives
v0 sin 
2Ze 2
dv

0 y 4 0 mv0b
Because
tan

2

 
 sin d  v0 sin  
0
2Ze 2
(1  cos  )
4 0 mv0 b
(5.5)
sin 

 tan we can write
1  cos 
2
2Ze 2
4 0 mv02 b
(5.6)
The scattering angle  is plotted as a function of b in figure 5.2 when v0  1.0  10 6 m s .

180,00
150,00
120,00
90,00
60,00
30,00
0,00
0,0E+00
2,0E-12
4,0E-12
6,0E-12
Mesafe b/m
8,0E-12
1,0E-11
Figure 5.2
If the only force effective between the alpha particle and the nucleus is the Coulomb force
then the calculated curve plotted in figure 5.2 should match experimental observations. But this is
not the case in reality. If b is very small and v0 is high enough the measured values of  are different
from the ones predicted by the calculation above. This means that there must be a force present in
the nucleus that is effective at very small distances. This force is called the strong nuclear force.
In actual real scattering experiments the target is a thin sheet of metal. A beam of alpha
particles is directed towards and through the metal sheet. The distance b for the many alpha
particles differs slightly for different alphas and the alphas thus scatter as shown in figure 5.3. We
therefore define a scattering cross section σ as
The number of scattered particles per unit solid angle and per unit time
The number of particles hitting the metal sheet per unit time
 ( ) 
(5.7)
metal sheet
d

alpha beam
nucleus
b
Figure 5.3
db
If N is the density of alphas in the beam hitting the metal sheet then the number of alphas
passing the sheet per unit time in a circular section of radius b and width db is n  v0 N (2bdb) .
When these alphas pass through the metal sheet they scatter into the solid angle d. Then the
scattering cross section is
 ( ) 
v0 N (2bdb)
db
 2b
v0 Nd
d
(5.8)
The distance b and db can be solved from equation 5.6 and expressed as a function of the scattering
angle  and d. The solid angle d can be written as d  2 sin d . We then get for the
scattering cross section (after some trigonometry)
 coulomb( ) 
Z 2e4
(4 0 ) 2 m 2 v04 sin 4 ( 2)
(5.9)
 coulomb( )
-
/rad

Figure 5.4
The scattering cross section is plotted in figure 5.4 as a function of the scattering angle .
The calculations above were carried out according to the classical theory of mechanics but the result
is essentially the same if quantum theory had been used. From the difference between the measured
cross section and the coulomb cross section coulomb of equation 5.9 it is possible to make
conclusions about the strong nuclear force.
Neutron-proton scattering
In neutron-proton scattering experiments neutrons from a nuclear reactor are guided through
a hydrogen target and the scattered neutrons are measured with a neutron detector. Since the
neutrons are neutral the scattering is caused only by the strong nuclear force. Using the neutron
wave functions from quantum theory it is possible to calculate the scattering cross section of
neutron-hydrogen (i.e., neutron-proton) scattering. Again, by comparing the theoretical results to
the experimental one can obtain an understanding of the nuclear strong force.
Electron-nucleus scattering
High energy electrons may pass through a nucleus. Then the scattering is effected not only
by the protons but also by the quarks of the nucleons. Electron-nucleus scattering experiments have
provided important knowledge on the inner structure of the nucleons (quarks).
Based on the various scattering experiments the following conclusions about the nuclear force have
been made:
1. The range of the nuclear force is very short. Two particles interact through the nuclear force
only if the distance between them is < 10-15 m.
2. The nuclear force is not dependent on the electrical charge.
3. The nuclear force is not completely central. This result has been concluded from the nonconstant angular momentum of the two nucleons of the deuteron ( 12 H ) nucleus.
4. The nuclear force becomes repulsive at very short distances (<<10 -15 m).
5. The nuclear force is affected by the spin of the nucleons.
Despite all this the exact form of the potential energy of the nuclear force is unknown. The Japanese
physicist Hideki Yukawa has suggested an exponential function to describe the nuclear potential:
E p (r )   E0 r0
e  r r0
r
(5.10)
The Yukawa potential falls rapidly to zero when r increases but it does not explain the repulsive
nature of the nuclear force at very short distances. The estimate of the nuclear potential based on
experiments is shown below in figure 5.4.
Ep
r
Figure 5.4
Examples
5.1 Proton-nucleus collision
Figure 5.5 shows the potential energy of a gold nucleus (Z = 79), i.e., the combination of the
energy from the nuclear and from the Coulomb force. Let us calculate the minimum kinetic energy
required for a proton to reach the effective range of the nuclear strong force.
Ep
Emin
proton
r0 = 10-15 m
r
Figure 5.5
The kinetic energy of the proton is E k 
Ep 
1 2
mv and the electric potential energy of the nucleus is
2
Ze 2
. The kinetic energy of the proton must exceed the Coulomb barrier:
4 0 r
Ek  E p (r0 ) 
79e 2
 1.823  10 11 J  114 MeV  vmin 
4 0 r0
2Ek
 148  10 6 m s .
m proton
5.2 Rutherford’s estimation of the radius of the Au nucleus
In Rutherford’s first experiments alpha particles were accelerated against gold foils. Some
of the alphas returned back at 180. Because the speed of these alphas were equal to the initial
speed one may estimate the the radius of the nucleus by writing
V
Electric potential
1
2  79e 2
79e 2
malphav 2 
 rAu 
2
4 0 rAu
 0 malphav 2
r0
Rutherford’s result was
r
rAu  3.2  10 14 m . This corresponded to the maximum speed
v  18  10 6 m s of the alphas returning at 180 (alphas faster than this did not return, why?). The
nuclear radius estimated by Rutherford is not exactly in agreement with equation 1.1 but was still a
very important result for the scientific community 100 years ago.
Problems
5.1 Calculate the scattering angle  for b  1  10 10 m and b  1  10 11 m (figure 5.1). If
v0  1 10 7 m s how do the values of  change?
5.2 Find the limit for b after which  > 90 when v0  1 10 7 m s in figure 5.1.
5.3 Find the limit for b after which  < 1 when v0  1 10 7 m s in figure 5.1. Compare the result to
the radius of the gold atom which is 1.2  10 10 m .
5.4 Find the radii of the silver and the aluminum nuclei based on Rutherford’s experiment in
example 5.2 and compare the results with the results given by equation 1.1 (Ag: Z = 47, Al: Z = 13
and v  18  10 6 m s is the maximum speed of the alphas returning at 180 degree).
5.5 a) Draw a graph of the Yukawa potential. b) Sketch a graph of the total nuclear potential
(including the Coulomb effect) according to experimental results.