Rutherford scattering experiments bombarded thin metal foils with alpha particles. Most alpha particles passed straight through, but some were scattered at large angles. This was inconsistent with the plum pudding model of the atom, but agreed with Rutherford's nuclear model. The scattering was analyzed using classical mechanics. For head-on collisions, large-angle scattering required the target mass be much greater than the alpha particle mass, as is the case for alpha scattering off atomic nuclei but not electrons. This supported the existence of a small, massive nucleus at the center of the atom.

1. Rutherford Scattering 3
1.2 Rutherford Scattering
The series of measurements performed by Hans Geiger and Ernest Marsden
under Rutherford's direction at Manchester provide a classic example of a
"fixed target" experiment. The target was a thin metal foil of relatively
large atomic number, while the projectiles consisted of a collimated beam
of low energy a-particles, which, as we will see in the next chapter, are
nothing more than the nuclei of helium atoms. The basic outcome of these
experiments was that most of the a-particles went straight through the foil
with very little angular deviation. Occasionally, however, the deflections
were quite large. A detailed analysis of these observations revealed the
structure of the target, which ultimately led to the nuclear model of the
atom.
To fully appreciate the beauty of these experiments, it is essential to
analyze the results in their proper historical context. Prior to this work,
the only popular model of the atom was due to Joseph Thomson, who visu-
alized the electrically neutral atom as a "plum pudding" where negatively
charged electrons were embedded, like raisins, within a uniform distribution
of positive charge. If this model were correct, one would expect only small
deviations in the a-particles' trajectories (primarily due to scattering from
the electrons), unlike what was found by Geiger and Marsden. To see this,
let us do a few simple kinematic calculations. Because the velocities of the
a-particles in these experiments were well below 0.1c (where c refers to the
speed of light), we will ignore relativistic effects.
Let us assume that an a-particle with mass ma and initial velocity vo
collides head-on with a target particle of mass mt, which is initially at
rest (see Fig. 1.1). After the collision, both particles move with respective
velocities va and vt- Assuming that the collision is elastic (namely, that no
kinetic energy is converted or lost in the process), momentum and energy
conservation yield the following relations.
Momentum conservation:
mav0 — mava + mtvt,
or v0 = va H vt. (1.1)

2. 4 Nuclear and Particle Physics
@ — — @
^ m,, w,
Fig. 1.1 Collision of a particle of mass ma and velocity vo with a target particle of
mass mt.
Energy conservation:
- mavl = - mavl + - mtv,
or vl =vl + ^ v l (1.2)
where we have labeled (Hi)2 = Vi-Vi as vf, for i = 0, a and i. Squaring the
relation in Eq. (1.1) and comparing with Eq. (1.2), we obtain
or »,2 (l-^)=2i?a .S1 . (1.3)
It is clear from this analysis that, if mt <C ma, then the left hand side of
Eq. (1.3) is positive and, consequently, from the right hand side we conclude
that the motion of the a-particle and the target must be essentially along
the incident direction. In other words, in such a case, one would expect
only small deviations in the trajectory of the a-particle. On the other hand,
if mt » ma, then the left hand side of Eq. (1.3) is negative, which implies
large angles between the trajectories of the a-particle and the recoiling
nucleus, or large-angle scattering. To get a feeling for the magnitude of the
numbers, let us recall that the masses of the electron and the a-particle
have the following approximate values

3. Rutherford Scattering 5
me « 0.5MeV/c2,
ma « 4 x 103 MeV/c2. (1.4)
Therefore, if we identify
m( = me,
then,
TTl
10~4. (1.5)
a
Now, from Eq. (1.3) it follows that ve = vt < 2va, and then Eq. (1.2) yields
va & v0. Therefore, meve = ma ^f- ve < 2 x 10~4 mava « 2 x 10~4 mavo,
and the magnitude of the momentum transfer to the electron target is
therefore < 10~4 of the incident momentum. Consequently, the change
in the momentum of the a-particle is quite small and, in the framework
of the "plum pudding" model of the atom, we would expect only slight
deviations in the a-trajectory after scattering from atomic electrons; thus,
the outcome of the experiments, namely the occasional scatters through
large angles, would pose a serious puzzle. On the other hand, if we accept
the nuclear model, wherein the atom has a positively charged core (the
nucleus) containing most of the mass of the atom, and electrons moving
around it, then the experimental observations would follow quite naturally.
For example, setting the mass of the target to that of the gold nucleus
mt = mAu « 2 x 105 MeV/c2, (1.6)
yields
™i«50. (1.7)
ma
A simple analysis of Eq. (1.3) gives vt < 2m7^v"., and from Eq. (1.2)
we again obtain that va « VQ- Therefore, mtVt < 2mava ftj 2mavo. This
means that the nucleus can carry away up to twice the incident momentum,
which implies that the a-particle can recoil backwards with a momentum
essentially equal and opposite to its initial value. Such large momentum

4. 6 Nuclear and Particle Physics
transfers to the nucleus can, therefore, provide large scattering angles. Con-
sequently, in the Rutherford picture, we would expect those a-particles that
scatter off the atomic electrons in gold to have only small-angle deflections
in their trajectories, while the a-particles that occasionally scatter off the
massive nuclear centers to suffer large angular deviations.
The analysis of the scattering process, however, is not this straight-
forward, and this is simply because we have completely ignored the forces
involved in the problem.1 We know that a particle with charge Ze produces
a Coulomb potential of the form
U[f) = ^ . (1.8)
We also know that two electrically charged particles separated by a distance
r = f experience a Coulomb force giving rise to a potential energy
V(r) = ^ - . (1.9)
Here Ze and Z'e are the charges of the two particles. An important point
to note about the Coulomb force is that it is conservative and central. A
force is said to be conservative if it can be related to the potential energy
through a gradient, namely
F{r) = -V^(r), (1.10)
and it is denned to be central if
V(f) = V(f) = V(r). (1.11)
In other words, the potential energy associated with a central force depends
only on the distance between the particles and not on their angular coor-
dinates. Because the description of scattering in a central potential is no
more complicated than that in a Coulomb potential, we will first discuss
the general case.
Let us consider the classical scattering of a particle from a fixed center.
We will assume that the particle is incident along the z-axis with an initial
xWe have also tacitly assumed, in the context of the Thomson model, that contribu-
tions to large-angle scattering from the diffuse positively charged nuclear matter can be
ignored. This is, in fact, the case, as discussed by Thomson in his historic paper.

5. Rutherford Scattering 7
velocity vo- (It is worth noting that, outside the foil, the incident and the
outgoing trajectories are essentially straight lines, and that all the deflec-
tion occurs at close distances of the order of atomic dimensions, where the
interaction is most intense.) If we assume that the potential (force) falls off
at infinity, then conservation of energy would imply that the total energy
equals the initial energy
E = - mvl = constant > 0. (1-12)
Equivalently, we can relate the incident velocity to the total energy
v0 = —. (1.13)
V m
Let us describe the motion of the particle using spherical coordinates with
the fixed center as the origin (see Fig. 1.2). If r denotes the radial coordi-
nate of the incident particle, and the angle with respect to the z-axis, then
the potential (being central) would be independent of x- Consequently, the
angular momentum will be a constant during the entire motion. (That is,
since r and F are collinear, the torque r x F vanishes, and the angular
momentum r x mv cannot change.) For the incident particle, the angu-
lar momentum is clearly perpendicular to the plane of motion and has a
magnitude £ = mvob, where b is known as the impact parameter. The im-
pact parameter represents the transverse distance that the incident particle
would fly by the source if there was no force acting. Using Eq. (1.13), we
can obtain the following relation
[2E
I = m — b = b v2mE,
V m
1 2mE
o r .fc2 = ~ p - - (L14)
From its definition, the angular momentum can also be related to the
angular frequency, x, as follows
— r + r-f- x ) I = mr2 -£• = mr2x, (1-15)
at at J at

6. 8 Nuclear and Particle Physics
__L i_iL_r^rA ^z
Fig. 1.2 The scattering of a particle of mass m, with initial (asymptotic) velocity vo,
from a center of force at the origin.
where, as usual, we have defined a unit vector x perpendicular to r = rf,
with v(r) = rf + rxx expressed in terms of a radial and an angular compo-
nent of the velocity, and the dot above a variable stands for differentiation
with respect to time. Equation (1.15) can be rewritten as
at mr2
The energy is identical at every point of the trajectory, and can be
written as
*-H£)>+Mt)"+™
- Hi)'=*->£?-"<••>•
or * . _ f » ( J S _ v ( r ) - 5 f T ) ] i . (I.X7)
dt [m 2mr2) J
The term ^~s is referred to as the centrifugal barrier, which for I ^ 0 can
be considered as a repulsive contribution to an overall effective potential
yeff(r) = V{r) + 2 ^ J - Both positive and negative roots are allowed in Eq.
(1.17), but we have chosen the negative root because the radial coordinate
decreases with time until the point of closest approach, and that is the time
(1.16)

7. Rutherford Scattering 9
domain we will be examining.2 Rearranging the factors in Eq. (1.17) and
using Eq. (1.15), we obtain
dr _ _ 2__P_ f 2mEr2 / _ V£) _ 11 *
dt ~ [m 2mr2 (? E ) J J
--iLH1-™)-*}'-
Prom Eqs. (1.16) and (1.18), we now obtain
A l A* l dl A
d-X = —2 dt = —_- — dr
I dr
or dX = — r- (1-19)
r[,2(l--M)-6f
Integrating this between the initial point and the point of closest ap-
proach, we obtain
2 The motion is completely symmetric about the point of closest approach (r = ro),
and consequently the positive and negative roots provide identical information. In fact,
if the a-particle approached the target with the velocity vo along the exiting trajectory
in Fig. 1.2, it would then emerge on the entering trajectory, with the same asymptotic
velocity. A simple way to see that this is true is to imagine the collision as observed
from both above and below the plane of scattering shown in Fig. 1.2. Viewed from these
two perspectives, the motion in Fig. 1.2 appears as the mirror image of the reversed
trajectory. This symmetry is a consequence of time-reversal invariance of the equations
of motion, a concept that will be discussed in Chapter 11.
(1.18)

8. 10 Nuclear and Particle Physics
fx° , fro bdr
/ d = - r>
Jo Joo r[r2(i_Y±riyb2y
/
°° dr
—— -r. (1.20).
o r^(i-Yg.yh2y
The point of closest approach is determined by noting that, as the par-
ticle approaches from infinity, its velocity decreases continuously (assuming
the repulsive potential for the case of an a-particle approaching a nucleus),
until the point of closest approach, where the radial velocity (^) vanishes
and subsequently changes sign. That is, beyond this point, the velocity of
the particle increases again. Therefore, at the distance of closest approach,
when r = ro, both the radial and the absolute velocities attain a minimum,
and we have
- = 0
which, from Eqs. (1.17) and (1.18), means that
or r g ( l - ! £ o > ) - * = o. (1.21)
Thus, given a specific form of the potential, we can determine r0, and
therefore xo> as a function of the impact parameter b.3 Defining the scat-
tering angle 6 as the change in the asymptotic angles of the trajectory, we
get
r°° dr
e = n-2Xo=7T-2b — — -r. (1.22)
^ r[r2(l-Vjrl)-b*y
3We note that, in general, with i ^ 0 and E > 0, that is, for 6 ^ 0 , -^ is maximum
at r = ro (see Eq. (1.16)). Also, for I ^ 0, even for an attractive Coulomb potential,
there will be a finite result for ro as determined from Eq. (1.21). This is because the
centrifugal barrier for I 7^ 0 acts as a repulsive potential that dominates over Coulomb
attraction at small distances.

9. Rutherford Scattering 11
Consequently, given an impact parameter b, and a fixed energy E, the
scattering angle of a particle in a potential can, at least in principle, be
completely determined.
As an application of the general result, let us now return to the scat-
tering of a charged particle from a repulsive Coulomb potential, for which
the potential energy is given by Eq. (1.9)
V(r) = ^ , (1.23)
where Z'e represents the charge of the incident particle and Ze the charge of
the scattering center. (The scattering of an a-particle from a nucleus would
then correspond to Z' = 2, with Ze representing the nuclear charge.) The
distance of closest approach can be obtained from Eq. (1.21)
2 ZZ'e2 l 2
ro ^ — ro-b2= 0,
^ * / ( y ) 2 + 4 f c 2 n „,.
or r0 = — — 2 (1.24)
Since the radial coordinate can by definition only be positive, we conclude
that
Consequently, from Eq. (1.22), we obtain
f°° dr
e = ir-2b r . (1.26)
Jr0 r [ r 2 ( 1 _ l ^ £ i ) _ 6 2 ] I
Let us define a new variable
x=1-, (1.27)
which gives
(1.25)

10. 12 Nuclear and Particle Physics
1 2E [ 4b2E2
Prom Eq. (1.27), we obtain
dr , dx
dx = —7 , or dr = -,
r x
and, in terms of this new variable, we can write
«,. f° ( dx x
r° dx
= 7r + 26/ — r . (1.29)
Now, using the following result from the integral tables
I , ^ = 4= cos"1 (- P+ 2lX ) , (1-30)
we obtain
, 1 i I T +2*>2x °6 = IT + 2b x - cos"1 . E
b W(V)2+4&v ,0
o l( 1 + ^ X °
= 7T + 2 COS"1 . ZZ e =
= 7r + 2cos"1 ( . 1 = I -2cos"1(l)
I /i I 4 ^ g 2 ~ / V ^
V y 1 + (ZZ'e!)5 /
= 7T + 2cos~1 ( . 1 I . (1.31)
V i + (ZZ>e*)*J
Equivalently, we can write
(1.28)

11. Rutherford Scattering 13
1 (9 TT
A , 46^~C O S l2~2J'
V (ZZ'e2)2
1 , / 0 7r . , 6» 1
or .., „, = cos = sin - = —^,
!+#fr V2 2/ 2 cosec2f
26£ 0
°r Z Z ^ = C O t 2 '
or b=-^- cot -. (1.32)
This relates the scattering angle, which is a measurable quantity, to the
impact parameter which cannot be observed directly. Note that, for fixed
b, E and Z the scattering angle is larger for a larger value of Z. This is
consistent with our intuition in that the Coulomb potential is stronger for
larger Z, and leads to a larger deflection. Similarly, for a fixed b, Z and
Z', the scattering angle is larger when E is smaller. Qualitatively, we can
understand this as follows. When the particle has low energy, its velocity
is smaller and, therefore, it spends more time in the potential and suffers a
greater amount of scattering. Finally, for fixed Z, Z1 and E, the scattering
angle is larger for smaller b. Namely, when the impact parameter is small,
the particle feels the force more strongly and hence the deflection is larger.
Equation (1.32) therefore incorporates all the qualitative features that we
expect of scattering in the Coulomb field.
1.3 Scattering Cross Section
As we have seen, the scattering of a particle in a potential is completely
determined once we know the impact parameter and the energy of the
particle; and, for a fixed incident energy, the deflection is therefore defined
by just the impact parameter. To perform an experiment, we prepare an
incident flux of beam particles of known energy, and measure the number
of particles scattered out of the beam at different 6. Because this number
is determined entirely by the impact parameters involved in the collisions,