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Rutherford scattering and hyperbola orbit

Authors:
1
Rutherford scattering and hyperbola orbit
Masatsugu Sei Suzuki and Itsuko S. Suzuki
Department of Physics, SUNY at Binghamton
(Date: June 15, 2016)
Ernest Rutherford, 1st Baron Rutherford of Nelson OM, FRS (30 August 1871 19
October 1937) was a New Zealand-born British chemist and physicist who became
known as the father of nuclear physics. In early work he discovered the concept of
radioactive half life, proved that radioactivity involved the transmutation of one chemical
element to another, and also differentiated and named alpha and beta radiation. This work
was done at McGill University in Canada. It is the basis for the Nobel Prize in Chemistry
he was awarded in 1908 "for his investigations into the disintegration of the elements,
and the chemistry of radioactive substances".
http://en.wikipedia.org/wiki/Ernest_Rutherford
CONTENT
1. Introduction
2. Historical Background
3. Nomenclature and feature of the hyperbola orbit
4. Property of hyperbola
5. Rutherford scattering experiment
6. Linear momentum for the elastic scattering
7. Newton’s second law for rotation: torque and angular momentum
8. Approach from conservation law of angular momentum and energy
(Born, Tomonaga)
9. The Kepler's First Law for the repulsive interaction
2
10. Differential cross section: d
d
(classical case)
11. Quantum mechanics (scattering due to the Yukawa potential)
12. Schematic diagram for the Rutherford scattering
13. The use of Mathematica for drawing the hyperbola
14. Experimental results by Rutherford’s group
15. Rough evaluation for the size of nucleus
16. Effective potential with angular momentum l (= 0, 1, 2, 3)
17. Possibility for the penetration of a particle into nucleus
18. -particle decay: quantum tunneling
19. Size of nucleus and evaluation of the kinetic energy K0
20. Summary: From Rutherford scattering to Bohr model of hydrogen
atom
REFERENCES
APPENDIX
I Illustration of the Rutherford scattering experiment (H.E. White)
II Mechanical model for Rutherford scattering by White
1. Introduction
Our knowledge of nuclear structure comes from the scattering of charged particles
such as
particles on nuclei. The information on the charge distribution of the nucleus
can be determined from their scattering intensities as a function of scattering angle. The
interpretation of such experiments requires quantum-mechanical calculation of the cross-
sections. However, it turns out that the quantum-mechanical calculation of scattering in a
1/r potential gives the same result as the classical Rutherford scattering. So the classical
Rutherford scattering can be used to interpret experiments using positively charged
particles whose energy is sufficiently low that they cannot penetrate inside the nucleus.
Under the instruction of Rutherford, Geiger and Marsden (1909) dis the experiment
on the scattering of scattering of
particles. They showed that a gold foil of thickness of
5
104
cm produced scattering over 90° or more for one
-particle in 20,000. This was
incompatible with Thomson’s model of the atom that the atom consists of a sphere of
uniformly distributed positive charge in which negatively charged electrons are
embedded (which he calls corpuscles. The total charge of the atom was zero. Rutherford
(1911) deduced from this that the atom should consist of a heavy central mass,
concentrated within a very small volume, surrounded by light masses of opposite charge,
occupying the outer parts of the atom.
One of us (M.S.) had an opportunity to teach Phys.323 (Modern Physics) in Fall 2011
and 2012 at the Binghamton University. Rutherford scattering is one of the most
significant experiments in the quantum mechanics. During this class, We prepared the
3
lecture note on the Rutherford scattering. We read a lot of textbooks on this matter,
including the textbooks of modern physics and quantum mechanics. We read a book
written by Segre (x-ray to quark). There is one figure (as shown below) of Rutherford
scattering which was published by Rutherford (1911). We realize that the definition of
the scattering angle (
) is different from that used for the conventional x-ray and neutron
scattering, except for
in the scattering (the quantum mechanics) and 2
in the x-ray and
neutron scattering (condensed matter physics). Using this angle, Segre shows that the
differential cross section is given by
)
2
(sin
1
4
d
d, (1)
in spite of the difference of the definition of the scattering angle. We realize that even for
a great scientist such as Segre, they had such an careless mistake for such as Rutherford
scattering which is so well-known and so well-discussed in the modern physics textbooks.
Here we start with the nomenclature of hyperbola orbit. Using Mathematica, we
examine the features of the hyperbola, which are closely related to the essential points of
the Rutherford scattering. There have been so many books on the Rutherford scattering.
In particular, the quantum mechanics textbook by Born, Tomonaga (geometrical
discussion) and the book by Longair (historical background) are very useful for our
understanding the physics. We also make use of the Mathematica to discuss the geometry
of hyperbola orbit. We also discuss the size of nucleus from alpha decay (tunneling
effect).
4
Fig.1 Original figure for the Rutherford scattering (Rutherford, 1911). Consider
the passage of a positive electrified particle close to the center of an atom.
Supposing that the velocity of the particle is not appreciably changed by
its passage through the atom, the path of the particle under the influence of
a repulsive force varying inversely as the square of the distance will be a
hyperbola with the center of the atom S as the external focus. The particle
to enter the atom in the direction PO, and that the direction of motion on
escaping the atom is OP'. OP and OP' make equal angles with the line SA,
where A is the vertex of the hyperbola. p = SN = perpendicular distance
from center on direction of initial motion of particle. The scattering angle
is related to the angle
SON as
2. So that
is not the
scattering angle in the conventional Rutherford scattering.
2. Historical Background
M. Longair, Quantum Concepts in Physics (Cambridge, 2013).
5
The discovery of the nuclear structure of atoms resulted from a series of experiments
carried out by Rutherford and his colleagues, Hans Geiger and Ernest Marsden, in the
period 1909–1912. Rutherford had been impressed by the fact that α-particles could pass
through thin films rather easily, suggesting that much of the volume of atoms is empty
space, although there was clear evidence for small-angle scattering. Rutherford persuaded
Marsden, who was still an undergraduate, to investigate whether or not α-particles were
deflected through large angles on being fired at a thin gold foil target. To Rutherford’s
astonishment, a few particles were deflected by more than 90◦, and a very small number
almost returned along the direction of incidence.
Rutherford realized that it required a very considerable force to send the α-particle
back along its track. In 1911 he hit upon the idea that, if all the positive charge were
concentrated in a compact nucleus, the scattering could be attributed to the repulsive
electrostatic force between the incoming α-particle and the positive nucleus. Rutherford
was no theorist, but he used his knowledge of central orbits in inverse-square law fields
of force to work out the properties of what became known as Rutherford scattering
(Rutherford, 1911). The orbit of the α-particle is a hyperbola, the angle of deflection
being
2
cot
20
K
b, (2)
where b is the impact parameter, 0
K is the kinetic energy of the α-particle, 2
2e
Zq
,
and Z
the nuclear charge. The eccentricity of the hyperbola is given by
2
sin
1
e, (3)
where
0
2K
a
and 2
cos
aeb . The hyperbola orbit can be expressed as
1
2
2
2
2
b
y
a
x. (4)
where
22 baae . (5)
6
It is straightforward to work out the probability that the α-particle is scattered through
an angle
. The differential cross section is given by
2
sin
1
16 4
2
0
2
K
d
d
. (6)
This famous 2
csc4
law derived by Rutherford, was found to explain precisely the
observed distribution of scattering angles of the α-particles (Geiger and Marsden, 1913).
Rutherford had, however, achieved much more. The fact that the scattering law was
obeyed so precisely, even for large angles of scattering, meant that the inverse-square law
of electrostatic repulsion held good to very small distances indeed. They found that the
nucleus had to have size less than about 10-14 m, very much less than the sizes of atoms,
which are typically about 10-10 m.
3. Nomenclature and feature of the hyperbola orbit
An alpha particle considered as a massive point charge, incident on the nucleus, is
repelled according to a Coulomb’s law, and, as Newton had already calculated, it follows
a hyperbolic orbit, with the nucleus, with the nucleus as one of the focal points of the
hyperbola. It seems that Rutherford had learned this as a student in New Zealand. Before
we discuss the physics of Rutherford scattering, we discuss the properties of the
hyperbola orbit.
7
Fig.2(a) Nomenclature of the hyperbola.
O
F
1
Ze
F
2
aaaeae
aeae
l
Hyperbola orbit
K
1
K
1
'
L
1
L
1
'
M
2
M
2
'M
1
'
M
1
N
2
'
N
2
N
1
'
N
1
N
1
'
B
2
B
1
Q
1
Q
1
'
C
1
'
C
1
8
Fig.2(b) Detail in the geometry of hyperbola. The point H1 is located on a circle
centered at the origin O with radius a. The point H1 is also located on a
circle centered at the focal point F1 with radius b. The angle 2
FOH 11
.
The points O, H1, F1, and H1’ are on the same circle with radius 2/ae .
e: eccentricity (e>1)
Line 21FF transverse axis
Line 'CC 11 conjugate axis
O
F
1
Ze
F
2
H
1
H
1
'
H
2
'
H
2
aa
b
a
ae
aeae
l
Hyperbola orbit
b
K
1
K
1
'
L
1
L
1
'
b
M
2
M
2
'M
1
'
M
1
N
2
'
N
2
N
1
'
N
1
N
1
'
A
2
A
1
A
1
'A
2
'
ae
b
B
2
B
1
Q
1
Q
1
'
9
O: center of the hyperbola
F1, F2 focal point
K1K1’, L1L1 asymptotes
B1, B2 vertrices
'MM 11 , 'MM 22 directrices
l semi-latus rectum
The hyperbola consists of the two red curves. The asymptotes of the hyperbola are
denoted by the lines K1K1 and L1L1’. They intersect at the center of the hyperbola, O.
The two focal points are labeled F1 (atom with Zqe) and F2, and the line joining them is
the transverse axis. The line through the center, perpendicular to the transverse axis is the
conjugate axis. The two lines parallel to the conjugate axis (thus, perpendicular to the
transverse axis) are the two directrices, M1M1’and M2M2’. The eccentricity e equals the
ratio of the distances from a point P on the hyperbola to one focus and its corresponding
directrix line. The two vertices (B1 and B2) are located on the transverse axis at ±a
relative to the center. θ is the angle formed by each asymptote with the transverse axis.
The length F2N2’ (l) is called the semi-latus rectum,
)1( 2
2
ea
a
b
l. (7)
4. Property of hyperbola
10
Fig.3 (a) Definition of hyperbpora. arr 2
21
. F1 and F2 are focal points. P is a
point on the hyperbola (particle trajectory)
O
F
2
F
1
P
r
2
r
1
B
2
B
1
11
Fig.3(b) a
1
OB , b'AB 11 .
aba 22
21 OFOF , a
1
OH ,
a'OH , b'HFHF 1111 . F1(Zqe): target nucleus (focal point). F2:
another focal point.
12
Fig.3(c) Hyperbola orbit. arr 2
21
. The target nucleus with e
Zq is located at the
focal point F1. The
particle is at the point P on the hyperbola
22 baae .
(a) Evaluation of the distance l
Suppose that the
particle is at the point N2’ of the hyperbola.
arr 2
21 ,
13
from the definition of hyperbola, and
lr
2.
Then we get

222
2
2
2
144)2(2 aallalarr .
Applying the Pythagorean theorem to the triangle 'NFF 221
with the right angle
122 FF'N, we have
222
2
14ealr .
From these two equations, we get
22222 444 ealaall ,
or
)1( 2eal .
l is called the semi-latus rectum,
(b) Equation of hyperbola with r1
We apply the cosine law for the triangle 12PFF
such that
2
1
2
1
2
111
222
1
2
244cos44)2( aarrraereaarr
,
or
arerae 111
2cos
,
leading to the result
1cos1cos
)1(
11
2
1
e
l
e
ea
r. (8)
14
when
where r
1
is the distance between the point P on the hyperbola and the focal point F
1
(the
target with the charge
e
Zq
. When 0
1
, we get the minimum distance

)1(
min
2ear
.
(c) Equation of hyperbola with r
2
For the same hyperbola, we apply the cosine law for the triangle
12
PFF such that
2
2
2
2
2
222
222
2
2
1
44cos44)2( aarrraereaarr
,
or
222
2cos)1( rerea
.
Then we have
22
2
2
cos1cos1
)1(
e
l
e
ea
r
, (9)
where r
2
is the distance between the point P and the focal point F
2
. When 0
2
, we get
the minimum distance

)1(
min
2ear
.
5. Rutherford scattering experiment
Fig.4 Experimental configuration of the Rutherford scattering [P.A. Tipler and
R.A. Llewellyn, Modern Physics 5-th edition (Fig.4.4)]
15
Rutherford scattering is the scattering of
-particle (light-particle with charge 2qe>0)
by a nucleus (heavy particle with charge Zqe). The mass of nucleus is much larger than
that of the
-particle. Thus the nucleus remains unmoved before and after collision.
There is a repulsive Coulomb interaction between the nucleus and the
particle, leading
to the hyperbolic orbit of the
-particle. The potential energy of the interaction
(repulsive) is given by
rr
Zq
Ue
2
2, (in cgs units) (10)
where 2
2e
Zq
. Here we use the charge e
q (>0) instead of e since we use e as the
eccentricity of hyperbola. The boundary conditions can be specified by the kinetic energy
0
Kand the angular momentum L of the
-particles, or by the initial velocity v0 and
impact parameter b,
2
00 2
1mvK , and bmvL 0
, (11)
where m is the mass of the
-particle.
((Note))
particle is He nucleus consisting of two protons and two neutrons (He2+)
6. Linear momentum for the elastic scattering
16
Fig.5 The momentum transfer (denoted by a purple circle) in the momentum
space. The hyperbolic Rutherford trajectory is also presented in the real
space. The angular momentum is conserved before and after the scattering.
The angular momentum: bmvL 0
. For the elastic scattering, b is kept
constant, where b is the impact parameter. The angle between the initial
and the final asymptote of the hyperbola, is related to the impact
parameter b. The hyperbola orbit is expressed in terms of r (distance)
between the point of the orbit and the focal point F1(Zqe) and angle (
).
The angle 0
corresponds to the position of closest approach.
p
i
p
i
p
f
p
2
2
O
F
1
Zq
e
F
2
H
1
H
1
'
aa
a
l
Hyperbola orbit
K
1
K
1
'
L
1
L
1
'
b
b
A
2
'
B
2
B
1
r
17
Fig.6 The elastic scattering for the Rutherford scattering.
2
sin2
2
sin2 0
mvpp i.
Qppp if , (Scattering vector) (12)
where
0
|||| mvp
if pp .
From the Ewald's sphere, we have
F
1
p
i
p
f
p
i
p
2
2
x
y
18
2
sin2
2
sin2 0
mvppQ .
7. Newton’s second law for rotation: torque and angular momentum
The torque is given by
dt
dL
Frτ,
where
is the torque, r is the position vector of the
-particle with charge 2qe (>0) and F
is the repulsive Coulomb force (the central force) between the
-particle and the nucleus
with charge Zqe. The direction of the Coulomb force is parallel to that of r. In other words,
the torque
is zero. The angular momentum L is conserved.
z
dt
d
mrzmrvvrvrrmm rˆˆ
)
ˆ
ˆ
()
ˆ
()( 2
vrprL .
or
bmv
dt
d
mr 0
2
or
2
0
r
bv
dt
d
, (13)
where b is the impact parameter.
((The impulse-momentum theorem))
F
p
dt
d,
or
f
i
f
i
f
i
t
t
t
t
t
t
if dtFdtFFdt )
ˆ
sin
ˆ
(cos)
ˆˆ
(
FppQ .
19
Since Q is parallel to the unit vector
ˆ, we get
f
i
t
t
dtFQ
cos ,
and
0sin
f
i
t
t
dtF
.
Using the relation 2
0
r
bv
dt
d
d
bv
d
bv
r
r
d
d
dt
F
dtFQ
f
i
f
i
f
i
f
i
t
t
t
t
t
t
cos
cos
cos
cos
0
0
2
2
where 2
2e
Zq
,
2
i, at t = ti,
and
2
f. at t = tf.
Here it should be noted that
20
0sin
sin
sinsin
0
0
2
2
d
bv
d
bv
r
r
d
d
dt
FdtF
f
f
f
i
f
i
f
i
t
t
t
t
t
t
Then we get
2
cos
2
sin
2
][sin2
cos2
cos
2
sin2
0
0
0
0
0
0
0
0
bv
bv
bv
d
bv
d
bv
mv
f
f
f
f
i
or
2
cot
22
cot
0
2
0
K
mv
b, (14)
and
0
2K
a
, (15)
where 0
K is the kinetic energy of the bombarding
-particle,
2
00 2
1mvK .
21
8. Approach from conservation law of angular momentum and energy (Born,
Tomonaga)
M. Born, M. Born, Atomic Physics, second edition (Blackie & Son, 1937).
S. Tomonaga, Quantum Mechanics I: Old Quantum Theory (North Holland,
1962). This book was written in Japanese. The English translation of this book
was made by Masatoshi Koshiba. Both Prof. Tomonaga and Prof. Koshiba got
Nobel Prize in 1965 (renormalization) and 1987 (observation of neutrino at
Kamiokande, Japan), respectively. Here we present a brief summary of the
Rutherford scattering based on the Tomonaga’s explanation. We think that the
original idea comes from that of Born.
Suppose an particle with a positive charge 2qe is passing by the positive charge e
Zq
concentrated in the center of the atom. The particle then moves on a hyperbola with F1 as
the outer focus. We take the x-axis through the focal point F1 (the line L
1-M-F1) and
parallel to the line, N2-O-N2’, along which the
particle is approaching the atom from
the left. The hyperbola then has this line, N2-O-N2’, as one of its asymptotes. If we denote
the other asymptote by L1-O-L1’, this gives the direction at infinity after the scattering.
The scattering angle is accordingly given by the angle
'ON'L 21 .
The distance from the x-axis of the
particle at infinity when it is approaching the
atom, i.e., the distance between the line N2-O-N2and the x-axis is denoted by b. This
distance b is a measure of how close the
particle comes to the atom and is an important
quantity in this kind of calculation. Hence this distance b is the impact parameter. When
b is very large, the
particle will pass the atom at a great distance and accordingly suffer
hardly any deflection. When, on the contrary, b is zero, the
particle will make a head-
on collision with the atom and suffer the maximum deflection which, from symmetry
considerations, amounts to 180°. The scattering angle
is in general a function of b, the
form of which we can determine in the following manner.
Let the velocities of the
particle at infinity and at the point of closes approach to the
atom, i.e., at the point P, be denoted by v0 and u, respectively. Then conservation of
angular momentum gives the relation,
22
Fig.7 Feature of the hyperbola. P is a point on the hyperbola orbit.
a2PFPF 21 . Since 21
F'P PF, the points P, N2, P’, and N2’ are on
the circle of radius a centered at the point O. a2'NNFN 2211 . bOM .
b2NF 12 .
min0 murbmvL ,
or
min0 urbv ,
N
1
M
N
2
N
2
'
F
1
Zq
e
F
2
P'
O
P
Hyperbola orbit K
1
K
1
'
L
1
L
1
'
23
where min
r is the length of the line 1
PF . On the other hand, the conservation of energy is
expressed by
mjn
r
mumv
2
2
02
1
2
1,
or
mjn
mr
uv
2
2
2
0, (16)
where
2
2e
Zq
. (17)
From Eqs.(16) and (17), we get
mjn
mr
r
bv
v
2
2
min
2
2
0
2
0,
or
)(
2
0
minmin
2
0
2
min
2
K
rr
r
mv
rb mjn
where
2
00 2
1mvK .
Introducing the angle
by
F1ON222
,
24
we get
cscOF
1bR .
Let the normal to the x axis from the other focus F2 be F2N1. Then from the known
feature of a hyperbola,
PFPF2FN 2111 a, (definition of the hyperbola).
or using the length b and the angle
2, we get
cot2)
2
tan(2tan2FN 11 bbb ,
where
12 NF =b2.
The distance 1
PF is given by
min
r1
PF = OP + 1
OF ,
or
2
cot
2
cos
2
sin2
2
cos2
)cos1(
sin
)cot(csc
2
min
b
b
b
br
Using this value of min
r, we have
25
2
cot
2
cot
0
22
0
2
min
2
b
K
b
r
K
rb mjn
or
2
cot
2
cot
0
2
K
bb ,
or
2
cot)1
2
(cot
0
2
K
b,
or
2
sin
2
cos
2
sin
cos
0
2
K
b,
or
sin
22
cos
2
sincos
00 KK
b.
Then we get the relation
2
cot
2
)
22
tan(
2
tan
2000
KKK
b.
Since 2
cot)
22
tan(tan
aaab , we have
0
2K
a
.
Note that a depends on the kinetic energy K0.
26
9. The Kepler's First Law for the repulsive interaction
We consider the central field problem for the repulsive interaction between the
nucleus ( e
Zq ) and the

particle ( e
q2).
Fig.8 Diagram of the deflection of an
particle by a nucleus: Rutherford
scattering. Repulsive force between the
particle (2qe) on the hyperbola
orbit and the atoms (Zqe) at the focal point F1. Two asymptotes: K1K1’ and
L1L1’.
The Lagrangian of the system is given by
r
P2q
e
O
F
1
Zq
e
F
2
Hyperbola orbitK
1
K
1
'
L
1
L
1
'
27
r
rrmL
)(
2
1222
, (18)
where
2
2e
Zq
.
The Lagrange equation is obtained as
r
L
r
L
dt
d
,
LL
dt
d
,
leading to the equations of motion as
2
2
r
mrrm
,
and
2
mrLz=const. (conservation of angular momentum)
Here we have
dmrLdt 2
.
Note that r depends only on
.
d
d
mr
L
dt
d
dt
d
2
,
)()( 22
d
d
mr
L
d
d
mr
L
dt
d
dt
d,
or
232
2
22 )( mrrm
L
d
dr
mr
L
d
d
mr
L
.
28
We define u as r
u1
,
d
du
rd
d
d
dr
r)
1
(
1
2.
Then we have
232
2
22
2
)( mrrm
L
d
du
d
d
rm
L
,
or
22
2
L
m
u
d
ud
. (19)
where
2
0
2
0
22)( bK
m
bmv
m
L
m
.
Then we have
2
cos L
m
Au
,
since u is an even function of
. So there is no term of
sin . When 0
,
)1(
11
min ear
u
.
When 2
, 0u. Then we get
)1cos(
1
2
e
L
m
r
u,
or
29
1cos1cos
)1(1 2
e
l
e
ea
u
r, (20)
where
e
1
2
sin)
2
cos(cos
.
We note that l is called the semi-latus rectum,
2
0
2
22
)1( bK
m
L
eal .
10. Differential cross section:
d
d
(classical case)
Let us consider all those particles that approach the target with impact parameters
between b and b +db. These are incident on the annulus (the shaded ring shape). This
annulus has cross sectional area
bdbd
2.
These same particles emerge between angles
and
+ d
in a solid angle given by
dd sin2.
The differential cross section d
d
is defined as follows.
bdbd
d
d
d
2
,
or
2
sin2
1
sinsin2
22 b
d
d
d
dbb
d
bdb
d
bdb
d
d
.
30
Fig.9(a) Repulsive scattering of particles with impact parameter b.
d
is the
element of solid angle.
31
Fig.9(b) The differential cross section
d
d
Note that
2
csc
2
cot
22
cot
2
2
2
0
2
2
0
2
Kd
d
Kd
db
.
Then we get
32
2
sin
1
16
2
cos
2
sin4
2
csc
2
cot
2
sin2
2
csc
2
cot
2
sin2
1
4
2
0
2
2
2
0
2
2
0
2
K
K
d
d
K
b
d
d
d
d
(21)
This is the celebrated Rutherford scattering formula. It gives the differential cross section
for scattering of
particle ( e
q2), with kinetic energy 0
K, off a fixed target of charge
(e
Zq )
((Mathematica))
11. Quantum mechanics (scattering due to the Yukawa potential
The Yukawa potential is given by
r
e
r
V
rV
0
)(
, (22)
where 0
V is independent of r. 1/
corresponds to the range of the potential. The
scattering amplitude (from the first order Born Approximation)
Clear
"Global`
"
;b1
kCot
2
;
f1
D
b1,
1
2k Csc
2
2
f2
1
2Sin
D
b1
2
,

TrigFactor
1
4k
2
Csc
2
4
33
0
'
0
2
0
'
0
2
)1(
)'sin('
1
2
)'sin(
'
''
1
2
)(
Qredr
V
Q
Qre
r
V
rdr
Q
f
r
m
r
m
Here we use m
for the reduced mass in order to avoid the confusion of the co-efficient
for the Yukawa potential with the reduced mass.
m
Mm
mM
m
,
where m is the mass of a particle and M is the mass of atom; mM 
Note that
22
0
')'sin('
Q
Q
Qredr r, (Laplace transformation)
22
0
2
0
)1( 12
)(
Q
V
f.
Since
)cos1(2)
2
(sin4 2222
kkQ .
so, in the first Born approximation,
222
2
2
0
2
)1(
])cos1(2[
1
2
)(
k
V
f
d
dm
. (23)
Note that as 0
, the Yukawa potential is reduced to the Coulomb potential, provided
the ratio
/
0
V is fixed.
2
02e
Zq
V,
)2/(sin16
1
)2()2(
)( 444
2
2
2
2
)1(
k
Zq
f
d
dem
.
34
Using
m
k
K
2
22
0
, we have
)2/(sin
1
16
1
)( 4
2
0
2
2
K
f
d
d
,
which is the Rutherford scattering cross section (that can be obtained classically). The
differential cross section can be obtained as follows.
2
2
2
2
2
0
2
)
42
(sin
1
16
k
K
d
d
.
The total cross section is
02
2
2
2
2
0
2
)
42
(sin
sin
16
sin2
k
d
K
d
d
d.
The change of variable 2
sin
2
k
x leads to xdx
k
d2
2
sin
. Then
222
0
2
42
2
2
2
2
2
0
2
22
/2
0
222
0
2
22
4
12
)
4
1
4
(
2
)1(
k
K
k
k
k
K
k
x
xdx
K
kk
12. Schematic diagram for the Rutherford scattering
35
Fig.10 Schematic diagram for the Rutherford scattering. b is the impact parameter
and
is the scattering angle. The hyperbolic orbit near the target (at the
point F1) is simplified by a straight line. ae
1
OF . The point O is the
intersection of the initial and final asymptotes (denoted by red lines) of the
hyperbola. 2
cot
ab . Geometry for the Rutherford scattering.
2
cos
aeb .
is the scattering angle. 2222 eaba .
As shown in the above figure, the impact parameter b is given by
2
cot
2
cos)
22
sin(sin
aaeaeaeb .
The impact parameter b is also expressed by
a
B
A
F
1
Zq
e
O
b
aeb
2
36
2
cot
20
K
b,
where
0
2K
a
. (units of length)
The differential cross section can be expressed by
2
4
4
2
4
2
0
2
4
)(
2
sin
1
4
2
sin
1
16
a
ae
a
K
d
d
where
0
Kis the kinetic energy. bmvL 0
(angular momentum). e
1
2
sin
,
)1()1(
2
22 22
22
2
0
22
2
0
22 eame
a
a
m
a
b
m
a
mbKmbbvmL
.
or
lmL
2.
where l is called the semi-latus rectum,
)1( 2
2
ea
a
b
l.
In general, a particle with impact parameters smaller than a particular value of b will have
scattering angles larger than the corresponding value of b will have scattering angles
37
larger than the corresponding value of
. The area
b2 is called the cross section for
scattering with angles greater than
.
((Note))
Here we discuss how to draw the diagram for the simplified Rutherford scattering. In
Fig.10, the length a
1
BF is given. The scattering angle is changed as a parameter. The
impact parameter b is 2
cot
ab . The length 1
OF is ae. The point O is expressed by
)
2
cos,
2
sin()sin,cos(OF
1
aeaeae ,
where 2/)(
. So we make a plot of the above diagram when the scattering angle
is changed as a parameter, with the value of a kept fixed. The diagram consists of the
initial and the final asymptotes of the hyperbola. For simplicity, the hyperbola is replaced
by the two asymptotes. The point O is the intersection of two asymptotes. Because of the
angular momentum conservation, the impact parameter b remains unchanged for both
initial and final asymptotes.
The eccentricity
of the hyperbola is
2
22
2
2
21a
ba
a
b
e
(>1).
So we have
eae
a1
2
sin
.
38
Fig.11(a) Schematic diagram for the Rutherford scattering where
is varied as a
parameter. The relation between the impact parameter b and the scattering
angle
. As b increases, the angle
decreases (smaller angle).
39
Fig.11(b) The
particles with impact parameters between b and b + db are scattered
into the angular range between
and
+ d
.
40
Fig.11(c) Rutherford scattering of
particles. The hyperbolic orbit near the target
(at the point F1) is simplified by a straight line. R
1
OF with two
asymptotes (straight lines). The point O is denoted by dots in the figure.
The value of a (related to the kinetic energy of the particle) is kept
constant.
13. The use of Mathematica for drawing the hyperbola
41
Fig.12 a = 0.5 (fixed, appropriately normalized). b (also normalized) is changed
as a parameter ( 51.0
b with .1.0
b). F1 is the focal point
(scatterer with the charge Zqe). The center of circle is at the point F1.The
detector is on the circle denoted by purple.
((Mathematica))
F
1
Zqe
42
14. Experimental results by Rutherford’s group
If the gold foil were 1 micrometer thick, then using the diameter of the gold atom
from the periodic table suggests that the foil is about 2800 atoms thick.
Density of Au:
= 19.30 g/cm3.
Atomic mass of Au:
Mg = 196.96654 g/mol.
The number of Au atoms per cm3;
A
g
N
molgM
cmg
n)/(
)/( 3
,
Clear "Global` " ;
Hy1 a,b:Module e1,1,J1,J2,J3,J4 ,e1 a
2
b
2
a;
1ArcTan b
a;
J1 ContourPlot x
2
a
2
y
2
b
2
1, x,6, 0 ,y,6, 6 ,
ContourStyle Hue b5, Thick ,Axes False ;
J2 Graphics Translate J1 1 , ae1,0 ,
Point 0, 0 ;
J3 Graphics Rotate J2 1 , 1,0, 0 ;
J4 Graphics Rotate J2 1 , 1,0, 0 ;ShowJ3,J4
G1 Graphics Black, Thin, Line 5, 0 ,5, 0 ,
Purple, Thick, Circle 0, 0 ,4.7 ,
Text Style "F
2
", Black, Italic, 12 ,0.5, 0 ;
G2 Show Table Hy1 0.5, b,b, 0.1, 5, 0.1 , G1,
PlotRange 5, 5 ,5, 5
43
where NA is the Avogadro number. Then we get the number of target nuclei in the
volume At (cm3) as
ntANs.
Fig.13 The total number of nuclei of foil atoms in the area covered by the beam is
ntA, where n is the number of foil atoms per unit volume, A is the area of
the beam, and t is the thickness of the foil.
[P.A. Tipler and R.A. Llewellyn, Modern Physics 5-th edition
(Fig.4.8)]
If
(= b2) is the cross section for each nucleus,
ntA is the total area exposed by the
target nuclei. The fraction of incident particles scattered by an angle of
or greater is
2
cot2
2
0
2
2
K
Zq
ntbntnt
A
ntA
fe.
The number of
particles which can be compared with measurements, is defined by
2
sin
1
4
)(
4
2
0
4
2
2
0
0
K
qZ
r
ntAI
d
d
ntIN esc
sc

,
44
where r is the distance between the target and the detector, I0 is the intensity of incident
particles, n is the number density of the target, and the solid angle sc
is defined by
2
r
Asc
sc  .
((Experimental results))
Fig.14 (a) Geiger and Marsden’s data for
scattering from thin gold and silver
foils. The graph is a log-log plot to show the data over several orders of
magnitude. Note that scattering angle increases downward along the
vertical axis. (b) Geiger and Marsden also measured the dependence of
N on t predicted by
2
sin
1
4
)(
4
2
0
4
2
2
0
0
K
qZ
r
ntAI
d
d
ntIN esc
sc

for
foils made from a wide range of elements, this being an equally critical
test. Results for four of the elements used are shown. Z = 79 for Au. Z =
47 for Ag, Z = 29 for Cu and Z = 13 for Al.
P.A. Tipler and R.A. Llewellyn, Modern Physics 5-th edition (Fig.4.9).
45
Fig.15 Original data presented by by H. Geiger and E. Marsden [Pjil. Mag. 24,
604, 1913]
Using the value of N(
=
), we have
2
sin
1
)(
)(
4
N
N,
where
2
0
2
4
2
0
4
)( Kr
qntZAI
Nesc
.
46
Fig.16 Plot of
2
sin
1
)(
)(
4
N
N as a function of scattering angle
.
15. Rough evaluation for the size of nucleus
We use the energy conservation law, we have
const
2
12r
mvUKEtot
,
where K is the kinetic energy and U is the potential energy. At r = ∞,
0
2
0
2
1KmvEtot .
At r = r0 (size of nucleus)
0
r
Etot
,
when 0v. Then we have
0
0
K
r
,
50 100 150
q
Degrees
5
10
15
20
25
30
N
q

N
q=
p
47
or
0
0K
r
.
We note that r0 is related to a as
0
2
1ra . (24)
Note that when r0 = 2a,
particle undergoes a head-on collision, during which the
velocity of the
particle becomes zero.
((Example)) Z = 79 for Au. K0 = 7.7 MeV.
r0 = 2.955 x 10-14 m = 29.5474 fermi
In conclusion, most of the mass and all of the positive charge of an atom, +Zqe, are
concentrated in a minute volume of the atom with a diameter of about 10-14 m.
1 fermi = 10-15 m
48
Fig.17 Rutherford scattering. a = 0.05 (kinetic energy fixed, appropriately
normalized). b is changed as a parameter between b = 0.01 and 0.15,
(01.0b). The circle centered at F1 (Zqe) has a radius ar 2
0
.
16. Effective potential for particles with angular momentum l (=0, 1, 2, 3)
We consider the reason why only the particles with small impact parameter b, either
penetrate or is close to the surface of nucleus. The angular momentum l is related to the
impact parameter b as
49
bmvl 0
.
When the momentum 0
mv kept fixed, the angular momentum l is proportional to b. When
b becomes zero, l becomes zero (in other words, only the s wave contributes to the
penetration of
particle into the nucleus. We note that the effective potential is given by
2
2
2
)1(
mr
ll
r
Veff
. (25)
The second term is dominant near r = 0. This term becomes zero (minimum) when l = 0.
So the probability that the s-waves can approach the surface of nucleus is much higher
than that for the p-wave, d-wave, and so on.
Fig.18 Effective potential depending on the angular momentum (which is
proportional to the impact parameter b). l = 0 (s), 1 (p), 2 (d), and 3 (f).
17. Possibility for the penetration of a particle into nucleus
We consider Au as target for the Rutherford scattering.
R
N
V r
r
l 0
l1
l2
l3
V
0
50
e
1
2
sin
,
0
2K
a
,
with
2
2e
Zq
.
Z = 79.
Suppose that 60
.
2
30sin
1
e
When
0
K 10 MeV, we have
a = 11.376 fm,
b = 19.703 fm, (impact parameter)
The shortest distance from the hyperbola to the focal point F1 is
)1( ea 34.127 fm.
MeV
afm
1 e a fm
60
10 20 30 40 50
20
40
60
80
51
Fig.19
= 60°. Au (Z = 79). The distance a [fm] and (1+e)a [fm] as a function of
the kinetic energy K0.
Fig.20 The impact parameter b [fm] as a function of the kinetic energy K0 [MeV].
= 60°. Z = 79 and A = 197 for Au. The blue line denotes the size of
nucleus for Au; 3/1
2.1 ARN = 7 fm. The red and blue lines intersect at
28
0K MeV.
MeV
bfm
60
10 20 30 40 500
10
20
30
40
50
52
Fig.21 Rutherford scattering. Paths of particles with different initial energies, all
scattered by
= 60◦. Au.
If the impact parameter b becomes smaller than RN, α particles penetrate into the nucleus.
For b < RN a new short range force, called the nuclear force, becomes important and the
F
1
Zq
e
60
1020304050fm
10
20
30
40
50
60MeV
53
Coulomb law is no longer valid. The attractive nuclear force is much stronger than the
repulsive Coulomb force and changes the deflection of α particles.
Fig.22 Deviation from Coulomb potential for paths with K0 > 25MeV for by
=
60◦. Au. (copied from the textbook of W. Demtröder).
A = 196.96657 ≈ 197 for Au
RN = 1.2 fm x 3/1
A
= 6.98 fm
Scattering of

particles by gold atoms, or more recent modern variants of Rutherford’s
initial experiments, using high energy electrons or protons, support the Rutherford’s
atomic model, which proposes the following structure of atoms. The major part of the
atomic mass is united in a very small volume, called the atomic nucleus with a typical
radius of 1-5 fm, which is about five orders of magnitude smaller than the electron orbit
surrounding the nucleus. Note that the radius of the electron orbit for the ground state of
hydrogen is 0.53Å.
54
18. -particle decay: quantum tunneling
All nuclei heavier than Pb (Z=82) exhibit alpha activity. Geiger and Nuttall (1911)
found an empirical relation between the half-life
2/1
T
of
decay and the energy E of the
emitted
particles. Using more recent data, the Geiger–Nuttall law can be written as
2/1
2/110
454.183.46log
ZET
where
2/1
T
is in seconds, E in MeV, and Z is atomic number of the daughter nucleus
George Gamow in 1928, just two years after the invention of quantum mechanics,
proposed that the process involves tunneling of an alpha particle through a large barrier.
The barrier is created by the Coulomb repulsion between the alpha particle and the rest of
the positively charged nucleus, in addition to breaking the strong nuclear forces acting on
the alpha particle. Gurney and Condon independently proposed a similar mechanism. A
plot of the nuclear potential also shows the alpha-particle wavefunction . The
amplitude of the transmitted wave is highly magnified
Fig.23 Gamov's model for the potential energy of an alpha particle in a
radioactive nucleus.
r
1
r
2
Coulomb repusion
Nuclear binding
V
r
r
O
E
55
Fig.24 The tunneling of a particle from the 238U (Z = 92). The kinetic energy 4.2
MeV.
http://demonstrations.wolfram.com/GamowModelForAlphaDecayTheGei
gerNuttallLaw/
For r1<r<r2.
ErVMr )(2
1
)(
At r = r2,
2
2
2
r
Zq
Ee
.
Using the WKB approximation, the tunneling probability is obtained as
])(2exp[
1
2
r
r
drreP
,
-
2
0
2
4
6
8
10
r
nuclea
r
pot ential
Me
V
R
e
y
a
r
E
a
56
where
])1([arccos
2
])(arccos[
2
1
2
)(
2
)(
2
1
2
1
2
1
2
121
2
1
2
2
2
1
2
1
2
1
r
r
r
r
r
r
r
EM
rrr
r
r
r
EM
dr
r
r
EM
drErV
M
drr
r
r
r
r
r
r
or
])1([arccos
2
2
2
1
2
1
2
1
2
r
r
r
r
r
r
E
Zq
EM e
where
M is the mass of
-particle (= 4.001506179125 u). fm = 10-15 m (fermi). We
note that
2
1
2
1
2
1
2
1
2
)1(arccos r
r
r
r
r
r
r
r
for 1/ 21 rr . Then we have
ZRqM
E
Z
Mq
Zq
ER
E
Zq
EM
e
e
e
e2
2
2
22
2
1
22
]
2
2
[
2
2
where
2
2
2
r
Zq
Ee
, N
Rr
1 (size of nucleus)
57
The quantity P gives the probability that in one trial an
particle will penetrate the
barrier. The number of trials per second could be estimated as
M
E
Rr
v
N
N
2
2
1
21
,
if it were assumed that a particle is bouncing back and forth with velocity v inside the
nucleus of diameter 2RN. Then the probability per second that nucleus will decay by
emitting a particle, called the decay rate
would be
2
2
e
R
v
N
.
or
2/1
2
10
2/1
2
10
2
1010
2
10
2/1
2
1010
101010
22
)(log
22
)(log]
4
)(log
2
[log
4
)(log
22
)(log
2
log
)(log2
2
loglog
ZE
Mq
eC
ZE
Mq
eZRqMe
R
v
ZRqMeZE
Mq
e
R
v
e
R
v
e
e
Ne
N
Ne
e
N
N
where C is defined by
NeNe
N
ZrqMeZRqMe
M
E
R
C2
10
2
1010
4
)(log
4
)(log)
2
2
1
(log
The leading term depends on energy as 2/1
and C is independent of energy. Using the
fine structure constant
036.137
1
2
c
e
,
2/1
10
log
BEC
,
58
where
2
22
10ln
1cMZB
,
001506466.4
M u,
37.3727
2cM
MeV.
For Z = 90 (Th), 735.154B
Note that
1
2/1 T.
19. Size of nucleus and evaluation of K0
The radius of the nucleus is given by
3/1
0ArRN, (26)
ArRV NN
3
0
3
3
4
3
4
,
where A is the atomic number; NPA
, P is the number of proton, N is the number of
neutron. 2.1
0r fm. The particle density is independent of the species; N and P.
We expect the Rutherford cross-section calculation to fail if a particle can penetrate
inside the nucleus. Classically, this will happen for head-on collisions if the initial kinetic
energy of the α particle is greater than the electrostatic potential at the nuclear surface:
3/2
3/2
2
02.1
2
2A
R
cA
R
cZ
R
Zq
R
K
NNN
e
N
[MeV],
or
3/13/1
0
4.2
2.1
2
A
Z
A
cZ
K
[MeV],
59
where 2
2e
Zq
, α is the fine structure constant, and we use the relation 3/1
2.1 ARN
[fm]. It is expected that the backward scattering may be suppressed for energies greater
than this value. For Au, Z = 79 and A = 197, K0 is evaluated as 33 MeV.
20. Summary: From Rutherford scattering to Bohr model of hydrogen atom
M. Longair, Quantum Concepts in Physics (Cambridge, 2013).
Rutherford attended the First Solvay Conference in 1911, but made no mention of his
remarkable experiments, which led directly to his nuclear model of the atom. Remarkably,
this key result for understanding the nature of atoms made little impact upon the physics
community at the time and it was not until 1914 that Rutherford was thoroughly
convinced of the necessity of adopting his nuclear model of the atom. Before that time,
however, Niels Bohr, the first theorist to apply successfully quantum concepts to the
structure of atoms. Niels Bohr spent four months with Rutherford in Manchester. Bohr
was immediately struck by the significance of Rutherford’s model of the nuclear structure
of the atom and began to devote all his energies to understanding atomic structure on that
basis. In the summer of 1912, Bohr wrote an unpublished memorandum for Rutherford,
in which he made his first attempt at quantizing the energy levels of the electrons in
atoms (Bohr, 1912).
In 1913 Niels Bohr proposed a model of the hydrogen atom that combined the work
of Planck, Einstein, and Rutherford and was remarkably successful in predicting the
observed spectrum of hydrogen. The Rutherford model assigned charge and mass to the
nucleus but was silent regarding the distribution of the charge and mass of the electrons.
Bohr made the assumption that the electron in the hydrogen atom moved in an orbit about
the positive nucleus, bound by the electrostatic attraction of the nucleus. Classical
mechanics allows circular or elliptical orbits in this system, just as in the case of the
planets orbiting the Sun. For simplicity, Bohr chose to consider circular orbits. Such a
model is mechanically stable
REFERENCES
E. Rutherford, Philosophical Magazine 21, 669-688 (1911). “The Scattering of
and
particles by Matter and the Structure of the Atom.”
H. Geiger and J.M. Nuttall (1911) "The ranges of the α particles from various radioactive
substances and a relation between range and period of transformation,"
Philosophical Magazine, Series 6, Vol. 22, No. 130, 613-621. See also: H. Geiger
and J.M. Nuttall (1912) "The ranges of α particles from uranium," Philosophical
Magazine, Series 6, Vol. 23, No. 135, 439-445.
H. Geiger and E. Marsden, Philosophical Magazine, 25, 604–623 (1913). “The laws of
deflexion of α-particles through large angles.”
G. Gamow (1928) "Zur Quantentheorie des Atomkernes" (On the quantum theory of the
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