1 Rutherford scattering Masatsugu Sei Suzuki Department of Physics, SUNY at Binghamton (Date: January 13, 2012) 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 ________________________________________________________________________ 1. Rutherford scattering experiment
23
Embed
13-2 Rutherford scattering - Binghamton Universitybingweb.binghamton.edu/.../13-2_Rutherford_scattering.pdf · Rutherford scattering is the scattering of -particle (light-particle
This document is posted to help you gain knowledge. Please leave a comment to let me know what you think about it! Share it to your friends and learn new things together.
Transcript
1
Rutherford scattering Masatsugu Sei Suzuki
Department of Physics, SUNY at Binghamton (Date: January 13, 2012)
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".
P.A. Tipler and R.A. Llewellyn, Modern Physics 5-th edition (Fig.4.4)
Rutherford scattering is the scattering of -particle (light-particle with charge 2e) by a nucleus (heavy particle with charge Ze). 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 is given by
r
ZeU
0
2
4
2
The boundary conditions can be specified by the kinetic energy K and the angular
momentum L of the -particles, or by the initial velocity v0 and impact parameter b,
202
1mvK , and bmvL 0
where m is the mass of the -particle.
((Note)) particle is He nucleus consisting of two protons and two neutrons (He2+) ((Ewald's sphere))
3
Fig. The hyperbolic Rutherford trajectory.
pi
p f
pi
Dpr
q
f
q2q2b
b
z
h
x
y
O
Opi
p f
pi
Dp
qq2q2
z
x
y
4
Fig. Ewald's sphere for the Rutherford scattering
Qppp if (Scattering vector)
where
0|||| mvpif pp
From the Ewald's sphere, we have
2sin2
2sin2 0
mvppQ
((Conservation of angular momentum))
dt
dLFrτ ,
where is the torque, r is the position vector of the -particle with charge 2e (e>) and F is
the repulsive Coulomb force (the central force) between the -particle and the nucleus with charge Ze. 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.
zdt
dmrzmrvvrvrrmm r ˆˆ)ˆˆ()ˆ()( 2 vrprL .
or
bmvdt
dmr 0
2
or
20
r
bv
dt
d
where b is the impact parameter.
5
((The impulse-momentum theorem))
Fp
dt
d,
or
f
i
f
i
f
i
t
t
t
t
t
t
if dtFdtFFdt )ˆsinˆ(cos)ˆˆ( FppQ
Since Q is parallel to the unit vector ̂ , we get
f
i
t
t
dtFQ cos
and
0sin f
i
t
t
dtF .
Using the relation 2
0
r
bv
dt
d
dbv
Ze
dbv
r
r
Ze
dd
dtF
dtFQ
f
i
f
i
f
i
f
i
t
t
t
t
t
t
cos2
4
1
cos2
4
1
cos
cos
0
2
0
0
2
2
2
0
where
6
2
i , at t = ti,
and
2
f . at t = tf.
Here it should be noted that
0sin2
4
1
sin2
4
1
sinsin
0
2
0
0
2
2
2
0
dbv
Ze
dbv
r
r
Ze
dd
dtFdtF
f
f
f
i
f
i
f
i
t
t
t
t
t
t
Then we get
2cos
4
4
1
sin4
4
1
][sin22
4
1
cos22
4
1cos
2
4
1
2sin2
0
2
0
0
2
0
00
2
0
00
2
00
2
00
bv
Ze
bv
Ze
bv
Ze
dbv
Zed
bv
Zemv
f
f
ff
i
or
2cot
4
1
2cot
2
4
1 2
02
0
2
0
K
Ze
mv
Zeb ,
where K is the kinetic energy of the bombarding -particle,
7
202
1mvK
2. Differential cross section: d
d
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.
bdbdd
dd 2
or
2
sin2
1
sinsin2
22b
d
d
d
dbb
d
bdb
d
bdb
d
d
8
Note that
9
2csc
2cot
42cot
42
2
0
22
2
0
22
K
Ze
d
d
K
Ze
d
db
.
Then we get
2sin
1
)8(
2cos
2sin4
2csc
2cot
4
sin2
2csc
2cot
4
sin2
1
422
0
42
2
2
0
2
2
2
0
2
2
K
eZ
K
Ze
d
dK
Ze
bd
d
d
d
This is the celebrated Rutherford scattering formula. It gives the differential cross section
for scattering of particle (2e), with kinetic energy K, off a fixed target of charge Ze. In general, this formula can be rewritten as
2sin
1
)16(
)(
422
0
2
K
qQ
d
d
for scattering of a charge q, with kinetic energy K, off a fixed target of charge Q. ((Mathematica))
10
3. Schematic diagram for the Rutherford scattering
Fig. 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 O) is
simplified by a straight line. ROA .
As shown in the above figure, the impact parameter b is given by
Clear"Global`"; b1 k Cot2;
f1 Db1,
1
2k Csc
22
f2 1
2 Sin Db12, TrigFactor
1
4k2 Csc
24
O
A
b Rq
ff
11
2cos)
22sin(sin
RRRb
The impact parameter b is also expressed by
2cot
2cot
4
1 2
0
kK
Zeb
where
K
Zek
2
04
1
Then we get
2sin
k
R
The differential cross section can be expressed by
42
sin
1
42
sin
1
)8(
2
4
2
422
0
42 Rk
K
eZ
d
d
In general, a particles 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 larger
than the corresponding value of . The area b2 is called the cross section for scattering
with angles greater than .
12
Fig. 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).
13
Fig. The particles with impact parameters between b and b + db are scattered into the
angular range between and + d.
14
Fig. Rutherford scattering of particles. The hyperbolic orbit near the target (at the
point O) is simplified by a straight line. ROA . The point denoted by OA is shown in the figure.
4. Experimental results
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
15
Atomic mass of Au;
Mg = 196.96654 g/mol The number of Au atoms per cm3;
Ag
NmolgM
cmgn
)/(
)/( 3
where NA is the Avogadro number. Then we get the number of target nuclei in the volume At (cm3) as
ntANs
Fig. 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
16
2cot
42
2
0
22
K
Zentbntnt
A
ntAf .
The number of particles which can be compared with measurements, is defined by
2sin
1
)8()(
422
0
42
20
0
K
eZ
r
ntAI
d
dntIN sc
sc
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
2r
Ascsc .
((Experimental results))
P.A. Tipler and R.A. Llewellyn, Modern Physics 5-th edition (Fig.4.9). Z = 79 for Au. Z = 47 for Ag, Z = 29 for Cu and Z = 13 for Al.
Using the value of N( = ), we have
2sin
1
)(
)(
4
N
N,
17
where
220
2
420
)8()(
Kr
entZAIN sc
5. Size of nucleus
We use the energy conservation law, we have
constr
ZemvUKEtot
0
22
4
2
2
1
where K is the kinetic energy and U is the potential energy. At r = ∞,
KmvEtot 202
1
At r = r0 (size of nucleus)
00
2
4
2
r
ZeEtot
Then we have
50 100 150q Degrees
5
10
15
20
25
30
NqNq=p
18
Kr
Ze
00
2
4
2
or
00
2
0 2 K
Zer
((Example)) Z = 79 for Au. K = 7.7 MeV.
r0 = 2.955 x 10-14 m. ((Mathematica))
6. CONCLUSION
Most of the mass and all of the positive charge of an atom, +Ze, are concentrated in a minute volume of the atom with a diameter of about 10-14 m. REFERENCES P.A. Tipler and R.A. Llewellyn, Modern Physics 5-th edition (W.H. Freeman, 2008).
H.E. White, Introduction to Atomic and Nuclear Physics (D. Van Nostrand Company, Inc.,
Princeton, NJ, 1964). _____________________________________________________________________ APPENDIX-I: Property of hyperbola
Clear"Global`";
rule1 eV 1.602176487 1019, qe 1.602176487 1019,
0 8.854187817 1012, MeV 1.602176487 1013, Z 79,
K0 7.7 MeV;
Z qe2
2 0 K0. rule1
2.954731014
19
The properties of hyperbola
22 bac
rPF 1 , 12 rPF ,
arr 21 , arr 21
or
22221 44)2( aarrarr
Cosine law:
cos44 2221 rccrr
Using the above two equations, we get
cos4444 222221 rccraarrr
OF1 F2
P
r r1
A1 A2
-4 -2 0 2 4
-4
-2
0
2
4
20
cos1 e
pr
,
where
a
ba
a
ce
22 >1
)1( 2 eap . The ParametricPlot:
cosrcx , sinry
with
cos1 e
pr
.
APPENDIX-II
From the book of H.E. White
Introduction to Atomic and Nuclear Physics (D. Van Nostrand Company, Inc., Princeton,
NJ, 1964).
Fig. Diagram of the Rutherford scattering experiment
21
Fig. Schematic diagram of a particles being scattered by the atomic nuclei in a thin
metallic film
22
Fig. Diagram of the deflection of an particle by a nucleus: Rutherford scattering
23
Fig. Mechanical model of an atomic nucleus for demonstrating Rutherford scattering