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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

 

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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))

 

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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

 

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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.

 

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((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

 

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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,

 

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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

 

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Note that

 

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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))

 

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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

 

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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 .

 

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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).

 

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Fig. The particles with impact parameters between b and b + db are scattered into the

angular range between and + d.

 

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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

 

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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

 

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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,

 

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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

 

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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

 

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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

 

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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

 

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Fig. Schematic diagram of a particles being scattered by the atomic nuclei in a thin

metallic film

 

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Fig. Diagram of the deflection of an particle by a nucleus: Rutherford scattering

 

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Fig. Mechanical model of an atomic nucleus for demonstrating Rutherford scattering