Download solutions manual for an introduction to the physics of nuclei and particles 1st

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Solutions Manual for An Introduction to the Physics of Nuclei and Particles 1st

edition by Richard A. Dunlap Link full download: http://testbankcollection.com/download/solutions-manual-for-an-introduction-

to-the-physics-of-nuclei-and-particles-1st-edition-by-dunlap/

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Chapter 3: Nuclear Composition and Size

3.1. (a) The initial kinetic energy, E0, is equal to the Coulombic potential at the point of closest approach, so

Zze2

E0

4 d

Then

Zze2

d 40 E0

(b) Using E0 = 8 MeV, Z = 79, z = 2 and the value of the Coulomb constant

e2

40

1.44

MeVfm

we find b = 28.4 fm.

3.2. Data may be analyzed on the basis of equation (3.12). This may be written in terms of the incident energy, E, as

d Zze2

2

1 2

csc4

. d 40 4E 2

Measurements are made for a fixed angle, , as a function of energy. Since the detector will subtend a sold angle,

, the total cross section for scattering into the detector will be

d Zze2

2

1 2

d 2 csc

4

sin d .

d 40 4E 2

The number of scattered particles observed will be proportional to and hence to E -2.

3.3. The minimum energy will occur for an impact parameter b = 0. In this case the initial kinetic energy, E0, is

equal to the Coulomb potential when the distance between the nuclei is such that their surfaces are just in contact.

This distance is

Using R0 = 1.2*A1/3 fm then

d R RAu

d 1.2 * 4

1/ 3 197

1/ 3 = 8.89 fm.

Equating the energies gives

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0

0 0

c

c c

E0 r

2 .

0 c

Zze2

E0

4 d

Using Z = 79 and z = 2 gives the energy E0 = 25.6 MeV.

3.4. (a) The scattering angle, , is related to the impact parameter, b, as

tan Zze

2

.

2 4 mv 2b

Solving for b and expressing the velocity in terms of the initial kinetic energy, E0;

e2

b 4

Zz .

Using Z = 79, z = 2, E0 = 8 MeV and , = 90° gives

0 2E0 tan 2

b 14.2 fm.

(b) Conservation of energy gives

E 1

mv 2

0 2

c

e2

Zz 40 rc

where the subscript c denotes the point of closest approach. Conservation of angular momentum gives

mv0b mvc rc .

Solving for vc and substituting into the expression for E0 gives

b2

e2

Zz E0 E0

r 2

4 r .

Substituting E0 = 8 MeV and b = 14.2 fm gives a quadratic in rc;

8r

2 228r 1618 0 .

Solving for rc gives rc = 34.4 fm.

(c) The kinetic energy will be

E 1

mv 2

c 2

c b2

c

Substituting numerical values gives Ec = 1.36 MeV.

3.5. The radius of a 208Pb nucleus is R0 = 1.2*A1/3 = 7.11 fm. The volume (assuming a well defined edge at R0) is



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V 4 R

3 1505 fm3

3 0

The number of nucleons is 208 so assuming a uniform density inside R0 the number of nucleons per unit volume is

0 A 0.138 nucleons/fm3.

V

This is consistent with the more detailed picture, Figure 3.9. Since one nucleon has a mass of 1 u then

0 0.138 u/fm3.

In more conventional units this is 0 2.310

25 g/fm3

or

0 2.31014

g/cm3

3.6. (a) All three nuclei have (0) 0.16 fm-3. The values of r90 and r10 are found so that

r90

r10

0.9 0.16 0.144 fm-3

0.1 0.16 0.016 fm-3.

The width of the surface region is then given as r10 – r90. Reading values from the graph gives values in the table.

nucleus r10 (fm) r90 (fm) (r10 – r90) (fm) 16O 3.6 1.2 2.4

118Sn 6.3 3.9 2.4 197Au 7.7 5.3 2.4

(b) Using equation (2.4) and assuming

r90 R0 1.2 2.4 R0 1.2

(fm)

we find that

r10 R0 1.2 (fm).

This gives

90

0.16

1 exp

1.2

a

and



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2

0.0144 .

10

1 exp

1.2

a

Either equation may be solved to give a = 0.55 fm.

3.7. (a) We write equation (3.9) as

Zze

2 2

1 2

cot2

40 4E 2

Using Z=79 and z=2. This gives 79 2 1.44

2

3.14

cot2

406 cot2

fm2

2 10 2 2

For the angles given we find

() (fm2)

1 5.3106

5 2.1105

20 1.3104

(b) We write equation (3.12) as

d Zze

2 2

1 2

csc4

d 40 4E 2

And substituting values as above d

32.4 csc4

d

fm2sr-1

This gives the following results

() (fm2sr-1)

1 5.6109

5 8.9106

20 3.6104

3.8. The relativistic scattering cross section is given as

d d

v2

d

d

1 c

2 sin

2 .

2



5




rel nonrel



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d

d d

v A Af

2

2

We define the relative size of the relativistic correction as

d d

f rel nonrel

d nonrel

For E = 0.1 MeV then mv2/2<<mc2 and we calculate

2 1

mv2

v sin

2 .

c2 2

v2 2

0.2

0.39 .

c2 c2 0.511

For E = 1 MeV or 100 MeV the E > mc2 so v c giving v2/c2 = 1. Using these values in the above expression gives

the results in the table.

E (MeV) v2/c2 (°) f

0.1 0.39 20 0.011

0.1 0.39 90 0.195

1.0 1.0 20 0.12

1.0 1.0 90 0.5

100 1.0 20 0.12

100 1.0 90 0.5

3.9. We define the central part of the nucleus as r < r90. The volume of the central region is therefore

4 r

3

and

the number of nucleons in this volume is

N

4 r

3

3 90

core 0 3

90

The fraction of surface nucleons will be f

Nsurf

Ntotal

Ncore

A Ncore

Ntotal

Ntotal

A

Reading values from the appropriate graphs gives the values in the table.

nucleus r90 (fm) Ncore f

16O 1.2 1.2 0.92 118Sn 3.9 40 0.63 197Au 5.3 100 0.52

3.10. (a) Conservation of energy gives

m (vi f

2 ) m v

2 .



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i i A Af

This may be rearranged to give m v v

v v

m v

2 .

Conservation of momentum gives m vi v f mAvAf .

Combining these expressions yields

vi v f

vAf .

Using this with the conservation of momentum equation gives

v 2 m vi .

mA

The recoil energy (kinetic energy) is obtained from this as

2m m 2v

2

R A i .

m mA

Using E

1 m v

2

i 2

i

this may be rewritten as

R 4m mA Ei .

m mA

(b) Using m = 4.0015 u and Ei = 10 MeV we obtain the following results:

nucleus mA (u) RA (MeV)

16O 15.995 6.4 118Sn 117.902 1.27 197Au 196.967 0.78

f f

m Af

A 2

A 2



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