16.451 Lecture 21: On to Finite Nuclei! 20/11/2003

Review:

1. Nuclear isotope chart: (lecture 1)

• 304 isotopes with t½ > 109 yrs (age of the earth)

• 177 have even-Z, even-N and J = 0+

• 121 are even-odd and only 6 are odd-odd

• N Z for light nuclei and N > Z for heavy nuclei

Z

N

2. Elastic scattering of electrons: (lecture 7)

• nuclei are approximately spherical

• RMS charge radius fitted to electron scattering data

• mass ~ A, and radius ~ A1/3 so the density M/V constant for nuclei ( 2 x 1017 kg/m3), implying that nuclear matter is like an incompressible fluid

fm3/1A2.1R

1

Recall from lecture 7 -- Nuclear charge distributions from experiment:

fm5.0

fm3/11.1

/)(1

:

a

AR

aRre

formFunctional

o

Approx. constantcentral density

2

continued...

3. Inelastic electron scattering: (lecture 9)

• Excited states can be identified, on a scale of a few MeV above the ground state, e.g.

E (MeV)

3

4. Quantum numbers for nuclear states:

• total angular momentum J, parity (see lecture 12, slide 10 for addition rules)

• isospin, T: (lecture 13)

for a nucleus, mT = ½ (Z-N) and T = |mT|, ie lowest energy has smallest T

• Example: “isobaric triplet” 14C, 14N, 14O:

14C: Z = 6, N = 8mT = -1, T = 1

14O: Z = 8, N= 6mT = + 1, T = 1

14N mT = 0, T = 0 (g.s) and T = 1 (8 MeV)

ppnn np

Extra T=0states athigher energyin the mT=0nucleus.

4

How do we understand the quantum numbers of nuclei and their excited states?

• integer or half-integer angular momentum depending on whether A is even or odd

• different systematics and energy level spacings for different nuclei

• some nuclei exhibit “single particle” and others “collective” excitations different models to describe this complementary behavior

5

What is the potential energy function V(r) that nuclei are eigenstates of?

This is not an easy question! The N-N interaction (lecture 20) is too complicated tosolve in a many-body system: state-of-the-art can go up to A = 3!

First approximation: a square well potential, width approx. equal to nuclear radius R:

)(rV

r0

oV

B

K

bound state

R

Assume somehow that we can treat the binding of neutrons and protons like electrons in atoms – individual nucleons have wave functions that are eigenstates of some average nuclear potential V(r).

Each nucleon has a binding energy B as shown (E = -B)

Kinetic energy K = (V0 – B)

R 1.2 A1/3 fm; most of the wave function is contained inside the well, so this should be approximately the right nuclear size...

6

Kinetic energy of bound nucleons:

Key point: once we specify the width of the well, the nucleons are confined, and so their kinetic energy is essentially determined by the uncertainty principle:

Simple estimate: Confining box of side 2 fm. px x ~ ħ

2422

22

MeV103~)/(3

/)(

0

xp

xpppp

p

xxxx

x

2422

22

MeV103~)/(3

/)(

0

xp

xpppp

p

xxxx

x

015.02 2

2

M

p

M

K

Conclusion: the motion is non relativistic; K 15 MeV

7

Now we can specify the potential parameters:

)(rV

r0

oV

B

K ~ 15 MeV

~ - 3 MeV

R ~ 2 fm

~ - 18 MeV

What next?

• We have a complicated system of A nucleons.

• About half of them are protons, so a repulsive (+ve energy) term has to be added to the square well to account for this (~ few MeV)

How to connect this model to something observable?

Independent particle model:

• Assume independent particle motion in some average nuclear potential V(r) as shown.

• Then we can fill the eigenstates of the potential to maximum occupancy to form a nucleus, as is done with electrons in atoms (to 1st order...)

8

Connection to average nuclear properties:

• The binding energy of each nucleon, in our model, is a few MeV.

• The potential energy of a bound nucleon is negative, by ~ 0.3% of its rest mass energy, which therefore has to show up as a decrease in its mass.

• For A nucleons, the total binding energy is:

MmBBA

ii

A

ii

11

MmBBA

ii

A

ii

11

mass of nucleus, M

The average binding energy per nucleon, B/A, can be determined from mass data and used to refine a model for V(r); it ranges systematically from about 1 – 9 MeV as a function of mass number for the stable isotopes.

9

Atomic Mass Units:

• By convention, we set the mass of the carbon-12 atom as a standard.

• Denote atomic masses with a “script” M, measured in atomic mass units, U

M (12C) 12.0000000000 ..... U (exact!) 1 U = 931.502 MeV (expt.)

Calculation for carbon-12:

MeV511.0

MeV6.939

MeV2.938

e

n

p

m

m

m

MeV0.178,1121

MeV8.269,116

U

mi

i

MeV8.91)(126 MmCB i

i

Binding energy per nucleon in 12C: B/A = 7.8 MeV;

Contrast to the deuteron 2H: B/A = 1.1 MeV

10

The famous Binding Energy per Nucleon curve: (Krane, fig. 3.16)

Most stable: 56Fe,8.8 MeV/ nucleon

gradual decreaseat large A due toCoulomb repulsion

very sharp riseat small A

almost flat, apart from Coulomb effects

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Implications of the B/A curve:

Greater binding energy implies lower mass, greater stability.

Energy is released when configurations of nucleons change to populate the larger B/A region nuclear energy generation, e.g.

Fission reactions at large A release energy because the products have greater binding energy per nucleon than the initial species:

MeV2003nYXnU 133100235 MeV2003nYXnU 133100235

distribution of final products

MeV17.6nHeHH

MeV4.03pH

MeV3.27nHeHH

432

3

322

MeV17.6nHeHH

MeV4.03pH

MeV3.27nHeHH

432

3

322

Binding energyof products is greater than the sum of thebinding energiesof the initial species.

Fusion reactions at small A release substantial energy because the B/A curve rises faster than a straight line at small A:

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A semi-empirical model for nuclear binding energies:

1. Volume and Surface terms:

First consider a 1-dimensional row of nucleons with interaction energy per pair =

total: A

ABA

i

221

ABA

i

221

each has 2 neighbors correctionfor the ends

By analogy, for a 3-d nucleus, there should be both volume and surface terms with the opposite sign, the surface nucleons having less binding energy:

3/13/2 AaaAaAaB SVSV AB 3/13/2 AaaAaAaB SVSV A

B

AAB 2 AA

B 2Approximately constant, with endeffects relatively smaller at large A.

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2. Coulomb term:

R

total charge + Ze

for a uniform sphere,

R

Ze

r

dqrqE

ooCoul 4

)(

4

)( 2

53

This effect increases the total energy and so decreases the binding energy.

Simple model: 3/12 AZaB C

But this is not quite right, because in a sense it includes the Coulomb self energyof a single proton by accounting for the integral from 0 to rp ~ 0.8 fm. The nucleushas fuzzy edges anyway, so we will have to fit the coefficient acto mass data.

3/43/1 )1()1( AZZaAZZaB CC AB 3/43/1 )1()1( AZZaAZZaB CC A

B

Solution: let B scale as the number of proton pairs and include a term:

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3. Symmetry Term:

So far, our formula doesn’t account for the tendency for light nuclei to have Z = N. The nuclear binding energy ultimately results from filling allowed energy levels in a potential well V(r). The most efficient way to fill these levels is with Z = N:

energ

y

0

E

2E

states full belowhere..neutrons protons

Z = N

Simplest model: identical nucleons as a Fermi gas, i.e. noninteracting spin- ½ particles in a box. Two can occupy each energy level. The level spacing ~ 1/A.

A mismatch between Z and N costs an energy price of E at fixed A as shown.

1212 )2()( AZAaANZaB AA1212 )2()( AZAaANZaB AA

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4. Pairing Term:

Finally, recall from slide 1 that for the case of even A, there are 177 stable nuclei with Z and N both even, and only 6 with Z and N both odd. Why?

Configurations for which protons and neutrons separately can form pairs must be much more stable. All the even-even cases have J = 0+, implying that neutrons and protons have lower energy when paired to total angular momentum zero.

Solution: add an empirical pairing term to the binding energy formula:

4/3

1

0

1

AaB ppair

4/3

1

0

1

AaB ppair

with +1 for even-even, 0 for even-odd, and -1 for odd-odd

Full expression:

123/13/2 )2()1(),( AZAaAZZaAaAaAZB ACSV 123/13/2 )2()1(),( AZAaAZZaAaAaAZB ACSV

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Fitting of coefficients to data:

123/13/2 )2()1(),( AZAaAZZaAaAaAZB ACSV 123/13/2 )2()1(),( AZAaAZZaAaAaAZB ACSV

MeV5.15Va

MeV8.16sa

MeV72.0Ca

MeV23Aa

MeV34pa (not shown)

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One more look at the Binding Energy per Nucleon curve:

Most stable: 56Fe,8.8 MeV/ nucleon

gradual decreaseat large A due toCoulomb repulsion

very sharp riseat small A

almost flat, apart from Coulomb effects

Solid line: fit to thesemi-empirical formula

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some largeoscillationsat small mass(next class)