s7 nuclear physics 2011

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SECTION 7: NUCLEAR PHYSICS

1. Introduction 2

2. Nucleons 2

2.1 Basic properties of the nucleon 2

2.2 Stability of nucleons 3

2.3 Intrinsic magnetic dipole moments 3

2.4 The structure of nucleons 4

3. Nuclear Forces 6

3.1 Properties of the strong force 6

3.2 Decomposition of the N–N potential 8

4. Nuclear Structure 9

4.1 Isotopes and isobars 10

4.2 Stability of nuclei 10

4.3 Nuclear masses and binding energies 11

4.4 Nuclear size and density 12

5. The Deuteron 13

5.1 Orbital angular momentum of the deuteron 14

5.2 The binding energy of the deuteron 15

5.3 Range and depth of the deuteron potential 16

5.4 Good quantum numbers for the deuteron 19

5.5 Angular momentum wave function of the deuteron ground state 20

References

A description of the basic properties of nuclei can be found in any first-year Physics textbook: see for example,

chapter 43 of Fundamentals of Physics, 6th edition, by D Halliday, R Resnick and J Walker, published by

Wiley, 2001. For a discussion of nuclear forces, see any introductory textbook on nuclear physics: see for

example Introductory Nuclear Physics, 1st edition, by K S Krane, published by Wiley, 1987. Specific references

for some sections of these notes are given as footnotes.

Original Version: 2000; Last Major Revision: November 2009

Revision Date: 26 October 2010

Printed: 26 October 2010

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1. INTRODUCTION

The substructure of the atom was investigated in the early part of the 20th century.

• The nuclear model of the atom was introduced by Rutherford in 1911 as a result of a series of experiments

done in 1909, at his suggestion, by his students Geiger and Marsden, who scattered alpha particles off gold

foil. Some particles were observed to be scattered backwards by the atoms in the foil, indicating an

extremely strong electric field in the atom; this led Rutherford to suggest that:

− the atom consists of negatively-charged electrons in orbit about a positively-charged nucleus;

− the nucleus itself is extremely small compared with the atom;

− the atom is therefore mostly a vacuum.

• The existence of positively-charged particles in the nucleus was suggested almost immediately after the

proposal of a nuclear atom. These particles are called protons.

• The presence of other particles in the nucleus was confirmed by the experimental discovery of an elec-

trically neutral particle, named the neutron, by Chadwick in 1932.

Therefore, in the simplest model first suggested by Heisenberg in 1932, all nuclei consist of a number of

particles called neutrons and protons.

2. NUCLEONS

Protons and neutrons are collectively referred to as nucleons; the table shows some of their properties.

Property Proton Neutron

Charge ep = +1.60 × 10−19 C en = 0

Mass mp = 938.26 MeV mn = 939.55 MeV

Lifetime stable 14.8 minutes

Magnetic dipole moment µp = +2.7928 µN µn = –1.9130 µN

Intrinsic spin sp = 1/2 sn = 1/2

2.1 Basic properties of the nucleon

• The neutron is electrically neutral (but see the discussion below concerning the structure of nucleons).

The proton has a charge exactly equal in magnitude but of opposite sign to the electron, thereby ensuring

that the atom is electrically neutral (since the atom contains equal numbers of protons and electrons).

• Like the electron, neutrons and protons have intrinsic spin s = 1/2. Nucleons are therefore fermions; they

obey the Pauli Exclusion Principle and the wave function of a collection of protons or neutrons must be

antisymmetric with respect to the exchange of identical particles.

• The masses of the neutron and proton are given in the table above in terms of the energy equivalent, the

MeV (which is approximately 1.602 × 10−13 J). They are both approximately 1000 MeV, with the neutron

more massive by just 1.29 MeV (i.e. by only 0.1%).

In summary, the neutron and proton are both spin-half fermions with almost the same mass. In addition, their

responses to the strong nuclear force, discussed below, are almost identical.

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For these and other reasons they are conveniently regarded simply as different charge states of the same

particle, the nucleon. They form part of a general classification of all subatomic particles.

2.2 Stability of nucleons

• The currently accepted model of particle physics, the Standard Model, indicates that the proton is stable;

recent experiments place a lower limit of about 2 × 1029 years on its lifetime (to be compared with the

estimated age of the Universe, which is about 1.5 × 1010 years).

• A free neutron (i.e. a neutron not bound within a nucleus) undergoes β – decay to a proton, an electron

and an anti-neutrino (the anti-particle of the neutrino ν) with the release of excess energy and a half-life

of about 10.2 minutes (or a mean lifetime of 14.8 minutes):

n p e ν−→ + +

• Neutrons inside stable nuclei are stable.

2.3 Intrinsic magnetic dipole moments

In order to explain the results of the Stern-Gerlach experiment (1922) two graduate students, Goudsmit and

Uhlenbeck, suggested in 1925 that the electron has an intrinsic spin angular momentum, and associated with

this is an intrinsic magnetic dipole moment. Classically, a spinning charged particle constitutes a magnetic

dipole, with dipole moment proportional to its spin:

sµ γ=� �

where γ is called the gyromagnetic ratio. For an object whose mass m and charge q are identically distributed

over its volume this ratio is given1 by 2q mγ = . Quantum mechanically, as shown by Dirac, the gyro-

magnetic ratio of the electron is in fact double this value, so that for an electron

es

mµ = −� �

(1)

where the charge of the electron is q = −e.

This equation can be generalised to any charged particle with intrinsic spin by introducing a dimensionless

quantity g called the spin g factor of the particle2:

2

qg sm

µ =� �

(2)

where m and q are the mass and charge of that particle.

Clearly, for the electron the Dirac prediction is g = −2, whereas the most accurate experimentally determined

value3 is −2.0023193043622 ± (1.5 × 10−12). The small discrepancy is known as the anomalous magnetic dipole

moment of the electron; it can be described in Quantum Electrodynamics (QED) as arising from the electron’s

interaction with virtual photons (QED is a relativistic theory in which the electromagnetic field is quantised

1 See Griffith page 178 (2nd edition). 2 Be aware that textbooks differ over the precise definition of the magnetic moment and the spin g factor, specifically about the placement of the negative sign for negative charges, and some textbooks are not internally consistent. 3 CODATA 2006.

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in terms of photons). In QED the electron g factor can be expressed as a power series in the fine structure

constant α::

2 12

gα

π

= − + + ⋯

The QED prediction agrees with the measured value to more than 10 significant figures, making it the most

accurately verified prediction in the history of physics.

By convention, the magnetic dipole moment measured for any quantum particle (including atoms and nuclei)

is the expectation value of the z component of the dipole moment operator in the state with maximum spin

projection:

2s z s s z s

qm s m s g m s s m s

mµ µ= = = = = =

For spin-half particles such as the electron, neutron and proton, we have q e= and

1

2s z sm s s m s= = = ℏ

so that

1

2 2

egm

µ =ℏ

. (3)

The combination of constants in this equation, with the mass being that of the electron, is defined as the Bohr

magneton:

55.79 10 eV/T2

B

e

e

mµ −= = ×

ℏ

to be contrasted with the nuclear magneton, defined using the proton mass:

83.15 10 eV/T2N

p

e

mµ −= = ×

ℏ (4)

The nuclear magneton is a convenient unit for magnetic dipole moments in nuclear physics and the Bohr

magneton is widely used as a unit in atomic and molecular physics.

• Note the implication that nuclear moments are typically about three orders of magnitude smaller than

those of atoms.

With these definitions the expression for the magnetic dipole moment of a particle becomes

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2gµ = (5)

when expressed in the appropriate magneton unit.

2.4 The structure of nucleons

The spin g factor of a fundamental particle can be found by solving the appropriate relativistic quantum

mechanical equation. The wave equation for particles with s = ½ is the Dirac equation, which predicts the

relationship eq.(2)

2

qg sm

µ =� �

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with g = −2 exactly for an electron.

• Since the proton is also a spin-half particle with charge of the same magnitude as the electron but opposite

sign, use of the Dirac equation would predict for the proton a g factor of g = 2.

• Since the neutron is electrically neutral, its g factor should be g = 0.

Measurement indicates that for a free proton g = +5.58 and for a free neutron g = −3.82.

• The signs of the magnetic dipole moments indicate that for the proton the dipole moment and spin vectors

are aligned whereas for the neutron they point in opposite directions.

• Note that the magnetic moment of the neutron is conventionally given in terms of the nuclear magneton,

which contains the proton mass and charge.

The fact that the measured g factors are far from the expected values indicates that protons and neutrons are

not actually fundamental, point-like particles but have a finite size and a substructure. The deviations from

the expected values are attributed to the motion of charged fundamental particles, called quarks, within the

nucleons.

• Nucleons are the ground states of bound systems of two types of quarks, up-quarks (u) with charge

2 /3e+ and down-quarks (d) with charge /3e− ; the neutron has structure (ddu) and the proton (uud).

Further evidence of nucleon substructure is provided by experiments in which electrons of sufficiently high

energy are scattered off protons or neutrons.

• These indicate that the neutron, although electrically neutral, does have an internal distribution of charge,

as shown in the diagram below4. There is an inner layer of positive charge surrounded by a layer of

negative charge and a very weak outer layer of positive charge; the net charge is zero.

• Similarly, the proton also has a charge distribution that is very roughly proportional to exp( )r a− where

a ≈ 0.23 fm. The measured charge radius of the proton is 0.875 fm.

These charge distributions reflect the average distribution of quarks within protons and neutrons. Note that

for both nucleons, the charge distribution extends well beyond the nominal radius of the particle.

4 These figures are adapted from RM Littauer et al, Phys. Rev. Lett. 7, 144 (1961). See Modern Physics, 2nd Edition, by HC Ohanian, published by Prentice Hall, 1995.

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3. NUCLEAR FORCES

Nucleons are subject to all four basic interactions, listed here in order of increasing strength.

• Gravitational force: This force acts between all pairs of particles in the universe. When acting between

nucleons, it is so weak compared with other forces that it can be ignored.

• Electromagnetic force: This force acts between all pairs of charged particles or between pairs of magnetic

dipoles.

• Weak nuclear force: The weak interaction is responsible for the β decay of many unstable nuclei (and of

the neutron and some other fundamental particles).

• Strong nuclear force: The strong interaction is responsible for the stability of nuclei; it acts between pairs

of particles called hadrons, which includes neutrons and protons. The other main class of fundamental

particles, the leptons (which includes electrons, neutrinos and muons) do not experience the strong

nuclear force.

3.1 Properties of the strong force

The fundamental strong interaction is actually the force between quarks; the N–N interaction (N = nucleon)

is a residual effect of this strong interaction, in the same way that the inter-molecular van der Waals force is

a residual effect of the electromagnetic interaction within molecules.

A proper theory of the N–N interaction requires a description in terms of the quark constituents of the

nucleons and the use of quantum chromodynamics (QCD); such a theory has yet to be fully developed. The

force between nucleons can however be described quite accurately as arising from the exchange of mesons,

mainly pions, which are bound states of quark-antiquark pairs such as ( ) ( )uu , ud etc.

We consider5 now properties of the strong interaction between two nucleons, which can be deduced from:

• the properties of the deuteron, the nucleus of deuterium which is an isotope of hydrogen with a single

neutron in its nucleus, and

• measurements of nucleon-nucleon scattering experiments.

Charge independence and charge symmetry

Analysis of N–N scattering experiments indicates that, provided the two nucleons are coupled to the same

spin-space state, the n – n interaction and the nuclear part of the p – p interaction are essentially equal; the

N–N interaction is said to be charge symmetric. It is also charge independent (the n – p interaction is the same

as the other two).

Strength of the interaction

Since the protons in the nucleus do not fly apart because of the repulsive Coulomb forces between them, we

deduce that:

5 See Section 4.4 of the book by Krane for a detailed description of the properties of the strong nuclear force, including its de-composition as discussed in the next subsection.

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− at short distances (a few fm), the nuclear force is strong compared with the electrostatic force between

two charged particles at the same separation;

− the force is basically attractive.

Range of the interaction

The force has a short range, being essentially zero for N–N separations greater than a few fm, to be compared

with the electrostatic and gravitational forces, both of which have an infinite range.

Evidence for the short-range is provided by the fact that the nuclear force plays no part in the interaction

between atoms in a molecule.

In addition, as we shall discuss below, the total binding energy of a nucleus is roughly proportional to the

number of nucleons it contains, implying that each nucleon can interact only with its nearest neighbours in

a nucleus and therefore sees the same environment independent of how many nucleons there are at large

distances. The nuclear force is said to “saturate” – a nucleon can interact only with a limited number of other

nucleons.

Repulsive core

At N–N separations somewhat less than about 1 fm, the force is repulsive, becoming strongly so at smaller

nucleon separations. This feature is called the repulsive nuclear core.

Evidence for the repulsive core:

− Nuclei do not collapse; as will be discussed below, the density of the nuclear interior is approximately

the same for all nuclei.

− The analysis of nucleon-nucleon scattering experiments conducted at sufficiently high nucleon beam

energies (so that small distances can be probed) also indicates the need for a repulsive core to the

nuclear force.

The nuclear force within nuclei

When discussing the effect of the strong interaction between nucleons within a nucleus there are a number of

additional considerations.

• In principle there could be many-body forces between nucleons (i.e. forces that require the presence of

more than two particles). However, study of three-nucleon systems such as the nuclei 3H and 3He indicate

that a three-body force, if present, must be much weaker than the two-body force. Nuclear binding and

other properties of nuclei can be understood in terms of two-body forces, and therefore study of two-body

systems is appropriate for a description of nuclei.

• However, when interpreting information extracted from two-body systems, it must be noted that in a

nucleus the effects of the force between any two nucleons are modified somewhat by the presence of the

other nucleons (partly because of the Pauli Exclusion Principle which hinders the exchange of energy in

collisions between nucleons).

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3.2 Decomposition of the N–N potential

The N–N strong interaction is conveniently regarded as being made up of several parts6, listed in order of

decreasing symmetry and increasing complexity:

( )1 2 1 2 3 4 12( ) ( ) ( ) ( )V V r V r s s V r L S V r S= + ⋅ + ⋅ +��

A central component V1(r)

Many of the properties listed above are evident in this component of the interaction (as modified by the

following two terms).

A spin-dependent term

This is also central in nature; its magnitude depends on whether the nucleon spins are aligned or anti-aligned.

It follows from

( )2 2 21 2 1 2 2s s S s s⋅ = − −� �

where 1s�and 2s

� are the spin operators of the two nucleons and S is the total spin of the two nucleons, that

( ) 2

2 1 2 2

22

1( ) ( ) for 1

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( ) for 04

V r s s V r S

V r S

⋅ = + =

= − =

� �ℏ

ℏ

Combining this term with the previous term gives a central potential whose magnitude depends on spin.

Evidence for such a term comes from the spin-dependence of N–N scattering cross-sections7 and also the fact

that there is only one bound state of the deuteron. If the nucleon-nucleon interaction were independent of spin,

we would expect to find deuteron bound states with S = 0 and S = 1 at more or less the same energy. In fact,

6 Other more complicated terms with different symmetries could be added to this equation. 7 See page 92 of the book by Krane for a detailed discussion.

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as we shall see below, the deuteron ground state has S = 1 whereas the S = 0 state is unbound, about 2.3 MeV

above the ground state.

A spin-orbit term 3( ) .V r L S⋅��

This is also central in nature. Evidence for such a term, and its sign, comes from the measurement and analysis

of N–N scattering cross-sections8.

Note that this is not a relativistic correction; it is large and of the opposite sign to the spin-orbit interaction

in atoms. In the atom, the spin-orbit term is due to a magnetic interaction, whereas in the nucleus it arises

from the strong interaction.

Note also that, since L r p= ×� � �

, the presence of such a component in the force indicates that the interaction

is velocity-dependent.

A non-central component or tensor term

This depends on how the spins are orientated relative to the vector n�, the unit vector along the line joining

the two nucleons:

( )( ) ( )12 1 2 1 23 .S s n s n s s= ⋅ ⋅ − ⋅� � � � � �

S12 is called the tensor operator.

• The term ( )1 2s s⋅� �

is conventionally included here for convenience; it ensures that if the tensor term is

averaged over all directions n� the result is zero.

This interaction has the same angular structure as the potential between two magnetic dipoles but its origin

is completely different; it can be understood as arising from the exchange of mesons between the interacting

nucleons. It is important because it is the only term that mixes states of different orbital angular momentum.

The best evidence for a non-central component comes from the non-zero quadrupole moment of the deuteron

ground state. If the deuteron potential were central in nature, the ground state would have zero angular

momentum and its wave function would be spherically symmetric. This implies a zero electric quadrupole

moment. Since the ground state has a non-zero quadrupole moment, it cannot therefore be pure l = 0. Wave

functions with mixed l can only result from a non-central potential.

4. NUCLEAR STRUCTURE

By convention, a nuclide (a specific species of nucleus) is referred to by the symbol AZ NX where

• X is the chemical symbol for the corresponding atom.

• Z is the number of protons, called the atomic number. It is equal to the number of electrons in the atom.

• N is the number of neutrons and is called the neutron number.

• A = N + Z is the total number of nucleons, called the mass number since the mass of a nucleus AZ NX is

approximately A in atomic mass units (defined below).

Note that N and Z are superfluous and may be omitted, so that the usual notation for a nuclide is AX.

8 See page 104 of the book by Krane for a detailed discussion.

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Calculating the properties of a nucleus accurately is a much more difficult task than the corresponding

calculation for an atom.

• In the latter case the electrons are bound by the central Coulomb potential due to the electric charge of

the nucleus while the residual effect of the electron-electron interactions can be treated as a perturbation.

In the case of the nucleus, the nucleons are bound by a potential that is due to the nucleons themselves;

there is no additional object to create a dominant central potential. The problem is therefore intrinsically

many-body in nature and becomes tractable only through the use of substantial approximation.

• The simplest atomic system, the hydrogen atom, has an infinite number of energy levels that permit a

comprehensive and quantitative comparison between experiment and theory.

In contrast, the simplest nuclear system, the deuteron, has only one bound state and can therefore provide

only limited guidance to the applicability of any theory to more complicated nuclear systems.

• Even after many decades of study, the nuclear force is not well-described; this is clearly not the case for

the corresponding force in the atomic case, the Coulomb interaction, which is well-understood.

This has lead to the development of a number of quite different models of the nucleus, each dealing with a

specific set of nuclear properties. In this course we do not attempt to describe any nuclear models, restricting

attention to a basic description of nuclear properties and a discussion of the deuteron.

4.1 Isotopes and isobars

• Atoms whose nuclei have the same number of protons but different numbers of neutrons are called

isotopes; examples are 12C, 13C and 14C which are all isotopes of the element carbon. Their chemical

properties are almost identical but their nuclear properties are quite different.

• Similarly, nuclei that have the same number of nucleons but different numbers of protons or neutrons are

called isobars. Examples are 14C, 14N and 14O; they have some nuclear properties in common.

• Nuclei that have the same number of neutrons but different numbers of protons are called isotones;

examples are 2H and 3He, both of which contain one neutron.

4.2 Stability of nuclei

More than 3000 isotopes are known; many are found in nature, but most have only been created (and studied)

in laboratories.

• Only 266 isotopes are stable, the heaviest being 20983Bi . The majority of isotopes are radioactive; the nuclei

of these are not stable but after a certain length of time will either fragment or emit particles to change

into a different element.

• The naturally occurring nuclei include the 266 stable isotopes plus a limited number of radioactive

isotopes that have extremely large lifetimes (typically billions of years); an example is 23892U which has a

half-life of 4.5 × 109 y for α decay.

• Traces of a few isotopes with short lifetimes also exist naturally since they themselves result from the

decay of nuclei with long lifetimes or from some other nuclear process. An example is tritium, an isotope

of hydrogen with two additional neutrons in the nucleus, which has a half-life of 12.3 y for β− decay.

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4.3 Nuclear masses and binding energies9

Nuclear masses are usually quoted in terms of the unified atomic mass unit u, which is defined such that the

mass of an atom of 12C is exactly 12 u.

• Thus neutrons and protons have masses of approximately 1 u.

• The conversion factor between this unit and the MeV can be calculated from E = mc2, yielding

1 u = 931.494 MeV/c2.

The mass m(Z,N) of a nuclide with Z protons and N neutrons differs from the mass M(Z,N) of the corre-

sponding atom by the total mass of Z electrons and the total electronic binding energy of the atom:

2 2 2

1

( , ) ( , ) ( )Z

e ii

m Z N c M Z N c Zm c B Z=

= − +∑

where B(Z)i is the energy required to remove the i’th electron from the atom.

Electronic binding energies are of the order of 10 – 100 keV in heavier atoms (and much smaller in lighter

atoms), whereas atomic masses are of the order of A × 1000 MeV. Therefore to a precision of about 1 part in

106 or better the last term in this equation can be neglected. In fact, in most instances we work with differences

in masses, and the effect of the atomic binding energies almost entirely cancels out in these differences.

A nucleus is a bound system – you would need to supply energy to split it into its constituent nucleons. The

amount of energy required to do this is called the binding energy of the nucleus; values are typically tens or

hundreds of MeV (to be compared with binding energies measured in eV or at most keV for atoms).

The binding energy of a nucleus containing Z protons and N neutrons is defined as

[ ] 2( , ) ( , )p nB Z N Zm Nm m Z N c= + −

where m(Z,N) is the mass of the nucleus, mp is the mass of the proton, and mn is the mass of the neutron.

Except for the very lightest nuclei, the nuclear mass m(Z,N) cannot be measured directly, since it is impossible

to strip off all the electrons from the atom. However, as we have just indicated, with almost no loss of accuracy

we can replace m(Z,N)c2 by 2 2( , ) eM Z N c Zm c− and mp by mH – me (where mH is the mass of the hydrogen

atom). Then we get

[ ] 2( , ) ( , )H nB Z N ZM Nm M Z N c= + −

Thus nuclear binding energies are usually calculated from measured atomic masses.

A useful quantity for comparing the stability of nuclei is the ratio B/A, the binding energy per nucleon, which

for A > 20 is roughly constant at about 7-9 MeV per nucleon as shown in the graph below.

9 See pages 65-6 of the book by Krane.

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4.4 Nuclear size and density

The nucleus, when compared with the atom, has a fairly sharp boundary. Experimentally it is found that the

radius R of a nucleus with mass number A is given approximately by

1/30R R A=

where the precise value of R0 depends of the type of experiment carried out to measure it (the distribution of

nuclear charge is not identical to the distribution of nuclear matter); typically R0 = 1.1 – 1.2 fm. Thus nuclei

have diameters of the order of 10−14 m, to be compared with atoms which have diameters of the order of

10−10 m. It should be noted however that nuclei are not necessarily spherical but are in general ellipsoidal; the

formula must then be interpreted as giving the average radius of the nucleus.

It follows from this formula that the volume of a nucleus is roughly proportional to A, the number of nucleons

it contains. This, coupled with the fact that the radius of a single nucleon is of the order 1 fm, has several

implications:

• The nucleons are tightly packed in a nucleus, and are themselves incompressible; adding more nucleons

does not squeeze the inner nucleons into a smaller volume. This obviously provides evidence for the

nucleon-nucleon interaction having a repulsive core.

• Nuclear matter has a constant density, with the same value for all nuclei. From the equation R = R0A1/3

with R0 = 1.15 fm this can be estimated as about 2.6 × 1017 kg/m3, to be compared with the density of

ordinary matter which is of the order of 103–104 kg/m3.

• The density of nucleons within any nucleus is found by dividing the density of nuclear matter by the mass

of a nucleon, giving approximately 0.16 nucleons/(fm)3.

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The diagram below shows the charge density for several nuclei as deduced from electron scattering experi-

ments10; the shape of the distribution was assumed in the analysis. The vertical dashed lines indicate the radius

predicted by the equation R = R0A1/3 with R0 = 1.2 fm.

The matter density of a nucleus can be found by dividing the charge density by the proton charge to find the

proton density, multiplying by the proton mass to give the proton mass density, and finally multiplying by

A/Z to find the nucleon matter density (assuming that the proton and neutron densities are the same).

Using information from the diagram:

• The density of nuclear matter is roughly the same for all the nuclei shown (this can be seen by multiplying

each curve by A/Z), and the value deduced agrees well with the value obtained from R = R0A1/3.

• The density is constant within a nucleus, except close to the nuclear surface where it decreases gradually

to zero. The skin thickness is a parameter that is defined as the distance over which the density decreases

from 90% to 10% of its central value; it is typically about 2.4 fm for all nuclei.

5. THE DEUTERON

The deuteron 2H, the nucleus of the atom deuterium, is a bound state of a proton and a neutron. It is the

simplest nuclear system.

• There is no bound system of two neutrons or two protons.

• Experiments show that the deuteron has only one weakly bound state, i.e. it has a ground state and no

bound excited states.

Various measured properties of the deuteron ground state are shown in the table below.

10 The figure is adapted from R Hofstadter, Ann. Rev. Nucl. Sci. 7, 231 (1957), © Annual Reviews Inc.

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Property Value

Charge ed = +1,60 × 10–19 C

Mass Md = 1875.5803 MeV

Magnetic dipole moment µd = +0.85741(2) µN

Electric quadrupole moment Qd = +0.282 fm2

Charge radius (rms) rC = 2.1 fm

Total angular momentum Jd = 1

With regard to these properties, the following should be noted.

• The charge of the deuteron is entirely due to the proton.

• The charge radius of the deuteron provides an indication of its size.

• The mass of the deuteron is not exactly equal to the sum of the proton and neutron masses. This is further

discussed below in connection with the deuteron’s binding energy.

• The magnetic dipole moment of the deuteron is not exactly equal to the sum of the proton and neutron

moments. This is discussed further in the next section.

• The fact that the quadrupole moment is not zero indicates that the deuteron is not spherical. A positive

moment indicates that it is shaped like a rugby ball.

• The total angular momentum of a nucleus is often rather confusingly referred to as the nuclear spin; it is

not the total intrinsic spin of the nucleons that make up the nucleus. The quantum number S for the

deuteron is discussed below.

5.1 Orbital angular momentum of the deuteron

By analogy with the ground state of the hydrogen atom, we might expect that the deuteron ground state has

orbital angular momentum zero; taken together with the measured angular momentum Jd = 1 it then follows

that the deuteron ground state has intrinsic spin S = 1. Since the deuteron has no other bound states, the

implication is that two nucleons will not be bound together in the S = 0 spin state. The nuclear force is thus

seen to be spin-dependent, as already discussed.

The diagram below shows schematically the four lowest-energy states of the two-nucleon system: two neutrons,

a neutron and a proton (the deuteron) and two protons (the nucleus 22He ). On the right of each state are

shown J π, its total angular momentum and parity. To the left are shown energies (in MeV) relative to the

ground state of the deuteron.

If the orbital angular momentum of the n-n and p-p states is zero, their spatial wave functions must be

symmetric with respect to interchange of the two particles, so the spin state must be antisymmetric, i.e.

S = 0. Thus these two-nucleon systems must have J = 0 and the J = 1 state cannot exist. These arguments

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do not apply to the deuteron since the two nucleons are not identical and J = 1 is therefore possible; this is

the ground state and the J = 0 state is an excited state.

The three J = 0 states shown in the diagram differ from one another only through the replacement of a proton

by a neutron, or vice versa. Because of the charge independence and charge symmetry of the strong interaction

they therefore have the same nuclear energy of interaction (i.e. the same energy after allowing for the small

neutron-proton mass difference and the Coulomb energy difference). Since the deuteron J = 0 state is known

to be unbound by about 70 keV, we would therefore not expect the two-neutron system to be bound. For the

two-proton system, the Coulomb repulsion between the two particles renders it even more unlikely to be

bound – the Coulomb energy of two protons about 2 fm apart is approximately 0.7 MeV. These observations

are confirmed by experiment – no bound states of two neutrons (or two protons) have been found.

One consequence of the assumption of zero orbital angular momentum for the deuteron is that its magnetic

dipole moment should just be the sum of the proton and neutron moments (since the orbital contribution,

which is proportional to the orbital angular momentum, must be zero). We should therefore expect:

µd = µp + µn = (2.7928 – 1.9130) µN = 0.8798 µN

The fact that this does not agree with the value in the table above, 0.8574 µN, casts doubt on the assumption

of zero orbital angular momentum. This would seem to indicate that the force between the nucleons has a

non-central component, as already mentioned – this will also be discussed more fully below.

Neither the proton nor the neutron has an electric quadrupole moment, so that the measured non-zero

quadrupole moment of the deuteron also points to the presence of non-zero angular momentum.

5.2 The binding energy of the deuteron

The binding energy of the deuteron ground state can be calculated from the general equation presented earlier,

using measured masses for the neutron, proton and deuteron:

2( )

938.256 939.550 1875.5803 MeV

2.226 MeV

d p n dB E m m m c= − = + −

= + −

=

The magnitude of the binding energy of a nucleus is an indicator of its stability. Most stable nuclei have

binding energies per nucleon in the range 6.5 to 8.5 MeV, indicating that the deuteron is not strongly bound.

In general it is not possible to measure the binding energy of a nucleus directly in an experiment, but the

deuteron is an exception. Its binding energy can be deduced from the kinematics of the photo-disintegration

reaction:

d n pγ + → +

where γ represents the photon. The process will occur only if the photon energy exceeds the deuteron binding

energy (there is a small correction for the recoil energy of the proton and neutron, necessary to conserve

momentum).

The deuteron binding energy can also be determined from the inverse of this reaction, by measuring the energy

of the γ-ray photon energy emitted when a neutron and proton at rest combine to produce a deuteron.

16

5.3 Range and depth of the deuteron potential11

The two body problem for the deuteron can be reduced to a one-body problem for the deuteron’s relative

motion by separating out the centre of mass motion in the usual way.

We consider a very simple model for the relative motion, in which the potential energy term in the Ham-

iltonian is approximated by a spherical square well of depth V0 > 0 and range R. This model, although only

a crude approximation to reality, is sufficient to give a qualitative description of certain properties of the

deuteron.

0( ) for

( ) 0 for

V r V r R

V r r R

= − <

= >

r is the separation of the neutron and proton, so that

R is a measure of the size of the deuteron.

The first step in solving the Schrödinger equation for the relative motion is the separation of variables for the

spatial wave function,

( )( )

( ) ,lm

u rr Y

rθ φΨ =

�,

where R(r) = u(r)/r is the radial wave function and ( , )lmY θ φ is a spherical harmonic representing the angular

part of the wave function.

We assume that the ground state of the deuteron is a bound state with orbital angular momentum l = 0 (an

assumption we know to be not quite accurate). The Schrödinger equation for the radial wave function then

reduces to

[ ]2

2 2

( ) 2( ) ( ) 0

d u r mE V r u r

dr+ − =ℏ

where

• m is the reduced mass of the deuteron; from 1 1 1

p nm m m= + we have m = 469.5 MeV, and

• E = −2.23 MeV is the experimentally determined ground-state energy of the deuteron.

For the square-well potential defined above, the Schrödinger equation becomes

11The original version of this section was based on problems 2.4 and 2.5 on page 70 of the textbook by Mandl. See also pp 82-3 of the book by Krane.

17

22

2

22

2

( )( ) 0 for

( )( ) 0 for

d u rk u r r R

dr

d u ru r r R

drκ

+ = <

− = >

where the constants k and κ are defined by

( )02 2

2 2 and .

mE mk V Eκ

−= + = + +

ℏ ℏ

For the wave function ( )u r r to be normalizable, we require boundary conditions12

( 0) 0

( ) 0

u r

u r

= =

→ ∞ =

The solution to the Schrödinger equation is then

( ) sin for

( ) exp( ) for

u r B kr r R

u r C r r Rκ

= <

= − >

where B and C are normalization constants. From the normalisation condition on the radial wave function,

it follows that

2( ) 1u r dr∞

−∞=∫

leading to

2 2 2

0sin exp( 2 ) 1

R

RB kr dr C r drκ

∞+ − =∫ ∫

This can in principle be used to give a relationship between the constants B and C.

From the condition that the wave function must be continuous at r = R we find

( )sin expB kR C Rκ= −

and from the condition that its derivative must be continuous at r = R we have

( )cos exp .kB kR C Rκ κ= − −

Two of these three equations can be used to fix the values of B and C. The existence of a third equation

imposes a relationship between k and κ; this implies a relationship between the parameters that define the

potential, assuming that the energy E is known. Dividing the second equation by the third yields a dispersion

relation:

tank

kRκ

= −

or

( )2

002

2tan .

mR V EV E

E

++ = −

−ℏ

This is a transcendental equation for which there is no exact analytic solution; it must be solved graphically

or numerically.

12 Remember that the radial wave function R(r) for l = 0 must be non-zero at the origin, which requires that u(r) vanish at the origin.

18

We can find an approximate solution by assuming 0E V≪ . Since the ground state of the deuteron is only

weakly bound and there are no bound excited states, this seems a reasonable assumption; it will be justified

below. Then the equation above reduces to

2

0 02

2tan

mR V V

E

−= − ≈ −∞

ℏ

Thus, to lowest order, the solution for the ground state is

2

02

2

2

mR V π=

ℏ

Or, rearranging,

2 2

0 2 .8V

mR

π=

ℏ

We see that the range and strength of the potential are directly related (this is generally the case, whatever

the form of the potential).

• Scattering experiments indicate that the range of the N–N force is approximately 1.5 fm; if we use this

value for R, the approximate equation yields V0 ≈ 50 MeV. We see that the assumption 0E V≪ is

justified since E = 2.23 MeV.

• A slightly more accurate solution of the transcendental equation indicates that kR = 116° rather than π/2

radians as found above, and the expression for V0 becomes

2 2

0 2 2

2

8

c EV

RmR mc

π= +

ℏ ℏ

• The approximate wave function u(r) calculated above peaks at r = R where the form of the solution

changes from a sine curve to exponential decay. The slightly more accurate calculation shows that the

wave function actually peaks just inside r = R, as shown in the diagram below.

There is evidently a large probability of measuring an n – p separation in the deuteron ground state that is

larger than the range of the potential. This partially explains why the charge radius of the deuteron

(2.1 fm) is larger than the range of the interaction.

• As already discussed, the N–N interaction is spin dependent. We shall see below that in the deuteron

ground state the neutron and proton are coupled to S = 1, so what we have calculated is the strength of

the S = 1 potential. Analysis of the scattering of neutrons by protons indicates that the strength of the

S = 0 potential is V0 ≈ 12 MeV, which is insufficient to produce a bound state (the S = 0 state of the

deuteron is unbound by 70 keV).

19

5.4 Good quantum numbers for the deuteron13

The form of the interaction between the neutron and proton, discussed earlier, imposes some restrictions on

the quantum numbers that can be used to label the states of the deuteron.

Central term V1(r)

As discussed in the section on atomic structure, the valid quantum numbers for a pure central potential are:

, , , , ,L p sp n snL M s m s m

where L� is the relative orbital angular momentum of the deuteron, and ML its projection onto the z axis, sp,

msp are the spin quantum numbers of the proton and sn, msn are the neutron’s spin quantum numbers.

Since the term V1(r) is clearly invariant under spatial inversion r r→ −� �

, parity π is also a good quantum

number.

Spin-dependent term ( )2( ) n pV r s s⋅� �

It is easily seen that

( ) ( )2 2 2 2 21 1 32 2 2n p p ns s S s s S⋅ = − − = −

� �ℏ

where p nS s s= +� � �

is the total intrinsic spin of the deuteron and we have put sp = sn = ℏ/2.

• The operator S 2 contained in n ps s⋅� �

does not commute with spz and snz so that msp and msn are no longer

good quantum numbers.

• However, S 2 does commute with S 2 and Sz; hence the corresponding quantum numbers S and MS can be

specified.

• Also, S 2 obviously commutes with L2 and Lz; hence the quantum numbers L and ML are still valid.

• Since this term in the Hamiltonian is invariant under spatial inversion14, parity π is still a good quantum

number.

Thus the valid quantum numbers are now

, , , , .L SL M S M π

Spin-orbit term 3( )V r L S⋅��

As we have shown previously

( )2 2 21

2L S J L S⋅ = − −��

where J L S= +� ��

is the total angular momentum of the deuteron.

• The operators J 2, L 2 and S 2 contained in the term L S⋅�� all commute with L2 and S 2, so that the

quantum numbers L and S are still valid.

• The operator J 2 does not commute with Lz and Sz so that ML and MS are no longer good quantum

numbers.

• However J 2 does commute with J 2 and Jz, hence the corresponding quantum numbers J and MJ can be

specified.

13The original version of this section was based on problem 5.8 on page 142 of the textbook by Mandl. 14 Remember that angular momentum is an axial vector and is therefore invariant under the parity operation.

20

• In addition, the spin-orbit interaction is invariant under spatial inversion, so that parity π is still a valid

quantum number.

The good quantum numbers are now

, , , , .JJ M L S π

Tensor term ( )( ) ( )4( ) 3T n p n pV V r s n s n s s = ⋅ ⋅ − ⋅ � � � � � �

It can be shown15 that, with r rn=� �

,

( )2

212 2

31.

2

S rS S

r

⋅ = −

� �

• The interaction VT does not commute with L2 since the dependence on r� destroys the rotational

symmetry. Thus L is no longer a valid quantum number.

• However VT does commute with J 2 and Jz; hence the corresponding quantum numbers J and MJ can still

be specified.

• The tensor term commutes with S 2 so that S is still a good quantum number. This also follows from the

fact that S12 is symmetric under interchange of sn and sp; consequently the spin states must be either

symmetric or antisymmetric in the labels n and p, i.e. they are either triplet or singlet states with

S = 1 or S = 0, respectively.

• S12 does not commute with Sz and the quantum number MS is therefore not valid.

• In addition, the interaction VT is clearly invariant under spatial inversion, so that parity π is still a valid

quantum number

The good quantum numbers are finally

, , ,JJ M S π

with all four terms included in the Hamiltonian.

5.5 Angular momentum wave function of the deuteron ground state16

From experiment, the deuteron ground state has total angular momentum J = 1. In addition, the total spin

angular momentum of the proton plus neutron must be either S = 1 or S = 0. From the fact that L J S= −� ��

we can deduce the possible values of the orbital angular momentum l, namely , ,( )l J S J S= − +⋯ , as

shown in the following table:

Spin Orbital angular momentum

S = 0 l = 1

S = 1 l = 0, 1, 2

No other combination of S and l can produce J = 1.

The parity of a state of orbital angular momentum l is given by π = (–1)l. Hence the four possible states for

the deuteron are:

15 See page 280 of the book by Mandl. 16The original version of these notes was based on pages 639 to 642 of Introductory Quantum Mechanics by R L Liboff, 2nd edition, published by Addison Wesley (1992).

21

3 31 1

1 31 1

and with

and with

D S

P P

π

π

= +

= −

• As we have discussed in the last section, l is not a valid quantum number; therefore the ground-state

cannot be a single configuration and must be a mixture of configurations with different values of l.

• Since parity is a good quantum number, the ground state must correspond to a mixture of configurations

with the same parity.

• We also determined in the last section that S is a valid quantum number, so that the ground state must

have definite S = 1 or S = 0 and is not a mixture of configurations with different S. Hence we can rule out

the possibility of a combination of the two wave functions with negative parity.

We therefore expect the deuteron ground state to be a mixture of the configurations 3S1 and 3D1 with S = 1

and π = +.

5.5.1 Magnetic dipole moment of the deuteron ground state

To proceed further we consider the magnetic dipole moment of the deuteron ground state. The magnetic

dipole moment of a nucleus arises from a combination of two sources:

• Firstly, each nucleon has an intrinsic magnetic dipole moment associated with its intrinsic spin (or more

fundamentally, it arises from the intrinsic spin and orbital motion of its quark constituents).

• Secondly, since a proton carries a net positive charge, its orbital motion constitutes a current loop which

produces a magnetic dipole moment.

The magnetic dipole operator for the deuteron is therefore given by

d p n Lµ µ µ µ= + +� � � �

The orbital contribution, due to the proton, is

1

2 2Lp

eL

mµ =

��

where we have assumed that each of the nucleons carries one-half the orbital angular motion associated with

the relative motion between them.

The contributions from the intrinsic moments of the nucleons are, from the table of deuteron properties,

with 5.5862

with 3.826.2

p p p pp

n n n np

eg s g

m

eg s g

m

µ

µ

= = +

= = −

� �

� �

Hence

( ) ( )

12

12

2

2 2 2

d p p n np

p n p np n p n

p

eg s g s L

m

g g g ges s s s L

m

µ = + +

+ − = + + − +

��

��

As discussed previously, the magnetic dipole moment measured for any nucleus is the expectation value of the

z component of the dipole moment operator in the state with maximum spin projection MJ = J.

22

Now it is easily shown17 that

( ) 0S SSM pz nz SMs sχ χ− =

for all two-particle spin states (1,2)SSMχ . Hence, the effective magnetic dipole operator is

( )1 12 22

0.880 0.5002

0.500 0.380 .2

z p n z zp

z zp

z zp

eg g S L

m

eS L

m

eJ S

m

µ = + +

= +

= +

where we have inserted the values of the g factors and used Jz = Lz + Sz. Therefore the predicted magnetic

dipole moment is

0.50 0 0.38 02

0.500 0.3802

d J z z Jp

J z J J z Jp

eM J J S M J

m

eM J J M J M J S M J

m

µ = = + =

= = = + = =

From the discussion above on valid quantum numbers for the deuteron, we know that the deuteron wave

function is an eigenfunction of the operator Jz but not of Sz. The first term is easily determined:

J z J JM J J M J M J= = = =ℏ ℏ

The expectation value of the operator Sz is easily found18.

( ) ( ) ( )( )

1 1 1, ,

2 1J z J J

J J S S l lLS JM J LS JM M

J J

+ + + − +=

+ℏ

Assuming for the moment that l is a valid quantum number, the result is

( ) ( ) ( )

( ){ }1 1 10.500 0.380

2 1d N

J J S S l lJ

J Jµ µ

+ + + − + = +

+

where we have introduced the nuclear magneton /2N pe mµ = ℏ . For the deuteron ground state J = 1 and the

equation becomes

( ) ( ){ }[ ]0.690 0.095 1 1d N S S l lµ µ= + + − +

The following table shows the calculated magnetic dipole moments for the various possible deuteron states

determined above.

3

1

3

3

For 0.880 with

For 0.500 with

For 0.690 with

For 0.310 with

d N

d N

d N

d N

S

P

P

D

µ µ π

µ µ π

µ µ π

µ µ π

= = +

= = −

= = −

= = +

Because the measured value is µd = +0.857 µN, which does not equal within the experimental uncertainty any

of the values calculated for pure configurations, we confirm that the ground state does not have good l.

17The proof is part of a tutorial problem. 18 Refer to the notes on the hydrogen atom in a weak magnetic field for a reference.

23

We concluded earlier that the deuteron ground state is a mixture of the configurations 3S1 and 3D1; its wave

function is therefore of the form:

( ) ( )3 3d S Da S a Dψ ψ ψ= +

where the quantities aS and aD are coefficients to be determined. The normalization requirement for the wave

function gives

2 2 1.S Da a+ =

With this wave function we find that:

( ) ( )2 3 2 3d S Da S a Dµ µ µ= +

where the quantities µ(3S) and µ(3D) represent the calculated dipole moments listed in the table above. The

cross terms that would otherwise appear in this equation are zero because of the orthogonality of the wave

functions ( )3Sψ and ( )3Dψ .

The coefficients aS and aD are now adjusted to reproduce the measured dipole moment µd = +0.857 µN. We find

2

2

0.96

0.04

S

D

a

a

=

=

We conclude that the deuteron ground state has definite S = 1 and positive parity, is mainly l = 0, but has

a 4% admixture of l = 2. It is this admixture that is responsible for the non-spherical shape of the deuteron.

Note that this argument based on the magnetic dipole moment is seriously flawed (there are other mechanisms

that could contribute to the measured dipole moment), but the conclusion is essentially correct!

s7 nuclear physics 2011

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