NuclearPhysics

Part 1: Nuclear Structure & Reactions

Last modified: 25/01/2018

Contents LinksThe Atomic Nucleus

NucleonsStrong Nuclear ForceNuclei Are Quantum SystemsAtomic Number & Atomic Mass NumberNuclidesIsotopesRadioisotopesLine of Stability

Binding EnergyMass DefectBinding Energy

Non-SI units: ’u’ and MeVNuclear vs Atomic MassesExample

Binding Energy CurveNuclear Reactions

Q-valueQ-value & Binding EnergyExothermic & Endothermic ReactionsExampleExample

Summary

The Atomic Nucleus Contents

Atoms consist of negatively charged electrons orbiting a small, positivelycharged nucleus.

Atomic nuclei vary in size, but always make up a tiny fraction of anatom’s volume. If we imagine that a nucleus were 1 mm in diameter,then the atom would be in the range of 30-50 metres across. An atom ismostly empty space.

∼ .1− .5 nm ∼ 2− 15 fm

proton (p)

neutron (n)

Despite its small size, the nucleus contains nearly all of the atom’s mass.

Nucleons Contents

The nucleus is made up of nucleons, of which there are two types: theproton, which is positively charged (equal and opposite to the electroncharge) and the neutron, which has zero charge.

mass (kg) electric charge (C)proton (p) 1.672623× 10−27 1.602× 10−19

neutron (n) 1.674928× 10−27 0

Apart from electric charge, the properties of the two types of nucleon arevery similar - in particular they have almost the same mass, with theneutron’s being only slightly larger.

Remember the mass of the electron is 9.109390× 10−31 kg, about 2000times smaller than the proton/neutron mass.

Strong Nuclear Force Contents

The positively charged protons inside a nucleus will experience repulsiveCoulomb forces. How does the nucleus stay together?

There is another force acting between nucleons, (imaginatively) calledthe strong nuclear force (or often just strong force).

Strong Force Properties:I It only acts between nucleons. Electrons are not affected by

the strong force.I It is always attractive.I The type of nucleon doesn’t matter. The force between a

neutron and a proton is the same as the force between twoneutrons or between two protons.

I It has a very short range - the force between two nucleonsdrops to zero at a separation of only a few femtometres(1 fm = 1015 m). This is very different to the familiarCoulomb and gravitational forces, which although theybecome weaker with distance, never completely disappear.

Contents

To form a stable nucleus, the strong forces holding the nucleus togethermust be greater than the Coulomb forces pushing the protons apart. Thiswill not always happen.

For instance, if we bring 2 protons close to-gether, then they will be attracted by thestrong force (red), but the repulsive Coulombforce (blue) will ‘win’ and so they do not forma nucleus.

But, after adding a neutron to the two pro-tons, additional strong forces will act, andin this case they can overcome the repul-sive Coulomb force and hold the nucleus to-gether.

p pCoulomb forces: repulsiveStrong forces: attractive

p pn

This pattern continues as larger numbers of nucleons are combined. Inonly some combinations will the strong force be able to overcome theCoulomb repulsion between the protons, and so form a nucleus.

Nuclei Are Quantum Systems Contents

We have already seen that atoms are quantum systems. Electrons in anatom must exist with one of a limited range of energies and can movebetween energy levels by absorbing or emitting photons of the appropriateenergy. These photons will typically have energies of a few eV or less.

Nuclei are also quantum systems. Like electrons in an atom, nucleonscan jump between energy levels by emitting or absorbing photons. Thesephotons are much more energetic than those seen with atoms and willtypically have energies measured in MeV (gamma rays).

Exactly the same terms are used as for atoms - the lowest energy level isthe ground state of the nucleus, and higher energy levels are theexcited states.

Different types of nuclei will each have a unique emission and absorptionspectrum, which can be measured experimentally and used to identifythem.

Atomic Number & Atomic Mass Number Contents

The number of protons in a nucleus is called the atomic number, Z ofthat nucleus.Z identifies elements in the periodic table, so determines what type ofatom the nucleus is part of. For example: Z = 6→ carbon, Z = 20→calcium, Z = 92→ uranium etc.

The number of neutrons is the neutron number, N.

The total number of nucleons is A = Z + N, the atomic massnumber. Since the mass of the proton and the neutron are very similar,the mass of the nucleus will be approximately A times the nucleon mass.

Nuclides Contents

The word nuclide is used to refer to the type of a nucleus (in the sameway as the word element is used to describe the type of an atom).A nuclide is represented using the following notation:

X chemical symbolAatomic mass number

Zatomic number

For example, the nuclide with 6 protons and 6 neutrons is:12

6C .

You may also see this be written as: 12C, carbon-12 or C12.

In speech this nuclide is usually referred to as ‘carbon 12’.

Isotopes Contents

The atomic number Z determines the number of electrons in the atom,and hence the atom’s chemical properties.

Adding or removing neutrons from the nucleus, will of course change themass of the nucleus, but not affect this chemistry (or at least only veryslightly).

Nuclides with the same atomic number, but different atomic massnumber, are known as the isotopes of the element.

Only a limited number of isotopes of each element are possible. Someisotopes can exist, but are unstable and will, over time transform intoother stable isotopes - usually of a different element. We will look atthese processes in more detail in the next lecture.

26 elements (including aluminium, sodium and gold) have only one stableisotope. Other elements have varying numbers up to the ten stableisotopes of tin.

Radioisotopes Contents

Naturally occuring elements will usually contain a mixture of stable andunstable isotopes, some of them being much less common than others.

For example, there are three possible isotopes of hydrogen: 11H, 2

1H(known as ‘deuterium’) and 3

1H (‘tritium’).

Of these, tritium is unstable and the others stable. Naturally occurringhydrogen atoms are 99.9885% 1

1H, the remainder 21H with only very small

amounts of tritium.

Some unstable isotopes (such as tritium) occur naturally as the result ofnuclear processes, while others are only seen when they are producedartificially in a lab.

As they transform into other nuclides, unstable nuclei will usually emitsome form of radiation, and are said to be radioactive.

Radioactive isotopes are known as radioisotopes.

Line of Stability Contents

The graph at right shows all observed iso-topes plotted with neutron number N vsatomic number Z and shaded to indicateeach isotope’s lifetime.

Stable isotopes (shown in black) form acurve known as the ‘line of stability’.

Some observations from this plot:I Lifetimes decrease (stability

decreases) further away from thisline

I The heaviest stable isotope hasZ = 82 (lead)

I For lighter nuclei (Z . 20) thestable isotopes have Z ≈ N

I Heavier stable isotopes haveincreasing numbers of neutrons,with N/Z ≈ 1.5 for the heavieststable nuclei

20 40 60 80 100

20

40

60

80

100

120

140

160

Z

N

long

erlif

e

stable

Isotope Lifetimes

Mass Defect Contents

The simplest multi-nucleon nucleus is 21H, also known as a ‘deuteron’,

consisting of one proton and one neutron:

p nImagine reaching into a deuteron at rest, grabbing hold of the twonucleons and pulling them apart, to a separation distance beyond therange of the strong force, where they are again at rest.

p nFF

⇒ p nrest rest rest

The applied forces have done work - adding energy to the deuteron.Where has this energy gone?

Contents

We began with the deuteron at rest, so it only had rest mass energy.The separated nucleons are also at rest, so also only have rest massenergy.

From this we conclude that the rest mass of the deuteron must beless than the combined rest masses of the two nucleons.

The difference between the total mass of the constituent nucleons andthe mass of a nucleus is called the mass defect of that nucleus.

For example, the mass defect of a deuteron is:

δm = mproton + mneutron −mdeuteron

= 1.672622× 10−27 + 1.674927× 10−27 − 3.343583× 10−27

= 0.003966× 10−27 kg

Binding Energy Contents

The energy equivalent of the mass defect (i.e. δm × c2) is calledthe binding energy of the nucleus.

The binding energy of a deuteron is:

0.003966× 10−27 × (2.99792× 108)2 = 3.564× 10−13 J = 2.225 MeV

The general formula for the binding energy BE of the nucleus AZX

with mass mX is:

BE = [Z ×mproton + (A− Z )×mneutron −mX ] c2

A larger binding energy indicates a smaller nuclear mass, and thus thatthe nucleus is more tightly stuck together.

Non-SI units: ‘u’ and MeV Contents

The masses of nucleons and nuclei expressed in kilograms are very smalland awkward to calculate with. For this reason, Nuclear Physicscalculations usually use a more convenient unit of mass - the unifiedatomic mass unit (u) (also called the dalton (Da))

Unified Atomic Mass Unit

1 u = 112 mass of a 12

6C atom = 1.660538921× 10−27 kg

Some example masses in u:

proton (p) 1.00727647 11H atom 1.00782504

neutron (n) 1.00866492 42He atom 4.002602

electron (e) 5.485799× 10−4 23892U atom 238.050788

The mass in u of the nuclide AZX is always very close to A. We will soon

see that these small differences are very important.

Contents

Of course you will not be expected to remember atomic masses (exceptpossibly 12

6C). In exams all necessary masses will always be given.If you ever need to look up a mass, Wikipedia is a good source. If forexample you needed the atomic mass of a gold isotope, search for‘Isotopes of gold’ .

The energy equivalent of 1 u is:

(1 u in kg)× c2 = 1.660538921× 10−27 × (2.99792× 108)2

= 1.49241× 10−10 J

= (1.49241× 10−10)(1.602× 10−19) eV = 931.5 MeV

Binding (and other) energies in Nuclear Physics are usually expressed inMeV.

Nuclear vs Atomic Masses Contents

The previous definition for the binding energy of the nucleus AZX used the

nuclear mass, which was represented as mX .

Especially for heavier nuclei, it is usually much easier to measure themass of an atom of A

ZX rather than this nuclear mass.

What is the difference? Z electrons!

If we represent the atomic mass of AZX by MX , then:

MX︸︷︷︸atom

= mX︸︷︷︸nucleus

+ Zme︸︷︷︸electrons

where me is the mass of an electron.

For example, the atomic mass MH of the hydrogen atom 11H is

MH = mproton + me

Contents

Returning to the definition of binding energy, and expressing the nuclearmass mX in terms of the atomic mass MX :

BE (in MeV) of AZX = [Z mproton + (A− Z ) mneutron −mX ]× 931.5

= [Z mproton + (A− Z ) mneutron − (MX − Zme)]× 931.5

= [Z (mproton + me) + (A− Z ) mneutron −MX ]× 931.5

= [Z MH + (A− Z ) mneutron −MX ]× 931.5

This has the same form as the original definition, but with thesubstitutions:

mproton →M(11H) and mX →MX

We will always use atomic masses so this is the required form of theequation.

Example Contents

Calculate the binding energy of 42He.

42He consists of 2 protons and 2 neutrons. Using atomic masses inu (shown earlier), the binding energy in MeV will be:

BE =(

2×M(11H) + 2×mn −M(4

2He)

)× 931.5

= (2× 1.00782504 + 2× 1.00866492− 4.002602)× 931.5= 0.03037792× 931.5 = 28.297 MeV

We might feel tempted to round these mass values off to one ortwo digits.In this case, that would give a very incorrect result:

BE = (2× 1.0 + 2× 1.0− 4.0)× 931.5 = 0 MeV

ALL the given decimal places in mass values must be used!

Contents

Calculate the binding energy per nucleon of 42He.

The previous calculation gave the binding energy for 42He to be

28.297 MeV. There are FOUR nucleons in total, so the bindingenergy per nucleon is calculated simply:

28.297 MeV/4 = 7.074 MeV

The average binding energy, or binding energy per nucleon, allowsus to compare the binding energies, and hence the stability, of differentnuclides.

Plotting the results of similar calculations to the above for all stablenuclides gives the following graph:

Binding Energy Curve Contents

BE/A (MeV)

A0 50 100 150 2000

2

4

6

8

10

42He

126C

168O 56Fe

Notes:I For small nuclei, as A increases, the average

binding energy also increases - indicating moretightly bound (i.e. more stable) nuclides.

I The curve peaks in the A = 50 − 60 range, sonuclei in this region are the most stable. 56

28Feis generally considered the most stable nuclide.

I The curve drops gradually for larger A,indicating less stable heavier nuclei.

I There are several unusually stable light nuclei:42He, 12

6C and 168O.

Nuclear Reactions Contents

A nuclear reaction occurs when a number of nuclei combine together toproduce other nuclei (very much like molecules combining in a chemicalreaction):

A1Z1

X1 + A2Z2

X2 + . . .→ A′1Z′1Y1 + A′2

Z′2Y2 + . . .

For such a reaction, there are two conservation laws which can be used toidentify an unknown nuclide:

Z1 + Z2 + . . . = Z ′1 + Z ′2 + . . .

A1 + A2 + . . . = A′1 + A′2 + . . .

For example, to identify the mystery particle X in:12

6C + 42He→ 14

7N + AZX

we know 6 + 2 = 7 + Z and 12 + 4 = 14 + A so we must have AZX ⇒ 2

1H

Q-value Contents

The Q-value for a reaction is the amount of energy released in thatreaction. It is the energy equivalent of δm - the difference between thetotal rest mass of the initial particles and the total rest mass of the finalparticles.

δm = (total mass at start)− (total mass at end)

= (mX1 + mX2 + . . .)− (mX ′1

+ mX ′2

+ . . .)

Q = δm c2

Using atomic masses expressed in u, then the Q-value in MeV will be:

Q = [(MX1 +MX2 + . . .)− (MY1 +MY2 + . . .)] 931.5

Q-value & Binding Energy Contents

This formula is very similar to that for binding energy, seen earlier. Infact binding energy can be thought of as a special case of the Q-value.

The Binding Energy of the nuclide AZX is equal to the Q-value of the

(hypothetical!) reaction forming the atom:

11H + 1

1H + . . .︸ ︷︷ ︸Z protons

+Z electrons

+ 10n + 1

0n + . . .︸ ︷︷ ︸A−Z neutrons

→ AZX︸︷︷︸

atom

Exothermic & Endothermic Reactions Contents

There are two possibilities for the Q-value of a reaction:

I If Q > 0 then energy is released. The reaction is exothermic.The released energy will appear as kinetic energy of the products.

I If Q < 0 then energy must be added (in the form of kinetic energyof the initial particles) before the reaction can proceed. The reactionis endothermic.

These words are used in exactly the same way as in Chemistry for achemical reaction.

Example Contents

Fluorine-18 (189F) is a radionuclide often used in medical procedures.

It does not occur naturally, but can be produced using a cylotron tocollide energetic protons with water containing 18

8O (Oxygen-18).(a) What is the full equation for this reaction?(b) Calculate the Q-value for the reaction.

(a) Determining the full equation for the reaction requires a littlesimple arithmetic to balance the numbers of protons andneutrons before and after:

11H + 18

8O→ 189F + 1

0n

To highlight the details of how the reaction is achieved, this equa-tion can also be written in a slightly different form:

188O︸︷︷︸

target

( 11H︸︷︷︸

beam

, 10n) 18

9F︸︷︷︸product

Contents

(b) The Q-value calculation is straightforward:

Q =[∑

initialmasses (in u)−

∑final

masses (in u)]× 931.5

= [(MH +MO)− (MF + mn)]× 931.5= (1.007825032 + 17.999161001− 18.000937956

− 1.00866492)× 931.5 = −2.44 MeV

What is the minimum kinetic energy of protons required forthis reaction to proceed?

A negative Q-value tells us that the reaction can only proceed if weadd energy. This can be achieved in the form of kinetic energy ofthe proton. How much is required? We need the total final energyto be positive, so any amount greater than 2.44 MeV will achievethis. The minimum energy required is of course 2.44 MeV.

Not Just Nuclei in Nuclear Reactions Contents

A nuclear reaction can include other particles in addition to nuclei.

For example, the capture of neutrons by Cadmium nuclei is important forthe operation of most nuclear reactors. This reaction will involve theproduction of a gamma ray (i.e. a high energy photon). One suchreaction is:

11148Cd + 1

0n→ 11248Cd + γ

This can be thought of as a two-stage process. The neutron is absorbed,resulting in a nucleus in an excited state, indicated by a *, which thendrops to its ground state by emission of a photon:

11148Cd + 1

0n→ 11248Cd∗ → 112

48Cd + γ

Remember, like an atom, a nucleus is a quantum system with quantizedenergy levels.

Example Contents

Find the Q-value for the reaction: 11148Cd

(10n, γ

) 11248Cd

This is the reaction we have just discussed, written to emphasizethat the moving neutron collides with the stationary Cadmium nu-cleus.

11148Cd + 1

0n→ 11248Cd + γ

The presence of non-nuclei makes no difference to the definitionor calculation of the Q-value. We need to remember that the restmass of the photon is zero:

Q = [(MCd-111 + mn)− (MCd-112 + mγ)]× 931.5= (110.9041781 + 1.00866492− 111.9027578− 0)× 931.5= 9.39 MeV

Most of this energy is carried by the photon.

Summary Contents

Nuclides are identified using the standard notation:

X X = chemical symbolA

A = atomic mass number (no. of nucleons)Z

Z= atomic number (no. of protons)

The binding energy of a nuclide is the difference in rest massenergy of the nucleus and its constituent nucleons:

BE =((mass of nucleons)− (mass of nucleus)

)c2

Usually, non-SI units are more convenient:

BE (in MeV) =(

(mass of nucleons)− (mass of nucleus)︸ ︷︷ ︸in u

)×931.5

Contents

The Q-value for a nuclear reaction is the energy released in thereaction, and is calculated as the difference between the total restmass energy of the initial reactants and the total rest mass energyof the final products:

Q =(∑

initialmass−

∑final

mass)

c2

Again, non-SI units are usually used:

Q (in MeV) =(∑

initialmass−

∑final

mass)

︸ ︷︷ ︸in u

×931.5

If Q < 0, the reaction is endothermic, and energy must be added(as kinetic energy of the initial reactants), in order for the reactionto proceed.