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23.1Mass defect and binding energy
Nuclear notation
We now move deep into the atom in order to describe the structure of its nucleus. Atomic nuclei are made up of smaller particles, called protons and neutrons. The word nucleon is used to denote a proton or a neutron.
The number of protons in a nucleus is denoted by Z and is called the proton number.
The total number of nucleons (protons + neutrons) is called the nucleon number and is denoted by A.
The number of neutrons in a nucleus is denoted by N and N = A â Z.
The electric charge of the nucleus is Ze, where e = +1.6 Ã 10â19 C is the charge of a proton. We use the proton and nucleon numbers to denote a nucleus in the following way: the symbol AZX stands for the nucleus of element X, whose proton number is Z and nucleon number is A. For example:
H11	
hydrogen nucleus with 1 proton and no neutrons
He24	
helium nucleus with 2 protons and 2 neutrons
Ca2040	
calcium nucleus with 20 protons and 20 neutrons
UÂ Â 92238	
uranium nucleus with 92 protons and 146 neutrons
A nucleus with a specific number of protons and neutrons is also called a nuclide.
CHECK YOURSELF 1
How many neutrons are there in the nucleus Ne1021?
We can apply this notation to the nucleons themselves. For example, the proton (symbol p) can be written as p11 and the neutron (symbol n) as n01. If we notice that the proton number is not only the number of protons in the nucleus but also its electric charge in units of e, then we can extend this notation to electrons as well. The charge of the electron in units of e is â1 and so we represent the electron by eâ1Â Â Â 0. The nucleon number of the electron is zero since the nucleon number is defined as the total number of neutrons plus protons.
The photon (the particle of light) can also be represented in this way: the photon has the Greek letter gamma (Î³) as its symbol. Since it has zero electric charge and is neither a proton nor a neutron, it is represented by Î³00.
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The neutrino (we will learn more about this later) is neutral and is represented by Î½00.
For every particle in nature, there exists an anti-particle: this is a particle with the same mass as the particle but all other properties, including electric charge, are opposite. Some neutral particles are identical with their antiparticle (the photon for example). but others are not (the neutron for example). Anti-neutrons are distinguished from neutrons by other properties that we will not discuss here. Ordinary matter is made of protons, neutrons and electrons. Table 23.1 gives a summary of these particles and their symbols.
Particle
Symbol
proton
anti-proton
p11
PÂ¯Â¯Â¯â1Â Â Â 1
neutron
anti-neutron
n01
nÂ¯01
electron
positron
eâ1Â Â Â 0
eÂ¯+1Â Â Â 0
photon
Î³00 (or just Î³)
neutrino
anti-neutrino
Î½00 (or just Î½)
Î½Â¯00 (or just Î½Â¯)
alpha particle
He24 or Î±24
Table 23.1: Some particles and their symbols
Fundamental forces
There are four fundamental interactions or forces in nature. (Knowledge of the weak nuclear force and the electroweak force is not required in this course.)
These are:
1the electromagnetic interaction: it acts on any particle that has electric charge. The force is given by Coulombâs law. It has infinite range.
2the weak nuclear interaction: it acts on protons, neutrons, electrons and neutrinos in order to bring about beta decay. It has very short range (10â18 m).
3the strong nuclear interaction: it acts on protons and neutrons to keep them bound to each other inside nuclei. It has short range (10â15 m).
4the gravitational interaction: it acts on any particle with mass. The small mass of atomic particles makes this force irrelevant for atomic and nuclear physics. It has infinite range.
(It is known that the electromagnetic interaction and the weak interaction are two sides of one forceâthe electroweak interaction.)
CHECK YOURSELF 2
Nuclei contain protons which repel each other electrically. Why then donât nuclei break apart?
Isotopes
Nuclei of the same element have the same number of protons but may have different number of neutrons; these are called isotopes of each other.
KEY POINT
Isotopes have the same proton number Z but different neutron number N and nucleon number A.
For example, H11, H12 and H13 are three isotopes of hydrogen, and UÂ Â 92235, UÂ Â 92236 and UÂ Â 92238 are just three (of many) isotopes of uranium. Since isotopes have the same number of protons, their atoms have the same number of electrons as well. This means that isotopes have identical chemical properties but different physical properties. The existence of isotopes is evidence for the existence of neutrons inside atomic nuclei.
CHECK YOURSELF 3
How many electrons are there in each of the neutral atoms of Ti2248 and Ti2250?
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The atomic mass unit
In atomic and nuclear physics, it is convenient to use a smaller unit of mass than the kilogram.
KEY POINT
The atomic mass unit is defined to be 112 of the mass of an atom of carbons-12, CÂ Â 612.
The symbol for the atomic mass unit is u. In terms of the kg, 1 u = 1.6605402 Ã 10â27 kg.
As we will see in the next worked example, this means that the mass of a proton and the mass of a neutron are both approximately equal to 1 u. In turn this means that a nucleus with nucleon number A has a mass that is approximately A u.
WORKED EXAMPLE 23.1
Determine to six decimal places, in units of u, the masses of the proton, neutron and electron. Use mp = 1.6726231 Ã 10â27 kg, mn = 1.6749286 Ã 10â27 kg, and me = 9.1093897 Ã 10â31 kg.
Answer
Using the relationship between u and kg, we find:
mp=1.6726231Ã10â271.6605402Ã10â27=1.007276Â u
Similarly:
mn=1.008665Â u
me=0.0005486Â u
(This shows that, approximately, mp â mn â 1 u).
The mass defect and binding energy
Protons and neutrons are very tightly bound to each other in a nucleus through the action of the strong nuclear force. To separate them, energy must be supplied to the nucleus. Conversely, energy is released when a nucleus is assembled from its constituent nucleons.
In Einsteinâs theory of relativity, a mass m is equivalent to an energy E according to the equation:
E=mc2
where c is the speed of light.
It is very convenient to find out how much energy corresponds to a mass of 1 u. Then, given a mass in u, we can easily find the energy that mass corresponds to. The energy corresponding to 1 u is:
1Â uÃc2=1.6605402Ã10â27Ã(2.9979Ã108)2Â J
=1.4923946316Ã10â10Â J
Changing this to eV, using 1 eV = 1.602177 Ã 10â19 J, gives an energy equivalent to a mass of 1 u of:
1.4923946316Ã10â10Â J1.602177Ã10â19JeVâ1=931.5Ã106eV=931.5Â MeV
So:
1Â uÃc2=931.5Â MeV or 1Â u=931.5Â MeVcâ2
Let us find the energy equivalent of the mass of a proton, a neutron and an electron whose masses in u are mp = 1.007276 u, mn = 1.008665 u and me = 0.0005486 u.
For the proton, the energy is mpc2: 1.007276 u Ã c2 = 1.007276 Ã 931.5 MeV câ2 Ã c2 = 938.3 MeV
KEY POINT
We multiply the mass in u by 931.5 to get the energy in MeV.
For the neutron, the energy equivalent is 1.008665 Ã 931.5 MeV câ2 Ã c2 = 939.6 MeV.
And for the electron: 0.0005486 Ã 931.5 = 0.511 MeV.
Notice that the mass-energy equivalence allows us to express mass in a new unit: for the proton for example we know that mpc2 = 938 MeV. We can then say that mp = 938 MeV câ2. Similarly, for the neutron, mn = 940 MeV câ2. The energy equivalent to the mass of a nucleus of helium-4 is 3727 MeV and so the mass of the nucleus may also be expressed as 3727 MeV câ2. To convert a mass in MeV câ2 to u we divide by 931.5. So if m = 1800 MeV câ2, the mass in u is 1800931.5â1.93Â u.
Since we must supply energy to a nucleus to separate the nucleons and since energy is equivalent to mass, it follows that the total mass of the separated nucleons is greater than the mass of the nucleus.
Take the nucleus of helium, He24, as an example. The mass of a neutral atom of helium is 4.0026 u. This includes the mass of two electrons. So the nuclear mass is:
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MnucleusÂ =(4.0026â2Ã0.0005486)u=4.0015Â u
The helium nucleus is made up of two protons and two neutrons. Adding their masses we find:
2mp+2mn=2Ã1.007276+2Ã1.008665=4.031882Â u.
This is larger than the mass of the nucleus by 0.030382 u. This leads to the concept of mass defect.
KEY POINT
The mass defect Âµ is the total mass of the nucleons minus the mass of the nucleus.
This can be written as:
Î¼=Zmp+NmnâMnucleusÂ 
where N = A â Z is the number of neutrons in the nucleus. Because Mnucleus = Matom â Zme where me is the mass of an electron, we can rewrite the mass defect as
Î¼=Zmp+Nmnâ(MatomÂ âZme)=Z(mp+me)+NmnâMatomÂ .
But mp + me = 1.007825 u is the mass of the hydrogen atom, MH, and so a more useful formula is
Î¼=ZMH+NmnâMatom
(This is more useful because tables usually give atomic masses rather than nuclear masses.)
WORKED EXAMPLE 23.2
Find the mass defect of the nucleus of gold, AuÂ Â 79197, whose nuclear mass is 196.923229 u and that of iron, Fe2654, whose atomic mass is 53.939611 u.
Answer
For gold, we have been given the nuclear mass directly; the nucleus has 79 protons and 118 neutrons, so:
Î¼=(79Ã1.007276Â u)+(118Ã1.008665Â u)â196.923229Â u=1.674Â u.
For iron we use the second formula in terms of atomic mass to get
Î¼=(26Ã1.007825Â u)+(28Ã1.008665Â u)â53.939611Â u=0.5065Â u.
The energy equivalent to the mass defect is called binding energy.
KEY POINT
The binding energy, Eb, of a nucleus is the minimum energy required to completely separate the nucleons of that nucleus: Eb = Âµc2.
The energy required to remove one nucleon from the nucleus is roughly the binding energy divided by the total number of nucleons. Thus, the binding energy per nucleon is a measure of how tightly bound the nucleus isâthe higher the binding energy per nucleon, the more tightly bound the nucleus.
The binding energy of gold-197 is 1.674 Ã 931.5 = 1559 MeV and so the binding energy per nucleon is b=1559197=7.914Â MeV. The binding energy for iron-54 is 0.5065 Ã 931.5 = 471.8 MeV with a binding energy per nucleon of b=471.854=8.737Â MeV.
WORKED EXAMPLE 23.3
Determine the binding energy per nucleon of the nucleus of carbon-12.
Answer
The atomic mass is 12.000000 u, and we will use Âµ = ZMH + Nmn â Matom.
The nucleus has 6 protons and 6 neutrons, so the mass defect is:
Î¼=(6Ã1.007825Â u)+(6Ã1.008665Â u)â12.000000Â u=0.098940Â u
Hence the binding energy is:
0.098940Ã931.5Â MeV=92.163Â MeV.
The binding energy per nucleon is then:
b=92.16312=7.68Â MeV.
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The binding energy curve
Figure 23.1 shows the variation with nucleon number A of the binding energy (Eb) per nucleon, b=EbA.
The main features of the graph are:
âThe binding energy per nucleon for hydrogen, H11, is zero because there is only one particle in the nucleus and there is nothing to separate.
âThe curve rises sharply for low values of A.
âThe curve has a maximum for A = 62 corresponding to nickel, which makes this nucleus particularly stable.
âThere are peaks at the position of the nuclei He24, CÂ Â 612 and OÂ Â 816, which makes these nuclei unusually stable compared to their immediate neighbours.
âThe curve drops gently from the peak at A = 62 and onwards.
âMost nuclei have a binding energy per nucleon between 7 and 9 MeV.

Figure 23.1: The binding energy per nucleon is almost constant for nuclei with A > 20.

Energy released in nuclear reactions
The reactions RaÂ Â 88226âRnÂ Â 86222+Î±24, NÂ Â 714+Î±24âOÂ Â 817+p11 and H12+Li36âHe24+He24 are all examples of nuclear reactions in which a nucleus may split into other nuclei or nuclei collide and form new nuclei and nucleons. Notice that the total proton number on the left of the reaction is equal to the total proton number on the right. This is a consequence of electric charge conservation. The nucleon numbers on the left and right also match. This is consequence of nucleon number conservation (formally known as baryon number conservation).
To decide whether energy is released in any nuclear reaction, we have to calculate the mass difference Îm:
Îm = total mass of reactants â total mass of products
If Îm is positive, then energy will be released and the decay will occur. If Îm is negative, the reactants will not react and the reaction can take place only if energy is supplied.
We can see how this works by looking at the reaction:
RaÂ Â 88226âRnÂ Â 86222+He24
(This is a type of reaction called alpha decay, which we will meet later on.)
The masses involved are:
Ra-226: 226.0254 u
Rn-222: 222.0176 u
He-4: 4.0026 u
These are atomic masses; we should be using nuclear masses. To get the nuclear mass from the atomic mass we must subtract the mass of the electrons from the atomic mass. We need to subtract 88 electron masses from the left of the reaction and 86 + 2 = 88 from the right, i.e. the same number. Since we will calculate the mass difference, using atomic or nuclear masses will give the same answer. When the number of electron masses that must be subtracted from each side is not the same, we need to use nuclear masses.
So the mass difference is:
Îm=226.0254Â uâ(222.0176+4.0026)Â u
Îm=226.0254Â uâ226.0202Â u
Îm=0.0052Â u
This is positive, so energy will be released. The quantity of energy Q released is the energy equivalent of the mass difference and is given by:
Q=0.0052Ã931.5Â MeV=4.84Â MeV
This energy is released in the form of kinetic energy, which is shared by the helium and radon nuclei. Helium, being much lighter than radon, has greater speed and greater kinetic energy as the next worked example shows.
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WORKED EXAMPLE 23.4
Calculate the ratio of the kinetic energies of the helium nucleus to that of the radon nucleus in the decay of radium (RaÂ Â 88226) to radon (RnÂ Â 86222). Assume that the radium nucleus decays at rest. Determine how much energy the helium nucleus carries.
Answer
The radium nucleus is at rest, so the initial momentum is zero. By conservation of momentum, the momenta of the products are opposite in direction and equal in magnitude. Thus pHe = pRn.
Kinetic energy is related to momentum by EK=p22m, so:
EHeERn=(p2He2MHe)(p2rn2Mrn)=MRnMHeâ2224â56
This means that the energy carried by helium is 5657Ã4.84â4.76Â MeV; in other words, practically all the energy released.
EXAM TIP
Momentum conservation applies to nuclear physics as well!
Consider the reaction in which an alpha particle collides with a nucleus of nitrogen producing a nucleus of oxygen and a proton:
NÂ Â 714+Î±24âOÂ Â 817+p11
Nuclear masses: mN = 13.9992 u; mÎ± = 4.0015 u; mO = 16.9947 u; mp = 1.0073 u
This is a famous reaction called the transmutation of nitrogen; the nucleus produced (oxygen) is heavier than the original nucleus (nitrogen). It was studied by Rutherford in 1909.
In this reaction the mass difference is negative:
Îm=18.0007â18.0020=â0.0013Â u
This reaction will only take place if the alpha particle has enough kinetic energy to make up for the deficit. The minimum kinetic energy needed is 0.0013 Ã 931.5 = 1.21 MeV. (But this is not quite correct. With this kinetic energy the total kinetic energy of the oxygen and the proton would be zero, and hence the total momentum of the products would also be zero. But this would violate conservation of momentum since the incoming alpha particle has kinetic energy and so must have non-zero momentum. A more careful calculation shows that the minimum kinetic energy required is about 1.53 MeV.)
THEORY OF KNOWLEDGE
The belief in unification
In the early nineteenth century there were three known forces: gravitational, electric and magnetic. Through the work of James Clerk Maxwell (1831â1879) physicists realised that electric and magnetic forces were two sides of the same force, the electromagnetic force. Thus began the notion (for some a belief, for others a prejudice) that all interactions, as more were being discovered, were part of the same âunifiedâ force. In the twentieth century two new forces were discovered: the weak nuclear force and the strong nuclear force. In the late 1960s the electromagnetic and the weak nuclear force were unified in the standard model of particles. All efforts to unify this electroweak force with the strong nuclear force in a grand unified force have failed. All attempts to unify any of these forces with gravity have also failed. Yet the dream of unification remains.
Links
Everything we did related to energy in nuclear reactions is based on the equivalence of mass and energy discovered by Einstein in his theory of relativity. A theory that rested on the postulate that the speed of light is a universal constant independent of the speed of the source of light has led to radical changes in Newtonian mechanics including the famous E = mc2 equation that we have used heavily in the preceding chapters.
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TEST YOUR UNDERSTANDING
1State the electric charge of the nucleus He23.
2State the number of protons and the number of neutrons in the nucleus PbÂ Â 82209.
3What is the energy equivalent
ain J to a mass of 65 kg and
bin MeV to a mass of 65 u?
4aState what is meant by the term isotope.
bState three ways in which the nuclei of the isotopes OÂ Â 816 and OÂ Â 818 differ from each other.
5What is the existence of isotopes evidence for?
6Why are the chemical properties of isotopes of the same element identical?
7Explain why a nucleus stays bound despite the fact that the protons in it repel each other.
8State the name of the dominant force between two protons separated by a distance of:
a1.0Ã10â15Â m
b1.0Ã10â10Â m.
9Why do we ignore the gravitational force in nuclear interactions?
10Find the mass defect and binding energy of the nucleus Li37. The nuclear mass is 7.0144 u.
11Find the mass defect and binding energy of the nucleus NÂ Â 712. The atomic mass is 12.0186 u.
12The binding energy per nucleon of the nucleus Fe2656 is 8.7903 MeV. What is the binding energy of this nucleus?
13The binding energy per nucleon of Zr4091 is 8.6933 MeV. What is the mass defect of this nuclide in MeV câ2 and in u?
14The binding energy of the nucleus He23 is 7.7180 MeV. What is the nuclear mass of He23 in MeV câ2 and in u?
15The binding energy per nucleon of XeÂ Â 54131 is 8.4237 MeV. What is the nuclear mass of XeÂ Â 54131 in MeV câ2 and in u?
16Find the binding energy and binding energy per nucleon of the nucleus Ni2862. The atomic mass is 61.9283 u.
17The isotopes CÂ Â 612 and CÂ Â 613 are both stable. Which nucleus is more tightly bound? The atomic mass of carbon-13 is 13.0033 u.
18Identify particle X by finding the missing numbers:
aAl1327+nâNa1124+Xâ¡â¡
bH11+Li37âBe47+Xâ¡â¡
cNa1122+nâNe1022+Xâ¡â¡
19How much energy is required to remove one proton from the nucleus of UÂ Â 92238? A rough answer to this question is obtained by giving the binding energy per nucleon (7.57 MeV). A better answer is obtained when we write a reaction that removes a proton from the nucleus. In this case UÂ Â 92238âp11+PaÂ Â 91237. Calculate the energy required for this reaction to take place, known as the proton separation energy. Compare the two energy values. (The nuclear mass of uranium is 238.0003 u; that of protactinium is 237.0012 u.)
20The reaction CfÂ Â 98254âPdÂ Â 46118+TeÂ Â 52132+4n01 is an example of spontaneous fission. What is the energy released? (Atomic masses: californium = 254.0873 u, palladium = 117.9190 u, tellurium = 131.9086 u.)
21What is the energy released in the reaction n01+NÂ Â 714âCÂ Â 614+p11? (Nuclear masses: nitrogen = 13.9992 u, carbon = 14.0000 u.)
22Calculate the energy released in the reaction H12+Li36âHe24+He24. (Atomic masses: deuterium = 2.0141 u, lithium = 6.0151 u, helium = 4.0026 u.)
