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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 a proton or a neutron 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 number of protons in a nucleus.

The total number of nucleons (protons + neutrons) is called the nucleon number the total number of protons and neutrons in a nucleus and is denoted by A.

The number of neutrons in a nucleus is denoted by N and N = A − Z.

The electric charge of the nucleus is Ze, where e = +1.6 × 10−19 C is the charge of a proton. We use the proton and nucleon numbers to denote a nucleus in the following way: the symbol Z A X stands for the nucleus of element X, whose proton number is Z and nucleon number is A. For example:

H 1 1 | hydrogen nucleus with 1 proton and no neutrons
He 2 4 | helium nucleus with 2 protons and 2 neutrons
Ca 20 40 | calcium nucleus with 20 protons and 20 neutrons
U 92 238 | uranium nucleus with 92 protons and 146 neutrons

A nucleus with a specific number of protons and neutrons is also called a nuclide a nucleus with a specific atomic and nucleon number.

Check yourself 1

How many neutrons are there in the nucleus Ne 10 21?

We can apply this notation to the nucleons themselves. For example, the proton (symbol p) can be written as p 1 1 and the neutron (symbol n) as n 0 1. 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 −1 and so we represent the electron by e−1 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 (γ) as its symbol. Since it has zero electric charge and is neither a proton nor a neutron, it is represented by γ 0 0.

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The neutrino (we will learn more about this later) is neutral and is represented by ν 0 0.

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 | p 1 1 P¯−1 1
neutron anti-neutron | n 0 1 n¯0 1
electron positron | e−1 0 e¯+1 0
photon | γ 0 0 (or just γ)
neutrino anti-neutrino | ν 0 0 (or just ν) ν¯0 0 (or just ν¯)
alpha particle | He 2 4 or α 2 4

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 force given by Coulomb’s law; acts on any particle that has electric charge; has infinite range: it acts on any particle that has electric charge. The force is given by Coulomb’s law. It has infinite range.

2the weak nuclear interaction force that acts on protons, neutrons, electrons and neutrinos in order to bring about beta decay; has very short range (10−18 m): it acts on protons, neutrons, electrons and neutrinos in order to bring about beta decay. It has very short range (10−18 m).

3the strong nuclear interaction force that acts on protons and neutrons to keep them bound to each other inside nuclei; has short range (10−15 m): it acts on protons and neutrons to keep them bound to each other inside nuclei. It has short range (10−15 m).

4the gravitational interaction force that acts on any particle with mass; has infinite range: 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 force—the electroweak interaction represents the unification of the electromagnetic and weak nuclear interactions.)

Check yourself 2

Nuclei contain protons which repel each other electrically. Why then don’t 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 nuclei of the same element containing different numbers of neutrons of each other.

KEY POINT

Isotopes have the same proton number Z but different neutron number N and nucleon number A.

For example, H 1 1, H 1 2 and H 1 3 are three isotopes of hydrogen, and U 92 235, U 92 236 and U 92 238 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 Ti 22 48 and Ti 22 50?

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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 1 12 of the mass of the neutral atom of carbon-12. is defined to be 1 12 of the mass of an atom of carbons-12, C 6 12.

The symbol for the atomic mass unit is u. In terms of the kg, 1 u = 1.6605402 × 10−27 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 m p = 1.6726231 × 10−27 kg, m n = 1.6749286 × 10−27 kg, and m e = 9.1093897 × 10−31 kg.

Answer

Using the relationship between u and kg, we find:

m p=1.6726231×10−27 1.6605402×10−27=1.007276 u

Similarly:

m n=1.008665 u

m e=0.0005486 u

(This shows that, approximately, m p ≈ m n ≈ 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 Einstein’s theory of relativity, a mass m is equivalent to an energy E according to the equation:

E=m c 2

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:

1 u×c 2=1.6605402×10−27×(2.9979×10 8)2 J

=1.4923946316×10−10 J

Changing this to eV, using 1 eV = 1.602177 × 10−19 J, gives an energy equivalent to a mass of 1 u of:

1.4923946316×10−10 J 1.602177×10−19 JeV−1=931.5×10 6 eV=931.5 MeV

So:

1 u×c 2=931.5 MeV or 1 u=931.5 MeV c−2

Let us find the energy equivalent of the mass of a proton, a neutron and an electron whose masses in u are m p = 1.007276 u, m n = 1.008665 u and m e = 0.0005486 u.

For the proton, the energy is m p c 2: 1.007276 u × c 2 = 1.007276 × 931.5 MeV c−2 × 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 × 931.5 MeV c−2 × c2 = 939.6 MeV.

And for the electron: 0.0005486 × 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 m p c 2 = 938 MeV. We can then say that m p = 938 MeV c−2. Similarly, for the neutron, m n = 940 MeV c−2. 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 c−2. To convert a mass in MeV c−2 to u we divide by 931.5. So if m = 1800 MeV c−2, the mass in u is 1800 931.5≈1.93 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, He 2 4, 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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M nucleus =(4.0026−2×0.0005486)u=4.0015 u

The helium nucleus is made up of two protons and two neutrons. Adding their masses we find:

2 m p+2 m n=2×1.007276+2×1.008665=4.031882 u.

This is larger than the mass of the nucleus by 0.030382 u. This leads to the concept of mass defect the difference between the mass of the nucleons and the mass of a nucleus.

KEY POINT

The mass defect µ is the total mass of the nucleons minus the mass of the nucleus.

This can be written as:

μ=Z m p+N m n−M nucleus

where N = A − Z is the number of neutrons in the nucleus. Because M nucleus = M atom − Z m e where m e is the mass of an electron, we can rewrite the mass defect as

μ=Z m p+N m n−(M atom −Z m e)=Z(m p+m e)+N m n−M atom .

But m p + m e = 1.007825 u is the mass of the hydrogen atom, M H, and so a more useful formula is

μ=Z M H+N m n−M atom

(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, Au 79 197, whose nuclear mass is 196.923229 u and that of iron, Fe 26 54, 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:

μ=(79×1.007276 u)+(118×1.008665 u)−196.923229 u=1.674 u.

For iron we use the second formula in terms of atomic mass to get

μ=(26×1.007825 u)+(28×1.008665 u)−53.939611 u=0.5065 u.

The energy equivalent to the mass defect is called binding energy the minimum energy required to separate the nucleons of a nucleus; E b = µc 2.

KEY POINT

The binding energy, E b, of a nucleus is the minimum energy required to completely separate the nucleons of that nucleus: E b = µc 2.

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 is—the higher the binding energy per nucleon, the more tightly bound the nucleus.

The binding energy of gold-197 is 1.674 × 931.5 = 1559 MeV and so the binding energy per nucleon is b=1559 197=7.914 MeV. The binding energy for iron-54 is 0.5065 × 931.5 = 471.8 MeV with a binding energy per nucleon of b=471.8 54=8.737 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 µ = ZM H + Nm n – M atom.

The nucleus has 6 protons and 6 neutrons, so the mass defect is:

μ=(6×1.007825 u)+(6×1.008665 u)−12.000000 u=0.098940 u

Hence the binding energy is:

0.098940×931.5 MeV=92.163 MeV.

The binding energy per nucleon is then:

b=92.163 12=7.68 MeV.

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The binding energy curve

Figure 23.1 shows the variation with nucleon number A of the binding energy (E b) per nucleon, b=E b A.

The main features of the graph are:

●The binding energy per nucleon for hydrogen, H 1 1, is zero because there is only one particle in the nucleus and there is nothing to separate.

●The curve rises sharply for low values of A.

●The curve has a maximum for A = 62 corresponding to nickel, which makes this nucleus particularly stable.

●There are peaks at the position of the nuclei He 2 4, C 6 12 and O 8 16, which makes these nuclei unusually stable compared to their immediate neighbours.

●The curve drops gently from the peak at A = 62 and onwards.

●Most 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 Ra 88 226→Rn 86 222+α 2 4, N 7 14+α 2 4→O 8 17+p 1 1 and H 1 2+Li 3 6→He 2 4+He 2 4 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 Δm:

Δm = total mass of reactants – total mass of products

If Δm is positive, then energy will be released and the decay will occur. If Δm 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:

Ra 88 226→Rn 86 222+He 2 4

(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:

Δ m=226.0254 u−(222.0176+4.0026) u

Δ m=226.0254 u−226.0202 u

Δ m=0.0052 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=0.0052×931.5 MeV=4.84 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 (Ra 88 226) to radon (Rn 86 222). 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 p He = p Rn.

Kinetic energy is related to momentum by E K=p 2 2 m, so:

E He E Rn=(p He 2 2 M He)(p rn 2 2 M rn)=M Rn M He≈222 4≈56

This means that the energy carried by helium is 56 57×4.84≈4.76 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:

N 7 14+α 2 4→O 8 17+p 1 1

Nuclear masses: m N = 13.9992 u; m α = 4.0015 u; m O = 16.9947 u; m p = 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:

Δ m=18.0007−18.0020=−0.0013 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 × 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 (1831–1879) 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 ‘unified’ 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 = mc 2 equation that we have used heavily in the preceding chapters.

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TEST YOUR UNDERSTANDING

1State the electric charge of the nucleus He 2 3.

2State the number of protons and the number of neutrons in the nucleus Pb 82 209.

3What is the energy equivalent

ain J to a mass of 65 kg and

bin MeV to a mass of 65 u?

4 aState what is meant by the term isotope.

bState three ways in which the nuclei of the isotopes O 8 16 and O 8 18 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:

a 1.0×10−15 m

b 1.0×10−10 m.

9Why do we ignore the gravitational force in nuclear interactions?

10Find the mass defect and binding energy of the nucleus Li 3 7. The nuclear mass is 7.0144 u.

11Find the mass defect and binding energy of the nucleus N 7 12. The atomic mass is 12.0186 u.

12The binding energy per nucleon of the nucleus Fe 26 56 is 8.7903 MeV. What is the binding energy of this nucleus?

13The binding energy per nucleon of Zr 40 91 is 8.6933 MeV. What is the mass defect of this nuclide in MeV c−2 and in u?

14The binding energy of the nucleus He 2 3 is 7.7180 MeV. What is the nuclear mass of He 2 3 in MeV c−2 and in u?

15The binding energy per nucleon of Xe 54 131 is 8.4237 MeV. What is the nuclear mass of Xe 54 131 in MeV c−2 and in u?

16Find the binding energy and binding energy per nucleon of the nucleus Ni 28 62. The atomic mass is 61.9283 u.

17The isotopes C 6 12 and C 6 13 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:

a Al 13 27+n→Na 11 24+X□□

b H 1 1+Li 3 7→Be 4 7+X□□

c Na 11 22+n→Ne 10 22+X□□

19How much energy is required to remove one proton from the nucleus of U 92 238? 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 U 92 238→p 1 1+Pa 91 237. 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 Cf 98 254→Pd 46 118+Te 52 132+4 n 0 1 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 n 0 1+N 7 14→C 6 14+p 1 1? (Nuclear masses: nitrogen = 13.9992 u, carbon = 14.0000 u.)

22Calculate the energy released in the reaction H 1 2+Li 3 6→He 2 4+He 2 4. (Atomic masses: deuterium = 2.0141 u, lithium = 6.0151 u, helium = 4.0026 u.)