Mass-Energy Relationships

Submitted by ChemPRIME Staff on Thu, 12/16/2010 - 16:02


In a nucleus the protons and the neutrons are held very tightly by forces whose nature is still imperfectly understood. When the nucleons are very close to each other, these forces are strong enough to counteract the coulombic repulsion of the protons, but they fall off very rapidly with distance and are essentially undetectable outside the nucleus. Because the energies involved in binding the nucleons together are very large, they give rise to an effect which makes it possible to measure them. According to Einstein’s special theory of relativity, when the energy of a body increases, so does its mass, and vice versa. If the change in energy is indicated by ΔE and the change in mass by Δm, these two quantities are related by the equation


ΔE = Δmc2      (1)


where c is the velocity of light (2.998 × 108 m s–1).

In ordinary chemical reactions this change in mass with energy is so small as to be undetectable, but in nuclear reactions we invariably find that products and reactants have different masses. As a simple example, let us take the dissociationThe breaking apart of one species into two or more smaller species; often applied to ions in a crystal lattice, which dissociate when the ionic solid dissolves in water. Dissociation refers to separation of particles that already exist; ionization refers to the formation of ions from neutral species, as in the ionization of a weak acid in aqueous solutoin. of a deuteron into a proton and a neutron:


{}_{\text{1}}^{\text{2}}\text{D }\to \text{ }{}_{\text{1}}^{\text{1}}p\text{ + }{}_{\text{0}}^{\text{1}}n


The molar massThe mass of a mole of substance; the same as molecular weight for molecular substances. of a deuteron is found experimentally to be 2.013 55 g mol–1 (see Table 1), but if we add the molar masses of a neutron and a proton,


TABLE 1 The Molar Masses of Some Selected Nuclei (Electrons Are Not Included in These Masses).

      Nucleus M/g mol–1       Nucleus M/g mol–1
      {}_{\text{0}}^{\text{1}}n (neutron)
1.008 67
      {}_{\text{26}}^{\text{56}}\text{Fe}
55.920 66
      {}_{\text{1}}^{\text{1}}p (proton)
1.007 28
      {}_{\text{27}}^{\text{59}}\text{Co}
58.918 37
      {}_{\text{1}}^{\text{2}}\text{D} (deuteron)      
2.013 55
      {}_{\text{36}}^{\text{84}}\text{Kr}
83.8917
      {}_{\text{1}}^{\text{3}}\text{T} (tritium)
3.015 50
      {}_{\text{50}}^{\text{120}}\text{Sn}
119.8747
      {}_{\text{2}}^{\text{4}}\text{He}
4.001 50
      {}_{\text{56}}^{\text{138}}\text{Ba}
137.8743
      {}_{\text{3}}^{\text{7}}\text{Li}
7.014 36
      {}_{\text{78}}^{\text{194}}\text{Pt}
193.9200
      {}_{\text{6}}^{\text{12}}\text{C}
11.996 71
      {}_{\text{83}}^{\text{209}}\text{Bi}
208.9348
      {}_{\text{8}}^{\text{15}}\text{O}
14.998 68
      {}_{\text{92}}^{\text{235}}\text{U}
234.9934
      {}_{\text{8}}^{\text{16}}\text{O}
15.990 52
      {}_{\text{94}}^{\text{239}}\text{Pu}
239.0006
      {}_{\text{8}}^{\text{17}}\text{O}
16.994 74



we obtain a somewhat higher value, namely, (1.007 28 + 1.008 67) g mol–1 = 2.015 95 g mol–1. The change in mass using the usual delta convention is thus (2.015 95 – 2.013 55) g mol–1 = 0.002 40 g mol–1. From Eq. (1) we then have


{\Delta\text{E}} = {\Delta\text{mc}}^2 = 0.002 40 \frac{\text{g}}{\text{mol}}\text{ }\times \text{ }\frac{\text{1 kg}}{\text{10}^{\text{3}}\text{ g}} \times (2.998 \times {\text{10}^{\text{8}}\frac{\text{m}}{\text{s}}})^2


                              = 2.16 × 1011 kg m2 s–2 mol–1 = 216 × 109 J mol–1 = 216 GJ mol–1


Since expansion workA mechanical process in which energy is transferred to or from an object, changing the state of motion of the object. or even electronic energies are negligible compared to this change in nuclear energy, we can equate the change in nuclear energy either to the change in internal energyA thermodynamic function corresponding to the energy of a system; represented by the symbol U or E. or the enthalpyA thermodynamic state function, symbol H, that equals internal energy plus pressure x volume; the change in enthalpy corresponds to the energy transferred as a result of a temperature difference (heat transfer) when a reaction occurs at constant pressure.; that is,


ΔE = ΔUm = ΔHm = 216 GJ mol–1


The energy needed to separate a nucleus into its constituent nucleons is called its binding energyThe energy released as particles are brought together to form an atomic nucleus; also called nuclear binding energy.. The binding energy of the 21D nucleus is thus 216 GJ mol–1. Notice how very much larger this is than the bond energyThe change in energy when a mole of chemical bonds of a particular type are broken; the molecules whose bonds are broken must be in the gas phase. Closely related to bond enthalpy. of an average moleculeA set of atoms joined by covalent bonds and having no net charge., which is about 200 or 300 kJ mol–1. Since a gigajoule is 1 million kJ, the energies involved in holding the nucleons together in a nucleus are something like a million times larger than those holding the atomsThe smallest particle of an element that can be involved in chemical combination with another element; an atom consists of protons and neutrons in a tiny, very dense nucleus, surrounded by electrons, which occupy most of its volume. together in a molecule.

Since the number of nucleons in a nucleus is quite variable, it is useful to calculate the average energy of each nucleon by dividing the total binding energy by the number of nucleons, A. This gives the binding energy per nucleon. In the case of the nucleus 5626Fe, for instance, we can easily find from Table 1 that Δm for the process


{}_{\text{26}}^{\text{56}}\text{Fe }\to \text{ 26}{}_{\text{1}}^{\text{1}}p\text{ + 30}{}_{\text{0}}^{\text{1}}n


has the value 0.528 72 g mol–1, giving a value for ΔHm, from Eq. (19.35)of 4750 GJ mol–1. Since A = 56 for this nucleus, the binding energy per nucleon has the value


\frac{\Delta H_{m}}{A} = \frac{\text{4750}}{\text{56}} GJ mol–1 = 848 GJ mol–1


The binding energy of a nucleus tells us not only how much energy must be expended in pulling the nuclei apart but also how much energy is released when the nucleus is formed from protons and neutrons. In the case of the 5626Fe, for instance, we have


\text{26}{}_{\text{1}}^{\text{1}}p\text{ + 30}{}_{\text{0}}^{\text{1}}n\text{ }\to \text{ }{}_{\text{26}}^{\text{56}}\text{Fe}      ΔHm = –4.75 × 103 GJ mol–1


that is, the energy of formation of 5626Fe is equal to the negative of the binding energy. In Fig. 1 the energy of formation on a per nucleon basis has been against the mass numberThe sum of the numbers of protons and neutrons in an atom; these two kinds of particles contain almost all of the mass of an atom. for the most stable isotopeOne of two or more samples of an element whose atoms differ in the number of neutrons found in the nucleus. of each elementA substance containing only one kind of atom and that therefore cannot be broken down into component substances by chemical means.. The zero energy axis in this plot corresponds to the energy of completely separated protons and neutrons, while the points on the graph correspond to the average energy of a nucleon in the nucleus in question. Obviously, the lower the energy, the more stable the nucleus.

As we can see from Fig. 1, the most stable nuclei are those of mass number close to 60, the nucleus with the lowest energy being the 5626Fe nucleus. As the mass number rises above 60, the nuclei become slightly higher in energy, i.e., less stable. Decreasing the mass number below 60 also brings us into a region of high-energy nuclei. With the exception of the 42He, nucleus, the nuclei of highest energy belong to the very lightest elements.

Figure 1 shows us that in principle there are two ways in which we can obtain energy from the nuclei of the elements. The first of these is by the splitting up or fissionA nuclear reaction in which large nuclei break apart to form smaller ones. of a very heavy nucleus into two lighter nuclei. In such a case each nucleon will move from a situation of higher to lower energy and energy will be released. Even more energy will be released by the fusionA nuclear reaction in which small nuclei are united to form larger ones. of two very light nuclei, each containing only a few nucleons, into a single heavier nucleus. Though fine in principle, neither of these methods of obtaining energy is easy to achieve in practice in a controlled way with due respect to the environment.


Figure 1 Energy of formation per nucleon (from protons and neutrons) as a function of mass number.