Conservation of Mass-Energy – Mass-Energy Equivalence

One of the striking results of Einstein’s theory of relativity is that mass and energy are equivalent and convertible one into the other (mass-energy). Equivalence of the mass and energy is described by Einstein’s famous formula E = mc2. In words, energy equals mass multiplied by the speed of light squared.

At the beginning of the 20th century, the notion of mass underwent a radical revision. Mass lost its absoluteness. One of the striking results of Einstein’s theory of relativity is that mass and energy are equivalent and convertible one into the other. Equivalence of the mass and energy is described by Einstein’s famous formula E = mc2. In words, energy equals mass multiplied by the speed of light squared. Because the speed of light is a very large number, the formula implies that any small amount of matter contains a very large amount of energy. The mass of an object was seen to be equivalent to energy, to be interconvertible with energy, and to increase significantly at exceedingly high speeds near that of light. The total energy of an object was understood to comprise its rest mass as well as its increase of mass caused by increase in kinetic energy.

In special theory of relativity certain types of matter may be created or destroyed, but in all of these processes, the mass and energy associated with such matter remains unchanged in quantity. It was found the rest mass an atomic nucleus is measurably smaller than the sum of the rest masses of its constituent protons, neutrons and electrons. Mass was no longer considered unchangeable in the closed system. The difference is a measure of the nuclear binding energy which holds the nucleus together. According to the Einstein relationship (E = mc2) this binding energy is proportional to this mass difference and it is known as the mass defect.

During the nuclear splitting or nuclear fusion, some of the mass of the nucleus gets converted into huge amounts of energy and thus this mass is removed from the total mass of the original particles, and the mass is missing in the resulting nucleus. The nuclear binding energies are enormous, they are of the order of a million times greater than the electron binding energies of atoms.

Generally, in both chemical and nuclear reactions, some conversion between rest mass and energy occurs, so that the products generally have smaller or greater mass than the reactants. Therefore the new conservation principle is the conservation of mass-energy.

Example: Mass-energy defect of a 63Cu

Calculate the mass defect of a 63Cu nucleus if the actual mass of 63Cu in its nuclear ground state is 62.91367 u.

63Cu nucleus has 29 protons and also has (63 – 29) 34 neutrons.

The mass of a proton is 1.00728 u and a neutron is 1.00867 u.

The combined mass is: 29 protons x (1.00728 u/proton) + 34 neutrons x (1.00867 u/neutron) = 63.50590 u

The mass defect is Δm = 63.50590 u – 62.91367 u =  0.59223 u

Convert the mass defect into energy (nuclear binding energy).

(0.59223 u/nucleus) x (1.6606 x 10-27 kg/u) = 9.8346 x 10-28 kg/nucleus

ΔE = Δmc2

ΔE = (9.8346 x 10-28 kg/nucleus) x (2.9979 x 108 m/s)2 = 8.8387 x 10-11 J/nucleus

The energy calculated in the previous example is the nuclear binding energy.  However, the nuclear binding energy may be expressed as kJ/mol (for better understanding).

Calculate the nuclear binding energy of 1 mole of 63Cu:

(8.8387 x 10-11 J/nucleus) x (1 kJ/1000 J) x (6.022 x 1023 nuclei/mol) = 5.3227 x 1010 kJ/mol of nuclei.

One mole of 63Cu (~63 grams) is bound by the nuclear binding energy (5.3227 x 1010 kJ/mol) which is equivalent to:

• 14.8 million kilowatt-hours (≈ 15 GW·h)
• 336,100 US gallons of automotive gasoline

Example: Mass defect of the reactor core

Calculate the mass defect of the 3000MWth reactor core after one year of operation.

It is known the average recoverable energy per fission is about 200 MeV, being the total energy minus the energy of the energy of antineutrinos that are radiated away.

The reaction rate per entire 3000MWth reactor core is about  9.33×1019 fissions / second.

The overall energy release in the units of joules is:

200×106 (eV) x 1.602×10-19 (J/eV) x 9.33×1019 (s-1) x 31.5×106 (seconds in year) = 9.4×1016 J/year

The mass defect is calculated as:

Δm = ΔE/c2

Δm = 9.4×1016 / (2.9979 x 108)2 = 1.046 kg

That means in a typical 3000MWth reactor core about 1 kilogram of matter is converted into pure energy.

Note that, a typical annual uranium load for a 3000MWth reactor core is about 20 tonnes of enriched uranium (i.e. about 22.7 tonnes of UO2). Entire reactor core may contain about 80 tonnes of enriched uranium.

Mass defect directly from E=mc2

The mass defect can be calculated directly from the Einstein relationship (E = mc2) as:

Δm = ΔE/c2

Δm = 3000×106 (W = J/s) x 31.5×106 (seconds in year) / (2.9979 x 108)= 1.051 kg

References:
Nuclear and Reactor Physics:
1. J. R. Lamarsh, Introduction to Nuclear Reactor Theory, 2nd ed., Addison-Wesley, Reading, MA (1983).
2. J. R. Lamarsh, A. J. Baratta, Introduction to Nuclear Engineering, 3d ed., Prentice-Hall, 2001, ISBN: 0-201-82498-1.
3. W. M. Stacey, Nuclear Reactor Physics, John Wiley & Sons, 2001, ISBN: 0- 471-39127-1.
4. Glasstone, Sesonske. Nuclear Reactor Engineering: Reactor Systems Engineering, Springer; 4th edition, 1994, ISBN: 978-0412985317
5. W.S.C. Williams. Nuclear and Particle Physics. Clarendon Press; 1 edition, 1991, ISBN: 978-0198520467
6. Kenneth S. Krane. Introductory Nuclear Physics, 3rd Edition, Wiley, 1987, ISBN: 978-0471805533
7. G.R.Keepin. Physics of Nuclear Kinetics. Addison-Wesley Pub. Co; 1st edition, 1965
8. Robert Reed Burn, Introduction to Nuclear Reactor Operation, 1988.
9. U.S. Department of Energy, Nuclear Physics and Reactor Theory. DOE Fundamentals Handbook, Volume 1 and 2. January 1993.