Nuclear Physics Tutorial 7 - Mass and Energy



The Atomic Mass Unit

Mass Defect

Binding Energy Per Nucleon

Radioactive Decay and Binding Energy



Fusion in Stars


The Atomic Mass Unit

Kilograms are useful for measuring large masses, but like many of the SI units, on the atomic scale they are far too big.  It’s a bit like a model maker using kilometres rather than millimetres.


The atomic mass unit is far more convenient to use with nuclear masses.  It uses carbon 12 as a reference and is defined as:


Exactly 1/12th the mass of a carbon 12 atom


The relative atomic mass of an atom is useful.  It is defined as:


___Mass of the atom       × 12

 Mass of carbon 12 atom


The relative atomic mass of atoms is usually very close to a whole number, which is consistent with the idea of a nucleus made up of whole numbers of nucleons.


1 atomic mass unit (u) = 1.661 × 10-27 kg.


Do not confuse the atomic mass units with the mass of a nucleon:

Mass of nucleon = 1.67 × 10-27 kg


The table shows particle masses in atomic mass units.  Note that the numbers are expressed to a large number of significant figures as the changes are quite subtle.



Mass (u)







Hydrogen atom (1p+ + 1 e-)


Helium atom (2p+ + 2 n + 2e-)


a particle (2p+ + 2 n)



We need to be careful to distinguish between the atomic mass and the nuclear mass.


Question 1

What is the nuclear mass of helium 3 (3He) of which the atomic mass is 3.016030 u?




Mass Defect

If we add together the mass of an electron and the mass of a single proton, we get the mass of a hydrogen atom. Let us do the same for a helium atom.



Mass (u)


Total (u)


















However, if we look in a data book, we see that the atomic mass is 4.002603 u.  There is a difference of 0.030377 u.


All atoms are lighter than the sum of the masses of the protons, electrons, and neutrons.  This is the mass defect, which is the difference between the total mass of the nucleons and the measured mass of the nucleus itself.


To extract a proton or a neutron from the nucleus, we have to pull pretty hard.  Then we find that it will regain its missing mass.  We can use the idea of binding energy to explain this.  The binding energy is defined as the energy released when a nucleus is assembled from its constituent nucleons.  It is equal to the energy needed to tear the nucleus apart into its nucleons.


So with our helium atom, the missing 0.030377 u is released when the nucleons come together.  That energy has to be put back to split the nucleus up again.


This brings us onto the strange idea that mass and energy at the nuclear level are interchangeable and can be related with Einstein’s simple equation:


E = mc2


[E – energy (J); m – mass (kg); c – speed of light (m s-1)]


It is important to convert the mass from atomic mass units to kilograms.  The answer we get is in joules.  We can convert this to eV by dividing by 1.6 × 10-19 J eV-1.


You may see the equation written as:


Question 2

What is the binding energy of the helium atom whose mass defect is 0.030377 u?



Joules are not convenient units to use at the nuclear level, so we convert to electron volts (eV) by dividing by 1.6 × 10-19 J eV-1.


A useful conversion factor between mass and energy is that 1 u = 931.3 MeV



Worked Example


Mass (u)


Total (u)










What is the mass defect in atomic mass units (u) and in kilograms for the lithium nucleus which has 7 nucleons, and a proton number of 3?  What is the binding energy in J and eV?  What is the binding energy per nucleon in eV?  The nuclear mass = 7.014353 u.  


Li has a nucleon number of 7 and a proton number of 3, which means there are 3 protons and 4 neutrons.


Now look up the masses for the proton and neutron from the data.  These will be given to you; you are not expected to remember them.


Add them together to get 7.056488 u


Now take away the nuclear mass from the number above to get the mass deficit.

7.056488 u - 7.014353 u = 0.042135 u  


Now convert to kilograms: 1 u = 1.661 × 10-27 kg

0.042135 u × 1.661 × 10-27 kg  = 6.9986235 × 10-29 kg  


Now use E = mc2 to work out the binding energy:


E = 6.9986235 × 10-29 kg × (3.0 × 108 m s-1)2 =  6.3 × 10-12 J


In electron volts, this is 6.3 × 10-12 J ÷ 1.6 × 10-19 eV J-1 = 3.9 × 107 eV = 39 MeV.  

There are 7 nucleons so the binding energy per nucleon = 3.9 × 107 eV ÷ 7 = 5.6 × 106 eV  



Question 3

What is the mass defect in atomic mass units (u) and in kilograms for the copper nucleus which has 63 nucleons, and a proton number of 29?  What is the binding energy in J and eV?  What is the binding energy per nucleon in eV?  The nuclear mass = 62.91367 u.




Binding Energy Per Nucleon

If we know the binding energy in a nucleus, and the number of nucleons, we can work out the binding energy per nucleon, which is the average energy needed to remove each nucleon.  The higher the binding energy per nucleon, the more stable is the nucleus.  For helium (4He) the binding energy per nucleon is:


                        Binding energy per nucleon = 28.38 MeV ÷ 4 = 7.1 MeV


We can plot a graph of binding energy per nucleon against nucleon number.


From this graph we can see the following:


Iron has the highest binding energy per nucleon so is the most stable nucleus.  If we look at large nuclei (greater than iron), we find that the further to the right (greater nucleon number) the less stable the nuclei.  This is because the binding energy per nucleon is getting less.  The explanation for this observation lies in that the strong nuclear force that binds the nucleus together has a very limited range, and there is a limit to the number of nucleons that can be crammed into a particular space. 

Radioactive Decay and Binding Energy

Radioactive decay happens when an unstable nucleus emits radiation.  It becomes more stable.  The daughter nuclei always have a higher binding energy per nucleon than the parent nucleus.


Let us look at alpha decay:  



Mass of the thorium nucleus = 227.97929 u

Mass of the radium nucleus = 223.97189 u

Mass of the alpha particle (helium nucleus) = 4.00151 u


Mass on the left hand side = 227.97929 u

Mass on the right hand side = 223.97189 u + 4.00151 u = 227. 97340 u


The right hand side has a mass defect = 227.97929 u  - 227. 97340  u = 0.00589 u


The mass defect can be written in kilograms and the energy can be expressed in joules, but nuclear physicists use a useful little dodge.  The energy equivalence of 1 u = 931.5 MeV.  So the energy given out by this decay is:

E = 931.5 × 0.00589 = 5.49 MeV.


This is a beta decay:

Question 4 

Given these data:

Mass of aluminium nucleus = 28.97330 u

Mass of silicon nucleus = 28.96880 u

Mass of beta particle (electron) = 0.00549 u

Mass of electron antineutrino = 0


What is energy given out by the above decay in MeV?  What form does it take?





If we look at the graph with binding energy per nucleon, we observe that the large nuclei have a lower binding energy per nucleon.  This means that they are less stable.  This lack of stability is usually shown by radioactive decay, which occurs in a predictable way.  Very rarely a large nucleus will split up spontaneously into fragments.  This splitting of the nucleus is called fission.


The easiest way to explain this is to consider the nucleus as a “wobbly drop”.  Nuclei are not tidy and neatly arranged rows of neutrons and protons; they are microscopic bedlam. 


The strong nuclear force acts between neighbouring nucleons. 



The nucleons are not linked with the same neighbours all the time.  Instead they are constantly swapping about.  However the enough of the nucleons linked to stop the repulsive electromagnetic force tearing the nucleus apart.


Now we imagine the nucleus as a wobbly drop:



Now if the nucleus gets to this shape



The nucleus flies apart in two fragments:



The detail of the mechanism that drives this process is complicated and is based on Heisenberg’s uncertainty principle.  A similar model can be used to explain how alpha decay works.


We can induce fission in large nuclei such as uranium-235.  The most common isotope of uranium, U-238, does not split easily, but the 235 isotope does.  We induce fission by “tickling” the nucleus with a “thermal” neutron.  The neutron has to have the right kinetic energy:



The tickled nucleus flies apart into a number of fragments, leaving on average three neutrons left over.  These too are able to tickle other nuclei and make them split.  Each neutron spawns three more neutrons in each fission, so we get a chain reaction.  



There is a mass defect in the products of the fission so energy is given out.   In an uncontrolled chain reaction, the energy is given out in the form of a violent explosion, which is many times more powerful than the explosive decomposition of TNT.  In an atomic bomb, the mass that is converted to energy is about 20 grams. 


Nuclear fission has NOTHING whatever to do with radioactive decay.  However the parent nucleus may decay by normal radioactive decay processes, and the daughter nuclei may well be radioactive.  This is a common bear trap.


The daughter fragments may well be highly unstable, and decay by radioactivity.  These form the dangerous fall-out of an atomic bomb detonation, or the waste from a nuclear power station.  Either way, they form some of the nastiest muck known to mankind.


Question 5

A thermal neutron is one that has a kinetic energy of about 1 eV.  How fast does a thermal neutron travel?



 Compared to the speed of many particles in nuclear and particle physics, this speed is pretty sluggish.


A common bear trap is to say that nuclei are smashed to pieces by neutrons.  The neutrons tickle the nucleus; they do not hammer it.


Some students confuse fission and fusion and use the word “fussion”.  It will be marked wrong in the exam, so don’t.




Fusion means joining nuclei together, every alchemist’s dream.  It is easier said than done.  The idea is that light nuclei are joined together, increasing the binding energy per nucleon.  This will result in lots of energy being given out.  A possible reaction is:



It is not simply a case of sticking some deuterium and tritium together and shaking it up.  Each nucleus has to have sufficient energy to:


This means that the gases have to be heated to a very high temperature, 100 million Kelvin.  As all matter at this temperature exists as an ionised gas (plasma), it has to be confined in a very small space by powerful magnetic fields.  A considerable amount of effort has been made to make fusion work to generate electricity.  A fusion reactor would be made to boil water to turn a turbine.  Fusion has occurred, but the energy put in to cook the gases enough to make them fuse is far greater than the energy got out by a fusion reaction.


The only use that fusion has been put to is in a thermonuclear device The third bomb from the left is a genuine thermonuclear device, now on display (with the nasty bits taken out).  The amount of hydrogen required in the bomb below (430 kilo-tonnes) would fill a small party balloon. 




Some scientists claim to have found fusion at low temperatures.  They had a strange chemical reaction, but it was not fusion.


Fusion, if it could be made to work, has a number of advantages over fission:


The downside is that materials that make up the reactor will be irradiated with neutrons which will make them radioactive.


Question 6

Data to use:

Mass of deuterium nucleus = 3.3425 × 10-27 kg

Mass of tritium nucleus = 6.6425 × 10-27 kg

Mass of helium nucleus = 6.6465 ×10-27 kg

Mass of a neutron = 1.675 × 10-27 kg


What is the energy in J and eV released in this reaction above?



Fusion in Stars

The process has three stages:

1. Proton + Proton ® Deuterium + positron + electron neutrino

2. Deuterium + proton ® Helium 3 + photon

3. Helium 3 + Helium 3 ® Helium 4 + proton + proton


Since two protons are left over, the reaction is self sustaining.


A more detailed further discussion on fission and fusion can be found at Physics 6 Tutorial 12.