 # Mass Defect

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## Mass Defect

Get two protons and measure their mass.

Now get two neutrons and find their mass too. (Tip: be careful where you put the protons whilst doing this - they are very easy to put down and lose.)

Now, very quickly and with all the violence you can manage, smack them together!

If you do it right you will end up with a nucleus of helium. Now weigh the result - and you should find that it's less than the mass of the four ingredients put together. Some of the mass has gone missing. Where is it? Well, did you notice the blinding flash of light and the enormous heat when you smacked the four together? That's a bit of a clue as to where the mass has gone. The missing mass has been converted into pure energy. You need to think of energy and mass as the same thing. Mass is just a solid form of energy. So you can convert one to the other and back without breaking the Principle of Conservation of Energy.

Einstein wrote the equation that allows you to find out how much energy will be released when you change mass into energy. It is:

E = mc2

Where:

E = the energy released (J)

m = the mass defect (kg)

c = the speed of light (ms-1)

Note: that the 'm' is the mass that goes missing (the mass defect). It is not the total mass involved in the reaction.

Let's work out the mass defect for a helium nucleus:

The mass of a proton is 1.6726486 x 10 -27 Kg and the mass of a neutron is 1.6749544 x 10 -27 Kg. A helium nucleus has 2 of each so the total mass of the particles needed to build it are:

The actual mass of a helium nucleus is actually 6.64477 x 10 -27 Kg. Notice how it is slightly less than expected.

The mass defect for a helium nucleus is:

Using Einstein's equation this corresponds to an energy equivalent of (remember: c = 3.0 x 108 ms-1)

This is quite a small amount of energy in the everyday world, but on an atomic scale it is significant.

Tip:

The numbers above are very small and hence difficult to deal with. Physicists often use atomic mass units to deal with the small masses, and electron volts to deal with small energies.

The atomic mass unit: u = 1.6605655 x 10 -27 Kg

The electron volt: eV = 1.6021892 x 10 -19 J

In these units you should be able to show that the mass defect for helium is around 0.03 u and this corresponds to an energy of around 28,000,000 eV or 28 MeV (M is Mega i.e. 1 million). The numbers are much easier to work with in this form. Make sure that you can express the numbers like this.

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