00:01
For this problem on the topic of nuclear physics, we want to show that the total binding energy of a given nucleide, ebe, is equal to z delta h plus n delta n minus delta.
00:11
Where delta h is the mass excess of h1, delta n is the mass excess of a neutron, and delta the mass excess of a given of the given nucleide.
00:21
Now, using this method, we want to find the binding energy per nucleon for gold 197.
00:28
Now, if a nucleus contains z protons and n neutrons, its binding energy, delta b .e is equal to the sum of mc squared over all the particles minus the mass of the neutral atom, capital m, times c squared.
00:54
And so this is equal to the number of proton z times the mass of a proton, plus the mass of a proton, plus the, number of neutrons n times the mass of a neutron minus the mass of the neutral atom capital m times c squared now if the masses are given in atomic mass units then the mass accesses are defined by delta h is equal to mh minus one times c squared delta n is equal to the mass of a neutron mn minus 1 times c squared and delta is equal to m minus a times c squared and so this implies that m h c squared is equal to delta h plus c squared mn c squared is equal to delta h plus c squared and m c squared is equal to delta plus a c squared and so the binding energy delta e be e is equal to z delta h plus n delta n minus delta plus z z plus n minus a times c squared.
02:58
So from above we get this to be z delta h plus n delta n minus delta as required...