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# Calculating Nuclear Masses + Binding Energies watch

1. I'm not too sure how my Physics book has come to this conclusion but
n + (235)U -> (135)Te + (97)Zr + 4 n + energy

Where n is a neutron, now I can see where they get the value of n from at 1.00867u (in terms of 'u', the conversion constant to the SI unit which is 1.6605x10^-27)

Basically the mass value in terms of u for Uranium-235 was 235.044u but I do not know how they get his value and also for Tellurium-135 they get 134.941u and Zr-97 they get 96.911u.

Can someone please like work through how to get the Uranium-235 mass and then I should be able to learn the method.

Thanks Guys.
2. I'm not too sure how my Physics book has come to this conclusion but
n + (235)U -> (135)Te + (97)Zr + 4 n + energy

Where n is a neutron, now I can see where they get the value of n from at 1.00867u (in terms of 'u', the conversion constant to the SI unit which is 1.6605x10^-27)

Basically the mass value in terms of u for Uranium-235 was 235.044u but I do not know how they get his value and also for Tellurium-135 they get 134.941u and Zr-97 they get 96.911u.

Can someone please like work through how to get the Uranium-235 mass and then I should be able to learn the method.

Thanks Guys.
This is a pretty complicated calculation you're talking about here. The info you need is:

mass of a proton = 1.0073 u

mass of neutron = 1.0087 u

Now in uranium-235 there are 143 neutrons and 92 protons, so the total mass SEEMS to be simply

However this is not the correct answer; it is slightly over. The 'missing mass' is known as the mass deficit. This is proportional to the decrease in energy that occurred when the nucleus formed from its constituent parts. A collection of 143 neutrons and 92 protons, all separate, are at a slightly higher energy than when bound together as U-235: if this were not the case then the bound state would not be energetically favorable and would not form.

The only real way to work out the mass deficit is to measure it experimentally, or to work it out by considering mass-energy conservation in a reaction where the masses and energies of all other reactants and products are known.

So in short the book looked up those values, which have been found by experiment. In theory I guess they are calculable by modeling the formation of the nucleus using quantum mechanics and determining the binding energy this way, but for a nucleus of the size of U-235 this is extremely difficult (certainly wont be done by hand - think supercomputer and many simplifying approximations!)

3. Ahh yes that it does answer my question, I was getting rather confused but now you mention it we were given a database sheet which I think must have these masses on. Thanks again!

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Updated: February 13, 2010
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