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In the above Alpha decay reaction, we get thorium by the alpha decay of uranium with an alpha particle. But we also get energy of amount Q by this process. And according to my course books, this Q energy comes from E = mc2 i.e. some of the mass is converted into energy.

My question is that which mass converted into energy? I mean if you add the nucleons of alpha particle (helium nucleus), you will get the total nucleons present in Uranium. But, where did the energy come from then?


You are correct that if you count the nucleons the number is the same. BUT if you add up the mass of thorium and the mass of the alpha particle you will find that there is an amount of mass missing which corresponds to the Q value. Contemplate this plot of the binding energy per nucleon versus the number of nucleons in the atom:

binding energy curve

There is a peak at the iron nucleus. The curve tells us that breaking a uranium atom into two there will be energy released because in the smaller nuclei each nucleon is more tightly bound.

This is the result of nuclear forces which obey the relativistic equations, all nucleons within a nucleus have smaller masses than the free mass given in the tables.

Edit: It should be noted though, that helium has less binding energy per nucleon than uranium, but thorium more than balances the energy budget , because of the great number of nucleons it has, 4, versus 234.

  • $\begingroup$ so, it is the binding energy that is release, right? $\endgroup$ – Rafique Feb 10 '13 at 11:57
  • $\begingroup$ yes, when the nucleons organize themselves into a thorium and an alpha they are in a higher binding and the excess energy is released. $\endgroup$ – anna v Feb 10 '13 at 12:03

You forget the binding energy, when you take for example a neutron and proton and combine them, the final composite particle's mass will be not equal to the sum of neutron and proton masses, because some mass will be converted into a binding energy, thus when you want to calculate the produced energy of your alpha decay, you should take in account the energy of binding, not just summing up neutron and protons masses, see Nuclear binding energy for details.

An intuitive way to think about above, is to think that nucleons when they are in the atom's nuclei (bidden) they are floating on water (that represents nuclear forces between them), and as you know the weight of the staff is less when they are in water, thus the "weight" of nucleons is less when they are bind.

So the exceed energy can be calculated by:

Binding Energy In of $B_{U_{92}^{238}}=92m_p+(238-92)m_n-m_{U_{92}^{238}}$

Binding Energy In of $B_{Th_{90}^{234}}=90m_p+(234-90)m_n-m_{Th_{90}^{234}}$

Binding Energy In of $B_{\alpha_{2}^{4}}=2m_p+(4-2)m_n-m_{\alpha_{2}^{4}}$

Finally the excess of energy is can be calculated as: $E_{\gamma}=m_{U_{92}^{238}}-m_{Th_{90}^{234}}-m_{\alpha_{2}^{4}} $

  • $\begingroup$ so, do you mean that the energy we obtain in the above example is the binding energy (which is present in nucleus due to conversion of some of mass of proton or neutron into energy), the protons or neutrons do not convert into energy (again)? $\endgroup$ – Rafique Feb 10 '13 at 11:49

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