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Everyone knows that the mass of a system is less than the mass of its components, with the equation:

$M = \sum_i m_i - BE(M) $

Now, if we consider a general decay, lets say

$A \rightarrow \sum_i B_i$

then, for the conservation of the first component of the four-vector momentum we obtain a necessary condition for the decay:

$ M(A) \ge \sum_i M(B_i) $

Isn't this last condition in contradiction with the first one? The only answer I get is a particle that can decay is not an eigenstate of the system. For this reason it shouldn't have a defined binding energy. So, for example, Tritium, that can decay, hasn't a BE? This appears so illogical if we think about nuclei with an half-life of years or more. Moreover I know this energy is experimentally defined and is bigger that 3He's BE.

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2  
what does $A \rightarrow \sum_i B_i$ and $BE(M)$ mean? specifically what is $A$ and $BE$?? –  Nikos M. Jun 13 at 23:51
    
that first equation is wrong if the symbols are interpreted normally. The mass of a system has to be larger than the summation of its component parts. The last relation one is correct. there is no contradiction. –  anna v Jun 14 at 4:26
    
@annav No, the question is right and your comment is wrong. For instance carbon-12 has mass 12.000 amu (by definition), but the masses of the proton and neutron are 1.007 u and 1.009 u. –  rob Jun 14 at 23:18

1 Answer 1

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The thing here is that when a particle decays (NB here I talk about point-like decays, I'll address decaying compound systems later), the products were not "in" the original particle in the first place.

That is $$A \to b + c + d$$ does not imply that $A$ was made up of $b$ and $c$ and $d$. It implies the combination of two facts

  • That $A$ and $b+c+d$ have compatible quantum numbers.
  • That $M_A$ is sufficiently larger than $M_b + M_c + M_d$ to allow for the products to escape from each other.

In other words a neutron is a particle distinct from the collection of a proton, an electron and a electron anti-neutrino, but those of the neutron's quantum numbers that are respected by the charged-current weak interaction are the same as the collection of lighter particles and the neutrons mass is high enough to produce them.

Compound systems with tree-level decays

Now consider beta decay of a non-trivial nucleus. Let's take tritium for the sake keeping the writing simple. The fact that the system is bound means $$M_T < M_p + 2M_n \,.$$

But it is still true that $$M_{^3\mathrm{He}^{+2}} + e^- + \bar{\nu}_e < M_T\,$$ and the quantum number of the resulting system are compatible with those of a Triton

Alpha-decay

Alpha decay is a little different. There is no tree-level transformation there, instead there are multiple masses to think about. I'm going to use $m$ to label the mother isotope, $d$ for the daughter isotope and $\alpha$ for the alpha particle.

Here we have a hierarchy of masses. The mother isotope is bound, so we know that $$ M_m < Z_m M_p + (A_m - Z_m) M_n \,.$$ (Here $Z_m$ and $A_m$ are the charge number and mass number of the mother isotope.)

But, because the decay is energetically allowed we also know that $$M_d + M_\alpha < M_m \,,$$ which is to say that the combined system of the daughter and an alpha is more bound than the mother isotope.

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