Reading "Fundamentals of Nuclear Physics" by Atam P. Arya, I understand that in a nuclear reaction, let say $x+X \to y+Y$ meaning that "when a particle $x$ strikes a target nucleus $X$, the outcome of the nuclear reaction is a recoil nucleus $Y$ and a particle $y$. In many cases more than one type of particle may be given out." Now, Arya applies the energy conservation to the nuclear reaction, writing $$E_i=E_f$$ where $$E_i=K_x+m_xc^2+K_X+M_Xc^2$$ and $$E_f=K_Y+M_Yc^2+K_y+m_yc^2$$ being $K_x, K_X, K_y, K_y$ the kinetic energies of the particle $x$, the target nucleus (parent), the recoil nucleus (daughter) and the particle $y$, respectively. And $M$ & $m$ its respective masses.

Rewriting energy conservation, he defines a quantity $Q$ (the disintegration energy): $$Q:=(K_Y+K_y)-(K_X+K_x)=(M_X+m_x)c^2-(M_Y+m_y)c^2 \qquad (1)$$ Now, my question is whether or not the equation (1) (and its interpretation) can be applied to alpha and beta decay (where there are no particle colliding with the nucleus).


1 Answer 1



Not only can it, but the values are tabulated for unstable isotopes.

Compare the data provided for $^{206}\mathrm{Pb}$ (stable) with the provided for $^{210}\mathrm{Pb}$ (unstable). Look in the section headed "Decay properties". Moreover, note that there is a separate $Q$ provided for each decay mode (but not for each channel).

In this instance, however you need to modify your understanding of what $Q$ means. It represents the energy excess of the parent over the total masses of all the products. In your notation:

$$ Q_\mathrm{channel} = (M_X - M_Y - \sum_\mathrm{products} m_i)c^2 \quad .$$

  • $\begingroup$ Your $Q_{channel}$ is my $Q$? If yes, then what is the difference between your equation and (1)? (Obviously there are no $m_x$). So you are saying that Q cannot be considered as the total change of kinetic energies? $\endgroup$
    – Ana S. H.
    Commented May 4, 2013 at 23:56

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