# 18.4.1 Alpha Decay

Radioactive decay occurs when an unstable nucleus spontaneously (and randomly) transforms into a different nucleus, emitting different kinds of radiation in the process. The H2 syllabus requires knowledge of just three kinds of decays: alpha, beta and gamma.

An alpha particle is actually a very energetic helium-4 nucleus. So an alpha decay is represented by the equation $\displaystyle {}_{Z}^{A}X\to {}_{{Z-2}}^{{A-4}}Y+{}_{2}^{4}He$

The parent nucleus X, transforms itself into the (smaller) daughter nucleus Y, after ejecting the helium‑4 nucleus. The alpha particle speeds off at very high speed in one direction while the daughter nucleus recoils in the opposite direction.

Sample Calculation

Let’s use the alpha decay of Radon-222 into Polonium-218 as an example. $\displaystyle {}_{{86}}^{{222}}Rn\to {}_{{84}}^{{218}}Po+{}_{2}^{4}He$

Atomic mass of Rn-222 $\displaystyle =222.0176u$

Atomic mass of Po-218 $\displaystyle =218.0090u$

Atomic mass of He-4 $\displaystyle =4.0026u$

Firstly, let’s consider the energy of the reaction. Obviously, the total kinetic energy of the decay products come from decrease in rest energy. From the mass-equivalence principle, this can be calculated from the decrease in mass.\ \displaystyle \begin{aligned}{l}K{{E}_{{total}}}&=\Delta m.{{c}^{2}}\\&=[222.0176-(218.0090+4.0026)](1.66\times {{10}^{{-27}}}).{{(3.00\times {{10}^{8}})}^{2}}\\&=8.964\times {{10}^{{-13}}}\text{ J}\\&=5.603\text{ MeV}\end{aligned}

Next, we bring in momentum considerations. Since the total momentum before the reaction is zero (Rn-222 was at rest), it should remain zero after the reaction. This means the momentum of He-4 pa in one direction must be balanced by the momentum of Po-218 ppo in the opposite direction (similar to the rifle and bullet scenario). $\displaystyle {{p}_{\alpha }}={{p}_{{po}}}=p$

If we write KE of the Po-218 and the alpha particle as $\displaystyle \frac{{{{p}^{2}}}}{{2{{m}_{{po}}}}}$ and $\displaystyle \frac{{{{p}^{2}}}}{{2{{m}_{\alpha }}}}$ respectively, we can very quickly work out their KE ratio. \displaystyle \begin{aligned}\frac{{K{{E}_{{po}}}}}{{K{{E}_{\alpha }}}}&=\frac{{{{{p}}_{{po}}}^{2}}}{{2{{m}_{{po}}}}}\div \frac{{{{{p}}_{\alpha }}^{2}}}{{2{{m}_{\alpha }}}}&=\frac{{{{m}_{\alpha }}}}{{{{m}_{{po}}}}}\\\frac{{K{{E}_{{po}}}}}{{K{{E}_{\alpha }}}}&=\frac{4}{{218}}\\K{{E}_{\alpha }}&=54.5K{{E}_{{po}}}\end{aligned}

Notice that the KE ratio is the inverse of the mass ratio. This explains why, in any alpha decay, the bulk of the energy released always goes to the alpha particle, while the more massive daughter nucleus only gets a small share.

Finally, since $\displaystyle K{{E}_{{total}}}=K{{E}_{\alpha }}+K{{E}_{{po}}}=(54.5+1)K{{E}_{{po}}}$, we get $\displaystyle \frac{{K{{E}_{\alpha }}}}{{K{{E}_{{total}}}}}=\frac{{54.5}}{{54.5+1}}\text{ }\Rightarrow \text{ }K{{E}_{\alpha }}=0.982K{{E}_{{total}}}=5.50\text{ MeV}$ $\displaystyle \frac{{K{{E}_{{po}}}}}{{K{{E}_{{total}}}}}=\frac{1}{{54.5+1}}\text{ }\Rightarrow \text{ }K{{E}_{{po}}}=0.018K{{E}_{{total}}}=0.10\text{ MeV}$

Why Alpha Decay is Mono-Energetic

Mathematically speaking, the outcome of an alpha decay is determined by equations (1) and (2), which arise from the Principle of Conservation of Momentum and Energy respectively. $\displaystyle \displaystyle \frac{{K{{E}_{\alpha }}}}{{K{{E}_{D}}}}=\frac{{{{m}_{D}}}}{{{{m}_{\alpha }}}}$                        —–(1) $\displaystyle \displaystyle K{{E}_{\alpha }}+K{{E}_{D}}=\Delta m.{{c}^{2}}$         —–(2)

(KED and mD denote the KE and mass respectively of the daughter nucleus)

Since both the mass ratio $\displaystyle \frac{{{{m}_{D}}}}{{{{m}_{\alpha }}}}$and the (increase in) mass defect $\displaystyle \displaystyle \Delta m.{{c}^{2}}$are fixed for a radioactive decay, we are looking at two equations with two unknowns, resulting in one and only one possible solution for KEα (and KED).

Or, since the total energy released in an alpha-decay is fixed, and the apportionment of KE between the alpha particle and daughter nucleus is fixed (by PCOM which constrains them to have equal but opposite momentum), the alpha particles (of a particular nuclide) can only be ejected with one value of KE.

Concept Test

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Alpha Sneeze