Why alpha-decay only occurs in elements with atomic number greater than 83?

Question

In my textbook, it is written that $α$-decay only occurs in heavier nuclei. But why? Why is that so? There is literally no explanation given in my textbook as to why this is true. Please explain. Would greatly appreciate it!

Answer 1

A general equation for the alpha decay is $$\ce{A -> B + ^4_2He + $\Delta E$}$$

You can check whether the decay is possible by calculating $\Delta E$ as $$\Delta E=\left(m_{\ce{A}}-m_{\ce{B}}-m_{\ce{He}}\right)c^2$$

The alpha particle has a relatively high binding energy of $28.296\ \mathrm{MeV}$ ($7.074\ \mathrm{MeV}$ per nucleon). However, when going down from $\ce{A}$ to $\ce{B}$, some binding energy is lost. The alpha decay is possible if this lost binding energy is less than $7.074\ \mathrm{MeV}$ per nucleon.

You should already know that the graph of the binding energy per nucleon has a maximum at around iron. For heavier nuclides, the binding energy per nucleon is decreasing. So the increase of total binding energy is not linear but levels out a bit. If this increase per nucleon falls below $7.074\ \mathrm{MeV}$, alpha decay becomes possible.

From the graph of total binding energy, we can see that alpha decay should be possible starting at a mass number of $A=142$. In fact, alpha decay was observed for $\ce{^142_58Ce}$.

Answer 2

Generally, lighter nuclei have too high nuclear bonding energy to allow alpha particles to escape.

For heavy elements, coulombic repulsive forces weakens the overall bonding, that leads to alpha radioactivity.

Exception are all nuclei with nucleon number 5 or 8 that quickly decay, forming an alpha plus a nucleon, resp. 2 alpha particles(eventually after beta decay )

Aside of the above exceptions, the lightest alpha emitter is $\ce{^{104}_{52}Te}$, probably because the target $\ce{^{100}_{50}Sn}$ nucleons have double magic numbers 50 + 50. Isotopes with proton or neutron count equal to magic numbers are often especially stable.