-2
$\begingroup$

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!

$\endgroup$
9
  • 5
    $\begingroup$ I have growing feeling you may confuse CH SE site with an online interactive course of chemistry. At doubts over your textbook, I advice you to think, search online and other sources well, read them, think again, and then think more. If not enough, ask with the explicit summary of your intermediate results of your prior effort. Without it, your questions may get frequently closed. $\endgroup$
    – Poutnik
    Apr 25 at 14:22
  • 1
    $\begingroup$ Define "heavier". $\endgroup$ Apr 25 at 14:27
  • $\begingroup$ Heavier means "with an atomic number greater than $83$" . $\endgroup$
    – Maurice
    Apr 25 at 14:45
  • 2
    $\begingroup$ @Pranita In practice, this mode of decay has only been observed in nuclides considerably heavier than nickel, with the lightest known alpha emitters being the lightest isotopes (mass numbers 104–109) of tellurium (element 52). Exceptionally, however, beryllium-8 decays to two alpha particles (Wikipedia) $\endgroup$ Apr 25 at 14:46
  • $\begingroup$ First, this is a better fit for Physics.SE. Second, why fixate on number 83? Is this a number you came up with, or is there a reference, since α-decay is also possible for lighter elements? Third, a nitpick: particle symbols are always upright and it makes no sense to use MathJax to denote them in text; there is no upgreek macro yet and in math mode all Greek letters are slanted, which is a wrong notation. $\endgroup$
    – andselisk
    Apr 25 at 15:57
6
$\begingroup$

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 $\ce{^142_58Ce}$.

$\endgroup$
3
$\begingroup$

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.

$\endgroup$
9
  • 1
    $\begingroup$ Yes. But why is this limit fixed at this number 83 ? Why 83 ? $\endgroup$
    – Maurice
    Apr 25 at 14:48
  • 1
    $\begingroup$ If it was 81 or 87, would you ask why 81 or 87? :-) At some proton number, the nuclear/EM force balance must have its threshold. $\endgroup$
    – Poutnik
    Apr 25 at 15:24
  • 2
    $\begingroup$ If it was 81, 87, 52, or 2000, I would ask the same question. Why is there a limit here ? Why should it have a threshold indeed ? Even Beryllium-8 emits two alphas ! $\endgroup$
    – Maurice
    Apr 25 at 16:45
  • $\begingroup$ Rather, 8Be cracks into 2 halves. Alpha has exceptionally low energy. 83 is the proton number of the first unstable element (not counting Tc and Pm) 209Bi has the halftime approx 10^19 years. $\endgroup$
    – Poutnik
    Apr 25 at 17:35
  • 1
    $\begingroup$ @Maurice I asked once long time ago, why all radioactive series terminate at lead? Why exactly lead? Why not other metals? In fact, I got 2 very beautiful answers. Turns out, there is no threshold. You can theoretically reach hydrogen from radioactive decay but it will take an immense amount of time for the decays to happen (more than the age of universe). $\endgroup$ Apr 26 at 3:15

Not the answer you're looking for? Browse other questions tagged or ask your own question.