6
$\begingroup$

A closed vessel with rigid walls contains 1 mol of $_{92}\ce{U}^{238}$ and 1 mol of air at 298K. Considering complete decay of uranium to $_{82}\ce{Pb}^{206}$, the ratio of final pressure to initial pressure of system at 298K is?

​The solution of this question is that during this decay, 8 moles of alpha particles will be emitted, so total moles of gas will be 9. At constant temperature and volume, pressure is proportional to moles of gas. So the answer is 9.

My question is: are helium nuclei considered gas molecules, and should we count them while finding $n$ in the ideal gas equation? My thought process for this question was to find out the number of $\beta^{-}$ particles (electrons) emitted (which is 6 here) and then figure out that they would combine with 3 helium nuclei to form 3 helium atoms. Then I would use the ideal gas equation and the answer would come out to be 7.

Improve this question
$\endgroup$
4
  • 2
    $\begingroup$ The net decay is neutral, with two electrons left behind around the decaying nucleus, so the alphas find electrons eventually as they slow down. $\endgroup$
    – Jon Custer
    Commented Jun 9, 2023 at 11:28
  • $\begingroup$ You seem to be in a grave misconception about the nature of radioactive decay. Sure, helium nuclei are most emphatically not considered gas molecules. But then again, helium nuclei do not exist. Let that sink in. They do not exist. Not for any significant amount of time, that is. They all instantly turn into helium atoms, and then you seem to know what to do. $\endgroup$
    – Ivan Neretin
    Commented Jun 10, 2023 at 1:02
  • $\begingroup$ A multi-MeV alpha particle most certainly does not ‘instantly’ turn into a neutral atom. It must slow down to where the possibility of grabbing an electron with reasonable probability can happen. One can happily propagate an MeV He++ beam for many cm in air without them becoming neutral (while creating many ions in the air along the track). $\endgroup$
    – Jon Custer
    Commented Jun 10, 2023 at 11:48
  • $\begingroup$ Ahh, relativity of terms! Multi-MeV alpha particle would pass these centimetres in less than nanosecond - much shorter time than duration of many "instant" processes. $\endgroup$
    – Mithoron
    Commented Jun 11, 2023 at 16:25

1 Answer 1

Reset to default
9
$\begingroup$

When it is an alpha particle, it is not a molecule(IUPAC), that is by definition neutral.

An electrically neutral entity consisting of more than one atom (n>1). Rigorously, a molecule, in which n>1 must correspond to a depression on the potential energy surface that is deep enough to confine at least one vibrational state.

In contrary, a molecular entity(IUPAC) covers ions too, so an alpha particle is a molecular entity:

Any constitutionally or isotopically distinct atom, molecule, ion, ion pair , radical, radical ion, complex, conformer etc., identifiable as a separately distinguishable entity.

When an alpha particle becomes a molecule, it is not an alpha particle anymore, as an alpha particle is a ion, a nucleus of helium-4. When they slow down enough, they are able to take electrons from literally any atom but helium. Well, in the first step even from helium, as

$$\ce{He^2+ + He -> 2 He+}$$

In the idealized case when no high energy particle leaves the idealized contained, the total charge balance remains neutral. There would be neutral atoms, molecules, ionic pair, ions and electrons that would eventually recombine to neutral $\pu{1 mol}$ of lead-206, $\pu{8 mol}$ helium-4 of , together with $\ce{1 mol}$ of air ($\ce{1 mol}$ of any aerial molecules).

Note that until all uranium decays (what would take much longer than the age of universe), lead would be already oxidized by oxygen.

Improve this answer
$\endgroup$
13
  • 1
    $\begingroup$ I suspect the question doesn't expect a full analysis of all the decay products (including electrons) and assumes that alpha particles will rapidly abstract electrons from somewhere. 8 moles of doubly charged ions would a hell of a driving force for electron abstraction even if the decay itself were not charge neutral. $\endgroup$
    – matt_black
    Commented Jun 9, 2023 at 15:21
  • 1
    $\begingroup$ Neither I suspect that. I also noted recombinations. $\endgroup$
    – Poutnik
    Commented Jun 9, 2023 at 15:26
  • 1
    $\begingroup$ @Poutnik Dammit, you beat me to that! ;-) $\endgroup$
    – matt_black
    Commented Jun 9, 2023 at 16:02
  • 1
    $\begingroup$ @matt_black Waiting for that last atom would be funny. :-) $\endgroup$
    – Poutnik
    Commented Jun 9, 2023 at 16:02
  • 1
    $\begingroup$ @Poutnik But decay would take infinite time if you were a mathematician and ignored the particulate nature of atoms. This generated a long, frustrating thread here a few days ago on a related topic. $\endgroup$
    – matt_black
    Commented Jun 9, 2023 at 16:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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