One question in the 2022 paper stood out for me, because it brought to light an issue that I think has caused some confusion to even the best prepared of students. In fact, given that the issue is on a fine point of detail, it might have caused issues only to the best prepared of students...the others would hardly notice!
The question in question was this one:
The issue that the section circled in red addresses also caused confusion in the 2021 paper with this section from question 11.
In both cases the possible confusion is caused by the fact the masses of the isotopes given on page 83 of the tables are the masses of entire atoms (including electrons), whereas masses listed on pages 46-47 are only those of the nucleons listed.
This probably wouldn't cause any confusion if nuclear physics wasn't inclined to use H as a symbol to represent a single proton, (instead of using a 'p') or to use He (rather than an alpha symbol) when discussing alpha particles. We all understand why this is done - the Hydrogen nucleus contains only a single proton, and so on - but it's obviously going to cause some confusion to a student trying to decide which mass to use, and whether to go with page 83 or page 47.
I imagine a good answer for the 2022 question would look something like this:
But this doesn't address the related issue that arose in 2021 - where students had to choose between using the page 47 value for an alpha-particle or the page 83 value for Helium.
The marking scheme offered the following information:
If you work through the details, you will see that the value given for the change in mass can only be found if you use the mass for Helium-4 given on page 83. But the problem with that is that the question clearly refers to an alpha-particle and not to Helium. The p.83 value will obviously include electrons, so why would we use it in this situation?
The answer - as was outlined by the SEC last year at the IoP review of the marking scheme - is because there are electrons involved.
The mass we would use for the carbon atom created in the reaction will include 6 electrons.
Before the reaction, the Beryllium atom involved will only include 4 electrons. So where have the extra two electrons come from?
The answer is, confusingly, that they don't exist, or at least that they're not factored into the reaction as described. Immediately after the reaction, the carbon will have only 4 electrons. It will undoubtedly acquire two more soon from the surroundings - but that is not within the scope of the question.
So to deal with this issue, we have a few choices:
We could subtract the mass of these two electrons from the mass of the carbon on the right hand side of the equation in order to find the correct mass immediately after the reaction.
Or we could subtract the masses of all electrons involved on both sides of the equation. That might be preferable from a purist point of view. After all, we are dealing with nuclear physics here and we only interested in what is in the nucleus.
But it is much easier to make a small change to the left hand side of the equation. If we use the mass of a Helium-4 atom from page 83 instead of using the mass of an alpha particle there, we have included the mass of six electrons on the left of the equation and six electrons on the right hand side. By doing that, they will balance each other out - and have no effect on the number we are actually looking for: the change of the nuclear mass during the reaction.
Is all of that a bit subtle for a Leaving Cert paper? Where students are not only trying to master this topic, but everything else on the syllabus too - and also everything they have to learn in six or seven other subjects? Perhaps it is. But we were reassured last year that they SEC didn't penalise students on this issue at all if they ignored the electron masses. Full marks were awarded regardless of whether they used the page 83 or page 47 values. Which seems fair enough to me.
But now that the issue has been highlighted in this way, I wonder will they be so forgiving in the future?