I thought specific activity was a property of the radionuclide?

Question

How can specific activity of Mo-99 be higher for Mo-99 produced from fission than Mo-99 produced from neutron capture? I thought specific activity would be a property of Mo-99 so it will always be the same? Can someone explain what I am understanding wrong please. Thanks.

Btw here's the website that says you can have higer specific activity Mo-99:

https://www.ncbi.nlm.nih.gov/books/NBK396163/#sec_2-5-4

Answer 1

As noted, there are two main paths (and several minor ones) for production of $^{99}$Mo, the parent of the heavily used medical radioisotope $^{99m}$Tc. The $^{99}$Mo isotope has a half-life of 2.75 days, and the $^{99m}$Tc has a half-life of only 6 hours.

One production path takes stable $^{98}$Mo and irradiates it with thermal neutrons to drive the $^{98}$Mo(n,$\gamma$)$^{99}$Mo reaction. Going to ENDF - Evaluated Nuclear Data Files this has a cross section that looks like:

Here the blue line is the total cross section for neutron scattering off of $^{98}$Mo, and the green line is the path to $^{99}$Mo. For low energy (i.e. thermal) neutrons such as found in a normal reactor, the cross section isn't horrible (slightly under 10 barns), so you will get production of $^{99}$Mo. There are then three issues to think about:

1. $^{98}$Mo, while the most abundant Mo isotope, still is only present at the 25% level in naturally occurring Mo. To get higher proportions of $^{98}$Mo you need to do isotope separation, a costly process.
2. When done irradiating, you have a bit of $^{99}$Mo in a lot of $^{98}$Mo (if isotope separated), or even worse in amongst all the other Mo isotopes. Ultimately the patient gets only a bit of $^{99}$Mo decaying to the $^{99m}$Tc useful stuff.
3. The $^{99}$Mo can also undergo neutron reactions during the irradiations to make it in the first place. This may result in formation of $^{100}$Mo, a much more stable (half-life of $10^{18}$years) isotope, wasting the $^{99}$Mo you were trying to make.

The second path is through fission of $^{235}$U and then picking through all the fission products to get the Mo. Seems kind of crazy, right? Well, as it turns out, the Fission Product Data nicely shows it might not be so crazy. Percentage yield vs mass number looks like:

Note that mass 99 is quite nicely near the top of the left (lower mass) hump. However, a mass of 99 does not mean it has to be $^{99}$Mo. Because $^{99m}$Tc is such an important isotope, the fission yield of mass 99 that results in $^{99m}$Tc is measured quite well (and easily through the characteristic gamma emitted through $^{99m}$Tc decay used in the medical diagnostics). The result is that about 6.1% of the low mass fission products are known to be $^{99}$Mo. It even turns out that the yield of mass 98 (and 97, another possible Mo isotope) are somewhat lower, roughly 5.7 and 5.8% respectively.

Further, as we well know, fission of $^{235}$U is easy. Really, really, really easy. In fact, the cross section for fission with thermal neutrons is about 30,000 barns which is larger than the $^{98}$Mo(n,$\gamma$)$^{99}$Mo reaction by a factor of 3000 or more. Now, of course, the problem is separating out the ${99}$Mo (and other Mo isotopes) from the debris, but that can be done chemically. You are left with Mo with a much higher proportion of $^{99}$Mo than you could possibly have obtained by simply irradiating pure $^{98}$Mo, much less natural Mo.

So, the $^{235}$U path has large advantages in both the cross-section (3000x) and the higher proportions of $^{99}$Mo vs other Mo isotopes (greater than 1:1 for $^{99}$Mo vs $^{98}$Mo). On a per-incident-thermal-neutron basis, you get roughly 180 time the yield of $^{99}$Mo through fission. The downside, of course, is the use of highly enriched $^{235}$U to get the easiest outcome, although some paths use low enriched uranium - just means more stuff to sift through to get the $^{99}$Mo in the end.