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This is not a question about politics, although the motivation is a political situation.

Iran, which enriches Uranium to a level of $60\%$ $\rm{^{235}U}$, is claiming it has civilian uses for this enriched material. I assume there could be more than one such use, but this question focuses on the use of HEU for producing $\rm{^{99m}Tc}$, meta-stable Technetium, used in medical imaging.

Iran had $\sim86.5$ Million inhabitants as of $2022$. I don't know how many SPECT scanners they have, so just assume whatever level of coverage of the population by scanners that you can quantify, even if it's not Iran's but the US' or an EU country's, and we can consider normalization ex-post-facto (or not normalize and have an overly-generous estimate). Assume also whichever distribution you like of technology age of deployed scanners w.r.t. the necessary dose per scan, assume any reasonable distribution of the kind of scans performed etc.

Under all of these assumptions - which I realize might not be easy to make - how much $60\%$ HEU would Iran need to produce (in $\mathrm{kg} / \text{day}$ or $\mathrm{kg} / \text{year}$) to satisfy its Technetium needs?

PM 2Ring
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einpoklum
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1 Answers1

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Speculative upper-bound answer based on @JonCuster's comment: No more than $6.5 ~\frac{\mathrm{kg}}{\text{year}}$.

Suppose for exaggeration that Iran can use $5\%$ of all Technetium used in the world (this circumvents the need to estimate things about dose policies, distribution of imaging scans etc.)

Suppose also that with HEU you need no more fuel than with LEU (this is a completely baseless assumption on my part, Physicists please correct me.)

Suppose that fuel use grows more-or-less linearly with reactor energy output.

@JonCuster reports that the two reactors which have so far provided the global Technetium needs were $135 ~\mathrm{MW}$ and $60 ~\mathrm{MW}$ respectively, and the smaller one used $40 ~\frac{\mathrm{kg}}{\text{year}}$ of Uranium.

By our assumptions, the two reactors together used $\left( 1 + \frac{135}{60} \right) \cdot 40 ~\frac{\mathrm{kg}}{\text{year}} = 3.25 \cdot 40 ~\frac{\mathrm{kg}}{\text{year}} = 130 ~\frac{\mathrm{kg}}{\text{year}}$ of Uranium. $5\%$ of this is $6.5 ~\frac{\mathrm{kg}}{\text{year}}$. So, under our assumptions, Iran should not need more than $6.5 ~\mathrm{kg}$ of $60\%$-enriched Uranium each year for Technetium production.

Edit: I neglected to consider a possible difference in process efficiency of the Iranian reactor vs. other reactors.

einpoklum
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