Nuclear Fuel Cycle

Q: Explain quantitatively why U-234, despite comprising only ~0.027% of enriched uranium by mass, dominates the total activity.

A:

The specific activity of a pure isotope is:

$a = \frac{\lambda N_A}{M} = \frac{\ln 2 \times N_A}{T_{1/2} \times M}$

Let us calculate this for each isotope:

U-234: $a_{234} = \frac{0.6931 \times 6.022 \times 10^{23}}{2.46 \times 10^5 \times 3.156 \times 10^7 \times 234} = \frac{4.172 \times 10^{23}}{1.816 \times 10^{15}} = 2.30 \times 10^{8} \text{ Bq/g}$

U-235: $a_{235} = \frac{0.6931 \times 6.022 \times 10^{23}}{7.04 \times 10^8 \times 3.156 \times 10^7 \times 235} = \frac{4.172 \times 10^{23}}{5.221 \times 10^{18}} = 7.99 \times 10^{4} \text{ Bq/g}$

U-238: $a_{238} = \frac{0.6931 \times 6.022 \times 10^{23}}{4.47 \times 10^9 \times 3.156 \times 10^7 \times 238} = \frac{4.172 \times 10^{23}}{3.357 \times 10^{19}} = 1.24 \times 10^{4} \text{ Bq/g}$

So the specific activity per gram of pure isotope is:

U-234 is approximately 18,500 times more active per gram than U-238. Even though it is present in a mass fraction roughly 3,600 times smaller than U-238 (0.027% vs 96.5%), it still contributes the most activity because 18,500/3,600 $\approx$ 5, meaning its activity contribution is still about 5 times larger.