Nuclear Fuel Cycle

Problem: Calculate the equilibrium $^{131}$I activity in a 100 MW(thermal) reactor.

Data:

• Fission yield of $^{131}$I: $Y$ = 2.9%
• 1 W = $3.1 \times 10^{10}$ fissions per second
• Half-life of $^{131}$I: $T_{1/2}$ = 8.02 days

Solution

Step 1: Determine which approximation to use.

The half-life of $^{131}$I is 8.02 days. A typical reactor operates for months to years, which is much longer than 8 days. Therefore:

$T_{1/2} \ll t_{\text{irradiation}}$

We can use the equilibrium (short-lived) approximation: $A_{\text{eq}} = F \cdot Y$

Step 2: Calculate the fission rate $F$.

$F = \text{Power} \times \text{Fissions per watt per second}$

$F = 100 \times 10^{6} \text{ W} \times 3.1 \times 10^{10} \text{ fissions/W/s}$

$F = 3.1 \times 10^{18} \text{ fissions/s}$

Step 3: Calculate the equilibrium activity.

$A_{\text{eq}} = F \cdot Y = 3.1 \times 10^{18} \times 0.029$

$\boxed{A_{\text{eq}} = 9.0 \times 10^{16} \text{ Bq} \approx 90 \text{ PBq} \approx 2.4 \times 10^{6} \text{ Ci}}$

Step 4: Sanity check.

Using the rule of thumb: total FP activity $\approx 10^{11} \times 10^{8} = 10^{19}$ Bq. One individual isotope with ~3% yield should be roughly $0.03 \times 10^{19} = 3 \times 10^{17}$ Bq. Our answer of $9 \times 10^{16}$ Bq is of the right order of magnitude (slightly lower because not all the yield goes to the ground state directly). This looks reasonable.

Common Mistakes --- Worked Example 1

Mistake	How to Avoid It
Forgetting to convert yield from percent to a fraction (using 2.9 instead of 0.029)	Always check: if $Y$ is given as a percentage, divide by 100 before multiplying
Using the long-lived formula when the short-lived formula should be used	Ask yourself: is $T_{1/2}$ much less than or much greater than the irradiation time?
Incorrect power of ten arithmetic	Write out each factor separately and count the powers: $10^{8} \times 10^{10} = 10^{18}$
Using $3.3 \times 10^{10}$ when the question gives $3.1 \times 10^{10}$	Always use the value specified in the question