Nuclear Fuel Cycle

Problem: Calculate the $^{137}$Cs activity in a 100 MW(thermal) reactor operated at constant full power for 1 year.

Data:

• Fission yield of $^{137}$Cs: $Y$ = 6%
• 1 W = $3.1 \times 10^{10}$ fissions per second
• Half-life of $^{137}$Cs: $T_{1/2}$ = 30.17 years

Solution

Step 1: Determine which approximation to use.

The half-life of $^{137}$Cs is 30.17 years. The irradiation time is 1 year. Since:

$T_{1/2} = 30.17 \text{ years} \gg t_{\text{irradiation}} = 1 \text{ year}$

We use the linear (long-lived) approximation: $A(t) = F \cdot Y \cdot \lambda \cdot t$

Step 2: Calculate the fission rate $F$.

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

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

Step 3: Calculate the decay constant $\lambda$ in compatible units.

Since the fission rate is in fissions per second, we need $\lambda$ in $\text{s}^{-1}$ and $t$ in seconds --- OR we can use the shortcut $\lambda \cdot t = 0.693 \times t / T_{1/2}$ where $t$ and $T_{1/2}$ are in the same units (years):

$\lambda \cdot t = \frac{0.693 \times t}{T_{1/2}} = \frac{0.693 \times 1}{30.17} = 0.02297$

Step 4: Calculate the activity.

$A(1 \text{ yr}) = F \cdot Y \cdot \lambda \cdot t$

$A(1 \text{ yr}) = 3.1 \times 10^{18} \times 0.06 \times 0.02297$

$A(1 \text{ yr}) = 3.1 \times 10^{18} \times 1.378 \times 10^{-3}$

$\boxed{A(1 \text{ yr}) = 4.3 \times 10^{15} \text{ Bq} \approx 4.3 \text{ PBq} \approx 1.2 \times 10^{5} \text{ Ci}}$

Step 5: Sanity check.

Compare with the $^{131}$I result: $^{137}$Cs has a higher yield (6% vs 2.9%) but the activity is about 20 times lower ($4.3 \times 10^{15}$ vs $9 \times 10^{16}$). This makes sense because $^{137}$Cs is far from equilibrium after only 1 year of a 30-year half-life. In contrast, $^{131}$I reached equilibrium quickly. This is physically consistent.

Common Mistakes --- Worked Example 2

Mistake	How to Avoid It
Using the equilibrium formula ($A = F \cdot Y$) for a long-lived isotope	Check: is $T_{1/2} \gg t$? If yes, use the linear formula with $\lambda \cdot t$
Mixing units of time (e.g. $t$ in years and $\lambda$ in $\text{s}^{-1}$)	Use the shortcut: $\lambda \cdot t = 0.693 \times t / T_{1/2}$ with both in the same units
Forgetting the factor of 0.693 (using $t/T_{1/2}$ instead of $0.693 \times t/T_{1/2}$)	Remember: $\lambda = 0.693 / T_{1/2}$, so the 0.693 is always present
Rounding too early and propagating errors	Keep at least 3 significant figures through intermediate steps