Nuclear Fuel Cycle

(a) Under what conditions may the following formula for the activity of a fission product in a reactor be used?

$A(t) = R \cdot \lambda \cdot t$

(b) Use the formula to calculate the $^{137}$Cs activity in a 1 kW research reactor operated for 6 months.

Data:

• 1 W = $3.3 \times 10^{10}$ fissions per second
• Half-life of $^{137}$Cs = 30.1 years
• Yield factor for $^{137}$Cs = 6%

(c) Using a $^{131}$I yield factor of 2.9%, calculate the equilibrium $^{131}$I activity in the same reactor.

Solution

(a) Conditions for use:

The formula $A(t) = R \cdot \lambda \cdot t$ may be used when:

1. The half-life of the fission product is long compared to the irradiation time (i.e. $T_{1/2} \gg t$). This is because the formula is a linear approximation of the full solution $A(t) = R(1 - e^{-\lambda t})$, valid when $\lambda t \ll 1$.
2. The simplified model assumptions hold: production is only from direct fission of $^{235}$U, and burn-up of the fission product by neutron capture can be neglected.

Examples of suitable isotopes: $^{137}$Cs ($T_{1/2}$ = 30.1 y) and $^{90}$Sr ($T_{1/2}$ = 28.8 y), when the irradiation time is on the order of months to a few years.

(b) $^{137}$Cs activity in a 1 kW reactor after 6 months:

Step 1: Identify the parameters.

Step 2: Confirm the approximation is valid.

$T_{1/2}$ = 30.1 years $\gg$ $t$ = 0.5 years. The ratio is 60:1, so the long-lived linear approximation is valid.

Step 3: Calculate the fission rate $F$.

$F = 1000 \text{ W} \times 3.3 \times 10^{10} \text{ fissions/W/s} = 3.3 \times 10^{13} \text{ fissions/s}$

Step 4: Calculate $\lambda \cdot t$ (using consistent time units).

$\lambda \cdot t = \frac{0.693}{T_{1/2}} \times t = \frac{0.693}{30.1} \times 0.5 = 0.02302 \times 0.5 = 0.01151$

Step 5: Calculate the activity.

$A = F \cdot Y \cdot \lambda \cdot t$

$A = 3.3 \times 10^{13} \times 0.06 \times 0.01151$

$A = 3.3 \times 10^{13} \times 6.906 \times 10^{-4}$

Combining step by step:

$3.3 \times 10^{13} \times 0.06 = 1.98 \times 10^{12}$

$1.98 \times 10^{12} \times 0.01151 = 2.28 \times 10^{10}$

Rounding:

$A = 2.28 \times 10^{10} \text{ Bq}$

Alternatively expressed:

$\boxed{A \approx 2.3 \times 10^{10} \text{ Bq} \approx 0.023 \text{ TBq} \approx 23 \text{ GBq}}$

Note: Some students may combine the calculation in a single line as: $A = 1000 \times 3.3 \times 10^{10} \times 0.06 \times 0.5 \times 0.02302 = 2.28 \times 10^{10}$ Bq This is acceptable provided each factor is clearly identified.

(c) Equilibrium $^{131}$I activity in the 1 kW reactor:

Step 1: Determine the appropriate approximation.

$^{131}$I has $T_{1/2}$ = 8.02 days. Even a research reactor operating for a few weeks will have run for much longer than 8 days. Therefore, $T_{1/2} \ll t_{\text{irradiation}}$ and we use the equilibrium (short-lived) approximation:

$A_{\text{eq}} = F \cdot Y$

Step 2: Calculate.

$A_{\text{eq}} = F \cdot Y = 3.3 \times 10^{13} \times 0.029$

$\boxed{A_{\text{eq}} = 9.6 \times 10^{11} \text{ Bq} \approx 0.96 \text{ TBq}}$

Step 3: Sanity check.

For a GW-scale reactor, we expect $^{131}$I activity of order $10^{18}$ Bq. Our reactor is $10^{6}$ times smaller in power (1 kW vs 1 GW), so we expect approximately $10^{18} / 10^{6} = 10^{12}$ Bq. Our answer of $9.6 \times 10^{11}$ Bq is consistent with this estimate.