Nuclear Fuel Cycle

A naval reactor operates at a thermal power of 50 MW for a period of 2 years.

(a) Calculate the equilibrium $^{131}$I activity in the reactor. (2 marks)

(b) Calculate the $^{90}$Sr activity at the end of the 2-year operating period. (3 marks)

(c) Using the rule of thumb, estimate the total fission product activity in the reactor at the end of operation, and estimate the total activity 24 hours after shutdown. (3 marks)

Data:

• 1 W = $3.1 \times 10^{10}$ fissions per second
• $^{131}$I: yield = 2.9%, $T_{1/2}$ = 8.02 days
• $^{90}$Sr: yield = 5.8%, $T_{1/2}$ = 28.8 years

Solution

(a) $^{131}$I equilibrium activity:

Step 1: $T_{1/2}$ of $^{131}$I = 8.02 days $\ll$ 2 years. Use the equilibrium approximation: $A_{\text{eq}} = F \cdot Y$.

Step 2: Calculate the fission rate.

$F = 50 \times 10^6 \times 3.1 \times 10^{10} = 1.55 \times 10^{18} \text{ fissions/s}$

Step 3: Calculate the activity.

$A_{\text{eq}} = 1.55 \times 10^{18} \times 0.029 = 4.5 \times 10^{16} \text{ Bq}$

$\boxed{A(^{131}\text{I}) = 4.5 \times 10^{16} \text{ Bq} = 45 \text{ PBq}}$

(b) $^{90}$Sr activity after 2 years:

Step 1: $T_{1/2}$ of $^{90}$Sr = 28.8 years $\gg$ 2 years. Use the linear (long-lived) approximation: $A(t) = F \cdot Y \cdot \lambda \cdot t$.

Step 2: $F = 1.55 \times 10^{18}$ fissions/s (from part (a)).

Step 3: Calculate $\lambda \cdot t$:

$\lambda \cdot t = \frac{0.693 \times 2}{28.8} = \frac{1.386}{28.8} = 0.04813$

Step 4: Calculate the activity.

$A(2\text{ yr}) = 1.55 \times 10^{18} \times 0.058 \times 0.04813$

$= 1.55 \times 10^{18} \times 2.791 \times 10^{-3}$

$\boxed{A(^{90}\text{Sr}) = 4.3 \times 10^{15} \text{ Bq} = 4.3 \text{ PBq}}$

(c) Total fission product activity (rule of thumb):

Step 1: At the end of operation:

$A_{\text{total}} \approx 10^{11} \text{ Bq/W} \times 50 \times 10^{6} \text{ W}$

$\boxed{A_{\text{total (at shutdown)}} \approx 5 \times 10^{18} \text{ Bq} = 5 \text{ EBq}}$

Step 2: After 24 hours, the activity decays by approximately a factor of 10:

$A_{\text{total (24 h)}} \approx \frac{5 \times 10^{18}}{10} = 5 \times 10^{17} \text{ Bq}$

$\boxed{A_{\text{total (24 h after shutdown)}} \approx 5 \times 10^{17} \text{ Bq} = 500 \text{ PBq}}$

Comment: Even 24 hours after shutdown, the total activity is still enormously high --- over 100 times the individual equilibrium $^{131}$I activity calculated in part (a). This illustrates why continuous cooling and shielding are essential for an extended period after shutdown.