Question 12.6.Q9: To date, out of over 3500 known radionuclides only four (ces......

Recalling that half-life \left(t_{1/2}\right)_{Co-60} = 5.26 a the decay constant of cobalt-60 is given as follows

The product \sigma_{\mathrm{Co}-59} \dot{\varphi} plays an important role in neutron activation theory and has, for our example that deals with cobalt-59 to cobalt-60 neutron activation in a nuclear reactor with thermal neutron fluence rate of \dot{\varphi}=1.2 \times 10^{13} \mathrm{~cm}^{-2} \cdot \mathrm{s}^{-1}, the following value

(a) Infinite number of parent nuclei: Saturation model
Activity \mathcal{A}_{Co-60}(t) as a function of activation time t is in the saturation model expressed as follows (T12.13)

\mathcal{A}_{\mathrm{Co}-60}(t)=\sigma_{\mathrm{Co}-59} \dot{\varphi} N_{\mathrm{Co}-59}(0)\left[1-e^{-\lambda_{\mathrm{Co}-60} t}\right]=\left(\mathcal{A}_{\mathrm{Co}-60}\right)_{\mathrm{sat}}\left[1-e^{-\lambda_{\mathrm{C} 0-60} t}\right]            (12.182)

where the saturation activity \left(\mathcal{A}_{Co-60}\right)_{sat} is defined as

\left(\mathcal{A}_{\mathrm{Co}-60}\right)_{\mathrm{sat}}=\sigma_{\mathrm{Co}-59} \dot{\varphi} N_{\mathrm{Co}-59}(0) \text {, }           (12.183)

with N_{Co-59}(0) the initial number of Co-59 nuclei placed at activation time t = 0 into the nuclear reactor with thermal neutron fluence rate \dot{\varphi} \text { of } 1.2 \times 10^{13} \mathrm{~cm}^{-2} \cdot \mathrm{s}^{-1}.

N_{\mathrm{Co}-59}(0)=\frac{N_{\mathrm{A}}}{A} m=\frac{6.022 \times 10^{23} \mathrm{~mol}^{-1}}{58.93 \mathrm{~g} \cdot \mathrm{mol}^{-1}} \times(10 \mathrm{~g})=1.022 \times 10^{23}            (12.184)

(1) After inserting (12.181) and (12.184) into (12.183), we now determine the saturation activity \left(\mathcal{A}_{Co-60}\right)_{sat} for our neutron activation example as follows

(2) After inserting (12.185) into (12.182), the activity \mathcal{A}_{Co-60}(t) of the sample as a function of activation time t is now given as

Activities \mathcal{A}_{Co-60}(t) of cobalt-60 calculated from (12.186) for the saturation model of neutron activation in steps of 1 year for activation times 0 ≤ t ≤ 30 a are listed in Table 12.14.

(3) To estimate the activation time t during which a given activity fraction f_{Co-60} of \left(\mathcal{A}_{Co-60}\right)_{sat} is attained we modify (12.182) as follows

\frac{\mathcal{A}_{\mathrm{Co}-60}(t)}{\left(\mathcal{A}_{\mathrm{Co}-60}\right)_{\mathrm{sat}}}=f_{\mathrm{Co}-60}=1-e^{-\lambda_{\mathrm{Co}-60} t}             (12.187)

Solving (12.187) for activation time t gives

t=-\frac{-\ln \left(1-f_{\mathrm{Co}-60}\right)}{\lambda_{\mathrm{Co}-60}}=\frac{\ln \frac{1}{1-f_{\mathrm{C} o-60}}}{\lambda_{\mathrm{Co}-60}}           (12.188)

Results of (12.188) for activity fractions λ_{Co-60} of 0.25, 0.50, 0.75, 0.90, and 1.0 are given in Table 12.15.

(b) Finite number of parent nuclei: Depletion model
Activity \mathcal{A}_{Co-60}(t) as a function of activation time t is in the depletion model expressed as follows (T12.22)

where \mathcal{A}_D is a parameter of the depletion model, dependent on neutron cross section σ_{Co-59} of the Co-59 nucleus, neutron fluence rate \dot{φ} in the reactor, and the initial number N_{Co-59}(0) of Co-59 nuclei placed into the reactor, as well as the decay constant λ_{Co-60}. Parameter A_0 of (12.189) is fortuitously related to the saturation activity \left(\mathcal{A}_{Co-60}\right)_{sat} of the saturation model given in (12.185) and is for our neutron activation example given as follows

(1) Following the depletion model of neutron activation, after inserting (12.180), (12.181), and (12.190) into (12.188), the activity A_{Co-60}(t) as a function of activation time t is expressed as

(2) For the depletion model, the activity \mathcal{A}_{Co-60} is 0 at activation time t = 0, rises with t until it reaches a maximum \left(\mathcal{A}_{Co-60}\right)_{max} at characteristic activation time t = \left(t_{Co-60}\right)_{max}, then it decreases with t and asymptotically approaches 0 at t → ∞. The characteristic activation time \left(t_{max}\right)_{Co-60}, at which activity maximum \left(\mathcal{A}_{Co-60}\right)_{max} occurs, is given by the following expression (T12.24), determined from setting to zero the derivative of (12.191) with respect to activation time t. For our specific example the characteristic time \left(t_{max}\right)_{Co-60} is given as

(3) The maximum activity \left(\mathcal{A}_{Co-60}\right)_{max} is calculated by inserting \left(t_{max}\right)_{Co-60} determined in (12.192) into (12.191) to get

It is interesting to note that, for our specific example, the saturation activity \left(\mathcal{A}_{Co-60}\right)_{sat} is 1226.3 Ci, as determined with the saturation model, while the maximum activity \left(\mathcal{A}_{Co-60}\right)_{max} is 940 Ci, as determined with the depletion model.
Activities \mathcal{A}_{Co-60}(t) of cobalt-60 calculated from (12.191) for the depletion model of neutron activation in steps of 1 year for activation times 0 ≤ t ≤ 30 are listed in Table 12.16.

(4) The activation time t, required for cobalt-60 activity \mathcal{A}_{Co-60}(t) to reach a fraction f of its maximum value (A_{Co-60})_{max}, is determined by inserting f \left(\mathcal{A}_{Co-60}\right)_{max} = \left(940\ Ci\right)\times f into (12.191) and solving for activation time t. The result is a transcendental equation of the form

\mathcal{A}_{\mathrm{Co}-60}(t)=(1372.19 \mathrm{Ci}) \times\left[e^{-\left(0.014 \mathrm{a}^{-1}\right) t}-e^{-\left(0.132 \mathrm{a}^{-1}\right) t}\right]=(940 \mathrm{Ci}) \times f           (12.194)

or

e^{-\left(0.014 \mathrm{a}^{-1}\right) t}=e^{-\left(0.132 \mathrm{a}^{-1}\right) t}+0.685 f          (12.195)

The transcendental equation (12.195) can be solved graphically for activation time t by plotting on the same graph the left side function y_1(t)=e^{-\left(0.014 \mathrm{a}^{-1}\right) t} of (12.195) separately from the right side function y_2(t)=e^{-\left(0.132 \mathrm{a}^{-1}\right) t}+0.685 f of (12.195) and determining the intercept of the two functions as the solution of (12.195).
A plot of the two functions y_1\ and\ y_2 is shown in Fig. 12.7 for five values of fraction f : 0.25, 0.50, 0.75, 0.90, and 1.0. For a given f the intercept between functions y_1\ and\ y_2 provides the solution to the transcendental equation (12.195). For all f (except for f = 1) the two functions y_1\ and\ y_2 exhibit two intercepts, one \left(t_{f-asc}\right) for the ascending portion of the activity \mathcal{A}_{Co-60}(t) curve and another \left(t_{f-denc}\right) for the descending portion of the \mathcal{A}_{Co-60}(t) curve. For f = 1 there is only one intercept between y_1\ and\ y_2 and it occurs at the maximum of the \mathcal{A}_{Co-60}(t) curve at the characteristic activation time \left(t_{max}\right)_{Co-60}, given in (12.192).
Activation times tf required for the cobalt sample to reach a given fraction f of the maximum activity \left(\mathcal{A}_{Co-60}\right)_{max} are given in Table 12.17 for 5 selected values of f (0.25, 0.50, 0.75, 0.90, and 1.0). In practice, for obvious reasons, one would only be interested in activation time t_{f−asc} to reach a desired activity fraction f .

(c) Radioactivation yield Y_{\mathrm{Co}-60}=\mathrm{d} \mathcal{A}_{\mathrm{Co}-60} /\left.\mathrm{d} t\right|_{t=0} is calculated as follows:

(1) For the saturation model (12.186) gives the following expression for \mathcal{A}_{Co-60}(t).

and the radioactivation yield Y_{\mathrm{Co}-60}=\mathrm{d} \mathcal{A}_{\mathrm{Co}-60} /\left.\mathrm{d} t\right|_{t=0} is then determined as follows

(2) For the depletion model (12.101) gives the following expression for \mathcal{A}_{Co-60}(t)

\mathcal{A}_{\mathrm{Co}-60}(t)=\left(5.077 \times 10^{13} \mathrm{~Bq}\right) \times\left[e^{-\left(0.444 \times 10^{-9} \mathrm{~s}^{-1}\right) \times t}-e^{-\left(4.179 \times 10^{-9} \mathrm{~s}^{-1}\right) \times t}\right]            (12.198)

and the radioactivation yield Y_{\mathrm{Co}-60}=\mathrm{d} \mathcal{A}_{\mathrm{Co}-60} /\left.\mathrm{d} t\right|_{t=0} is then determined as follows

As expected, both activation models give identical results for the radioactivation yield Y_{\mathrm{Co}-60} \text { as a result of identical initial }(t=0) \text { slopes of } \mathcal{A}_{\mathrm{Co}-60}(t) curves plotted against activation time t.

(d) Figure 12.8 plots the cobalt-60 activity \mathcal{A}_{Co-60}(t) against activation time t for our example of neutron activation of a 10 g sample of cobalt-59 into cobalt-60 in a nuclear reactor with thermal neutron fluence \dot{\varphi} \text { of } 1.2 \times 10^{13} \mathrm{~cm}^{-2} \cdot \mathrm{s}^{-1} \cdot \mathcal{A}_{\mathrm{Co}-60}(t) is shown with dashed curve for the saturation model (12.186) and with solid curve for the depletion model (12.191). Both curves exhibit the same initial slope at t = 0 (i.e., same radioactivation yield Y_{\mathrm{Co}-60}=1.9 \times 10^5 \mathrm{~s}^{-2}); however, with increasing activation time t the curves diverge and the saturation model curve increases exponentially and saturates at \left(\mathrm{A}_{\mathrm{Co}-60}\right)_{\mathrm{sat}}=1226.3 \mathrm{Ci} for t → ∞, while the depletion model curve attains a broad maximum \left(\mathcal{A}_{Co-60}\right)_{max} = 940 Ci at \left(t_{max}\right)_{Co-60} = 19.03 a and then gradually drops to 0 as t → ∞.