Question 12.6.Q4: In discussions of neutron activation one usually assumes tha......

For the saturation and depletion models of neutron activation the general variables x, y_P,\ y_D as well as the activation factor m are defined as follows

x=m \frac{t}{\left(t_{1 / 2}\right)_{\mathrm{D}}} ;             (12.92)

y_{\mathrm{P}}=\frac{N_{\mathrm{P}}(t)}{N_{\mathrm{P}}(0)}           (12.93)

y_{\mathrm{D}}=\frac{N_{\mathrm{D}}(t)}{m N_{\mathrm{P}}(0)}=\frac{\mathcal{A}_{\mathrm{D}}(t)}{\sigma_{\mathrm{P}} \dot{\varphi} N_{\mathrm{P}}(0)} ;           (12.94)

m=\frac{\sigma_{\mathrm{P}} \dot{\varphi}}{\lambda_{\mathrm{D}}} .           (12.95)

In the “parent depletion–daughter activation model” we use the following additional parameters:\lambda_{\mathrm{D}}^*, \varepsilon^* \text {, and } m^* \text { as well as variable } y_{\mathrm{D}}^*(x). The additional parameters are defined as follows

\lambda_{\mathrm{D}}^*=\lambda_{\mathrm{D}}+\sigma_{\mathrm{D}} \dot{\varphi} ;         (12.96)

\varepsilon^*=\frac{\lambda_{\mathrm{D}}^*}{\lambda_{\mathrm{D}}}=1+\frac{\sigma_{\mathrm{D}} \dot{\varphi}}{\lambda_{\mathrm{D}}} ;             (12.97)

m^*=\frac{\sigma_{\mathrm{P}} \dot{\varphi}}{\lambda_{\mathrm{D}}^*}=\frac{m}{\varepsilon^*} ;            (12.98)

y_{\mathrm{D}}^*=\frac{\mathcal{A}_{\mathrm{D}}^*(t)}{\sigma_{\mathrm{P}} \dot{\varphi} N_{\mathrm{P}}(0)} .                (12.99)

(a) The standard form for the number of parent nuclei N_P(t), undergoing nuclear activation in particle fluence rate \dot{φ}, is expressed by the following equation

N_{\mathrm{P}}(t)=N_{\mathrm{P}}(0) e^{-\sigma _{ \mathrm{P} }\dot{\varphi} t},         (12.100)

that, after incorporating (12.92), (12.93), and (12.95), takes up the following form giving the number of parent nuclei N_P(t) normalized to the initial number of parent nuclei N_P(0) as

y_{\mathrm{P}}=\frac{N_{\mathrm{P}}(t)}{N_{\mathrm{P}}(0)}=e^{-\sigma_{\mathrm{P}} \dot{\varphi} t}=e^{-m \lambda_{\mathrm{D}} t}=e^{-m \frac{t}{\left(t_{1 / 2}\right) \mathrm{D}} \ln 2}=e^{-x \ln 2}=\frac{1}{2^x} \equiv 2^{-x}       (12.101)

Equation (12.101) is valid in general irrespective of the activation model used, and is thus valid for the saturation, depletion, as well as depletion–activation model.

(b) The standard equation used for describing the depletion–activation model is in Prob. 251 [see (12.51)] expressed as (T12.47)

N_D(t) = N_P(0)  \frac{\sigma_P \dot φ}{\lambda^*_D  –  \sigma_P \dot φ} [e^{-\sigma_P \dot φ t}  –  e^{-\lambda^*_D t}]          (12.51)

\frac{\mathrm{d} N_{\mathrm{D}}(t)}{\mathrm{d} t}=\sigma_{\mathrm{P}} \dot{\varphi} N_{\mathrm{P}}(t)-\lambda_{\mathrm{D}} N_{\mathrm{D}}(t)-\sigma_{\mathrm{D}} \dot{\varphi} N_{\mathrm{D}}(t)=\sigma_{\mathrm{P}} \dot{\varphi} N_{\mathrm{P}}(t)-\lambda_{\mathrm{D}}^* N_{\mathrm{D}}(t),          (12.102)

resulting in the following expression [Prob. 251, see (12.52)] for the daughter activity \mathcal{A}_{\mathrm{D}}^*(t).

\mathcal{A}_{\mathrm{D}} (t) = N_P(0) \frac{\lambda_D  \sigma_P \dot φ}{\lambda^*_D  –  \sigma_P \dot φ}[e^{-\sigma_P \dot φ t}  –  e^{-\lambda^*_D t} ]=  \sigma_{\mathrm{P}} \dot{\varphi} N_{\mathrm{P}}(0) \frac{\frac{\lambda_{\mathrm{D}}}{\lambda_{\mathrm{D}}^*}}{1-\frac{\sigma_{\mathrm{P}} \dot{\varphi}}{\lambda_{\mathrm{D}}^*}}\left[e^{-\sigma_{\mathrm{P}} \dot{\varphi} t}-e^{-\lambda_{\mathrm{D}}^* t}\right]          (12.52)

\mathcal{A}_{\mathrm{D}}^*(t)=\lambda_{\mathrm{D}} N_{\mathrm{P}}(t)=\sigma_{\mathrm{P}} \dot{\varphi} N_{\mathrm{P}}(0) \frac{\frac{\lambda_{\mathrm{D}}}{\lambda_{\mathrm{D}}^*}}{1-\frac{\sigma_{\mathrm{P}} \dot{\varphi}}{\lambda_{\mathrm{D}}^*}}\left[e^{-\sigma_{\mathrm{P}} \dot{\varphi} t}-e^{-\lambda_{\mathrm{D}}^* t}\right]          (12.103)

where \lambda_{\mathrm{D}}^*=\lambda_{\mathrm{D}}+\sigma_{\mathrm{D}} \dot{\varphi} is the modified decay constant for the daughter accounting for the radioactive daughter decay (through λ_D) as well as the daughter activation \text { (through } \sigma_{\mathrm{D}} \text { and } \dot{\varphi} \text { ). }
Combining (12.103) with (12.96), (12.97), (12.98), and (12.99) as well as recalling that \sigma_{\mathrm{P}} \dot{\varphi} t=x \ln 2 \text { and } \lambda_{\mathrm{D}}^* t=\left(x / m^*\right) \ln 2 we obtain the following expression (T12.57) for the normalized daughter activity y_{\mathrm{D}}^*.

y_{\mathrm{D}}^*(x)=\frac{\mathcal{A}_{\mathrm{D}}(t)}{\sigma_{\mathrm{P}} \dot{\varphi} N_{\mathrm{P}}(0)}=\frac{1}{\varepsilon^*\left(1-m^*\right)}\left[e^{-x \ln 2}-e^{-\frac{x}{m^*} \ln 2}\right]=\frac{1}{\varepsilon^*\left(1-m^*\right)}\left[\frac{1}{2^x}-\frac{1}{2^{\frac{x}{m^*}}}\right]          (12.104)

(c) Normalized daughter activity for the depletion model was determined in (12.64) of Prob. 252 as

(1) A comparison of y^∗ _D given by (12.104) for the depletion–activation model with y_D(x) given by (12.105) for the depletion model shows that the two expressions are similar except for the factor ε^∗ which is equal or larger than 1 (ε^∗ ≥ 1) and depends on fluence rate \dot{φ}. A closer look at expressions (12.104) and (12.105) shows that (12.105) of the depletion model actually follows from (12.104) of the depletion– activation model, since, for the depletion model, λ^∗ _D = λ_D\ and\ ε^∗ = 1 as a result of σ_D = 0. We conclude that the depletion model is a special case of the depletion– activation model when the activation cross section σ_D of the daughter nucleus is zero. Thus, we expect the behavior of y^∗ _D(x) of the depletion–activation model to be similar to that of y_D(x) of the depletion model: rise from 0 at x = 0 to reach a maximum \left(y^∗ _D\right)_{max} at x = \left(x^∗ _D\right)_{max}, and then asymptotically decrease to 0 as x → ∞.

(2) The characteristic normalized time \left(x_{\mathrm{D}}^*\right)_{\max } \text { at which } y_{\mathrm{D}}^* exhibits its maximum \left(y_{\mathrm{D}}^*\right)_{\max } can be determined as a function of the activation factor m^* by setting \mathrm{d} y_{\mathrm{D}}^* /\left.\mathrm{d} x\right|_{x=\left(x_{\mathrm{D}}^*\right)_{\max }} \text { and solving for }\left(x_{\mathrm{D}}^*\right)_{\max } as follows

Solving (12.106) for (x^∗_D)_{max} we now get

2^{-(x^*_D)_{max}} =  \frac{1}{m^*}2^{-(x^*_D)_{max} / m^*}           or           -(x^*_D)_{max}  \ln  2 = \ln  \frac{1}{m^*}  –  \frac{(x^*_D)_{max}}{m^*}  \ln  2             (12.107)

and finally the following expression for \left(x^∗ _D\right)_{max}

\left(x_{\mathrm{D}}^*\right)_{\max }=\frac{m^*}{\left(m^*-1\right)} \frac{\ln m^*}{\ln 2} .             (12.108)

For m^*=1 (12.108) is not defined; however, since it gives \left(x_{\mathrm{D}}^*\right)_{\max }=0 / 0, we can apply the l’Hôpital rule to get \left.\left(x_{\mathrm{D}}^*\right)_{\max }\right|_{m \rightarrow 1} as follows

\left.\left(x_{\mathrm{D}}^*\right)_{\max }\right|_{m^* \rightarrow 1}=\lim _{m^* \rightarrow 1} \frac{\frac{\mathrm{d}\left(m^* \ln m^*\right)}{\mathrm{d} m^*}}{\frac{\mathrm{d}\left(m^*-1\right)}{\mathrm{d} m^*} \ln 2}=\lim _{m^* \rightarrow 1} \frac{1+\ln m^*}{\ln 2}=\frac{1}{\ln 2}=1.443 .            (12.109)

Thus, \left(x_{\mathrm{D}}^*\right)_{\max } is given by (12.108) for any positive m^* \text { except for } m^*=1 for which \left(x_{\mathrm{D}}^*\right)_{\max }=1.443, as determined in (12.109).

(3) The maximum daughter activity \left(y_{\mathrm{D}}^*\right)_{\max } can be determined by inserting \left(x_{\mathrm{D}}^*\right)_{\max } \text { of }(12.108) \text { into } y_{\mathrm{D}}^*(x) given by (12.104) as follows

Components F_1^* \text { and } F_2^* \text { of }(12.110) \text {, after insertion of }\left(x_{\mathrm{D}}^*\right)_{\max } given by (12.108) and using the following identity e^{z \ln 2}=e^{\ln 2^z}=2^z, yield the following expressions

F_1^*=\frac{1}{2^{\left(x_{\mathrm{D}}^*\right)_{\max} }}=\frac{1}{2^{\frac{m^*}{\left(m^*-1\right)} \frac{\ln m^*}{\ln  2} }}= 2^{\frac{m^*}{(1  –  m^*)}\frac{\ln  m^*}{\ln 2}} =e^{\ln 2^{\frac{m^*}{\left(1-m^*\right)} \frac{\ln m^*}{\ln 2}}}=e^{\frac{m^*}{1-m^*}} \ln m^*         (12.111)

and

F_2^*=\frac{1}{2^{\left(x_{\mathrm{D}}^*\right)_{\max }\left[\frac{1-m^*}{m^*}\right]}}=\frac{1}{2^{\frac{m^*}{\left(m^*-1\right)} \frac{\ln m^*}{\ln 2}\left[\frac{1-m^*}{m^*}\right]}}=2^{\frac{\ln m^*}{\ln 2}}=e^{\ln m^*}=m^* .             (12.112)

The maximum normalized daughter activity \left(y^∗ _D\right)_{max} of (12.109) after incorporating F_1^* \text { of (12.111) and } F_2^* of (12.112) can now be written as follows