Question 12.6.Q2: For better understanding of neutron activation processes we ......

The general variables x, y_P, y_D as well as the activation factor m are for neutron activation defined as follows

x=m \frac{t}{\left(t_{1 / 2}\right)_{\mathrm{D}}}         (12.53)

y_{\mathrm{P}}=\frac{N_{\mathrm{P}}(t)}{N_{\mathrm{P}}(0)} ;          (12.54)

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)} ;          (12.55)

m=\frac{\sigma_{\mathrm{P}} \dot{\varphi}}{\lambda_{\mathrm{D}}}           (12.56)

(a) The standard form for the number of parent nuclei N_P(t), undergoing neutron activation in neutron 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},          (12.57)

which, after incorporating (12.53), (12.54), and (12.56), takes up the following form (T12.27), giving the number of parent nuclei N_P(t) normalized to the initial number of parent nuclei N_P(0).

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} \text {. }           (12.58)

(b) The number of daughter nuclei N_D(t) is proportional to the number of initial parent nuclei N_P(t) but differs for the two neutron activation models, as derived in Prob. 251:

(1) For the saturation model given in (12.43) in Prob. 251 N_D(t)\ and\ \mathcal{A}_D(t) are

N_{\mathrm{D}}(t)=\frac{\sigma_{\mathrm{P}} \dot{\varphi} N_{\mathrm{P}}(0)}{\lambda_{\mathrm{D}}}\left[1-e^{-\lambda_{\mathrm{D}} t}\right]         (12.59)

N_D(t) = \frac{σ_P\dot{φ}N_P(0)}{λ_D}\{1 −e^{−λ_Dt}\}.              (12.43)

and

\mathcal{A}_{\mathrm{D}}(t)=\lambda_{\mathrm{D}} N_{\mathrm{D}}(t)=\sigma_{\mathrm{P}} \dot{\varphi} N_{\mathrm{P}}(0)\left[1-e^{-\lambda_{\mathrm{D}} t}\right]=\mathcal{A}_{\mathrm{sat}}\left[1-e^{-\lambda_{\mathrm{D}} t}\right]          (12.60)

where \mathcal{A}_{sat} is the saturation activity defined as the product σ_P\dot{φ}N_P(0) and attained at time t → ∞.
Combining (12.60) with (12.55) and (12.56) we now write the normalized daughter activity z_D(x) for the saturation model as

(2) For the depletion model given in (12.47) in Prob. 251, N_D(t)\ and\ \mathcal{A}_D(t) are

N_{\mathrm{D}}(t)=\frac{\sigma_{\mathrm{P}} \dot{\varphi} N_{\mathrm{P}}(0)}{\lambda_{\mathrm{D}}-\sigma_{\mathrm{P}} \dot{\varphi}}\left[e^{-\sigma_{\mathrm{P}} \dot{\varphi} t}-e^{-\lambda_{\mathrm{D}} t}\right]            (12.62)

N_D(t) = N_P(0)\frac{σ_{P\dot{φ}}}{λ_D −σ_P\dot{φ}}[e^{−σ_P\dot{φ}t} −e^{−λ_Dt}].     (12.47)

and

\mathcal{A}_{\mathrm{D}}(t)=\lambda_{\mathrm{D}} N_{\mathrm{D}}(t)=\frac{\lambda_{\mathrm{D}} \sigma_{\mathrm{P}} \dot{\varphi} N_{\mathrm{P}}(0)}{\lambda_{\mathrm{D}}-\sigma_{\mathrm{P}} \dot{\varphi}}\left[e^{-\sigma_{\mathrm{P}} \dot{\varphi} t}-e^{-\lambda_{\mathrm{D}} t}\right] .          (12.63)

Combining (12.63) with (12.55) and (12.56) we now write the normalized daughter activity y_P for the depletion model as

(c) Equation (12.64) for normalized daughter activity y_D as a function of normalized time x is valid for all positive values of the activation factor m from 0 to ∞ with the exception of m = 1 for which y_D is not defined. However, since for m = 1 (12.64) gives y_D = 0/0, we can apply the l’Hôpital rule and determine the function that governs y_D at m = 1 as follows (T10.46)

\left.y_{\mathrm{D}}\right|_{m=1}=\lim _{m \rightarrow 1} \frac{\frac{\mathrm{d}}{\mathrm{d} m}\left[\frac{1}{2^x}-\frac{1}{2^{x / m}}\right]}{\frac{\mathrm{d}}{\mathrm{d} m}(1-m)}=\lim _{m \rightarrow 1} \frac{-2^{-\frac{x}{m}}(\ln 2) \frac{x}{m^2}}{-1}=(\ln 2) \frac{x}{2^x} \text {. }          (12.65)

(d) Equation (12.64) of (b) gives the normalized daughter activity y_D [see (12.55)] as a function of normalized activation time x [see (12.53)] for 0 ≤ m ≤ ∞ with the exception of m = 1 for which y_D is simplified and given by (12.65) in (c). Normalized daughter activities y_D of (12.64) and (12.65) are equal to zero for x = 0 (initial condition) and x = ∞ (when all nuclei of parent P and daughter D have decayed). This suggests that y_D (in conjunction with \mathcal{A}_D) passes through a maximum at a specified characteristic normalized time \left(x_D\right)_{max} between 0 and ∞ for all m except for m = 1. The characteristic time \left(x_D\right)_{max} can be determined as a function of activation factor m by setting \mathrm{d} y_{\mathrm{D}} /\left.\mathrm{d} x\right|_{x=\left(x_{\mathrm{D}}\right)_{\max }}=0 \text { and solving for }\left(x_{\mathrm{D}}\right)_{\max } as follows

Solving (12.66) for \left(x_D\right)_{max} we now get

2^{-\left(x_{\mathrm{D}}\right)_{\max }}=\frac{1}{m} 2^{-\left(x_{\mathrm{D}}\right)_{\max } / m} \quad \text { or } \quad-\left(x_{\mathrm{D}}\right)_{\max } \ln 2=\ln \frac{1}{m}-\frac{\left(x_{\mathrm{D}}\right)_{\max }}{m} \ln 2           (12.67)

and finally

\left(x_{\mathrm{D}}\right)_{\max }=\frac{m}{m-1} \frac{\ln m}{\ln 2}          (12.68)

For m = 1, (12.68) is not defined, however, since it gives \left(x_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}(m \ln m)}{\mathrm{d} m}}{\frac{\mathrm{d}(m-1)}{\mathrm{d} m} \ln 2}=\lim _{m \rightarrow 1} \frac{1+\ln m}{\ln 2}=\frac{1}{\ln 2}=1.4427 .           (12.69)

Thus, \left(x_D\right)_{max} is calculated from (12.68) for any positive m except for m = 1 for which \left(x_D\right)_{max} = 1.44, as determined in (12.69).

(e) The maximum daughter activity \left(y_D\right)_{max} can be determined by inserting \left(x_D\right)_{max} of (12.68) into y_D given by (12.64) as follows

Components F_1\ and\ F_2 of (12.70), after insertion of \left(x_D\right)_{max} given by (12.68) 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}{(m-1)} \frac{\ln m}{\ln 2}}}=2^{\frac{m}{(1-m)} \frac{\ln m}{\ln 2}}=e^{\ln 2^{\frac{m}{(1-m)} \frac{\ln m}{\ln 2}}}=e^{\frac{m}{1-m} \ln m}          (12.71)

and

F_2=\frac{1}{2^{\left(x_{\mathrm{D}}\right) \max \left[\frac{1-m}{m}\right]}}=\frac{1}{2^{\frac{m}{(m-1)} \frac{\ln m}{\ln 2}\left[\frac{1-m}{m}\right]}}=2^{\frac{\ln m}{\ln 2}}=e^{\ln m}=m .            (12.72)

The maximum normalized daughter activity \left(y_D\right)_{max} of (12.70), after incorporating F_1 of (12.71) and F_2 of (12.72), can now be written as follows

\left(y_{\mathrm{D}}\right)_{\max }=\frac{1}{1-m} F_1\left[1-F_2\right]=\frac{1}{1-m} e^{\frac{m}{1-m} \ln m}[1-m]=e^{\frac{m}{1-m} \ln m} .           (12.73)

Since \left(x_D\right)_{max} not only defines the maximum in the normalized daughter activity y_D(x) but also defines the point of ideal equilibrium where the y_D(x) curve crosses over the y_P(x) curve that is given in (12.58), we can state the following relationship

y_{\mathrm{D}}\left[\left(x_{\mathrm{D}}\right)_{\max }\right]=\left(y_{\mathrm{D}}\right)_{\text {max }}=y_{\mathrm{P}}\left[\left(x_{\mathrm{D}}\right)_{\text {max }}\right],          (12.74)

suggesting a much simpler calculation of y_D\left[\left(x_D\right)_{max}\right] than that carried out in (12.70). In (12.58) we saw that y_P(x) is given by a very simple expression as y_P(x) = 2^{−x} = e^{−x\ ln\ 2}, allowing us to determine y_D\left[\left(x_D\right)_{max}\right] directly from (12.74) as follows

Equation (12.73) is valid for all positive m except for m = 1 in which case \left(y_D\right)_{max} can be determined by applying the l’Hôpital rule to (12.73) as follows

\left.\left(y_{\mathrm{D}}\right)_{\max }\right|_{m=1}=\lim _{m \rightarrow 1} \exp \frac{\frac{\mathrm{d}(m \ln m)}{\mathrm{d} m}}{\frac{\mathrm{d}(1-m)}{\mathrm{d} m}}=\lim _{m \rightarrow 1} \exp \frac{\ln m+1}{-1}=e^{-1}=\frac{1}{e}=0.368 .           (12.76)

(f) It is easy to show that the relationship for positive m but m ≠ 1 between \left(y_D\right)_{max} given by (12.73) and \left(x_D\right)_{max} given by (12.68) is a simple exponential expression

\left(y_{\mathrm{D}}\right)_{\max }=e^{\frac{m}{1-m} \ln m}=e^{-\frac{m}{m-1} \ln m}=e^{-(\ln 2)\left(x_{\mathrm{D}}\right)_{\max }}=2^{-\left(x_{\mathrm{D}}\right)_{\max }}=\frac{1}{2^{\left(x_{\mathrm{D}}\right)_{\max }}},            (12.77)

while for m=1,\left(y_{\mathrm{D}}\right)_{\max }=1 / e=0.368 \text {, as shown in }(12.76) \text {, and }\left(x_{\mathrm{D}}\right)_{\max }= 1 / \ln 2=1.4427, as shown in (12.69).