Question 10.1: Prove that for a reactor operating at a very high flux level......

The Xenon concentration following shutdown is given by Eq. (10.19):

X(t)=\overline{{{\Sigma}}}_{f}\phi\!\!\left[\frac{(\gamma_{I}+\gamma_{X})}{\lambda_{X}+\sigma_{a X}\phi}e^{-\lambda_{X}t}+\frac{\gamma_{I}}{\lambda_{I}-\lambda_{X}}\!\left(e^{-\lambda_{X}t}-e^{-\lambda_{I}t}\right)\right]

To find the maximum we set its derivative equal to zero:

For a very large value of the flux, the \phi in the denominator causes the first term to vanish. Hence

\left(-\lambda_{X}e^{-\lambda_{X}t}+\lambda_{I}e^{-\lambda_{I}t}\right)=0

or

t={\frac{1}{\lambda_{I}-\lambda_{X}}}\ln(\lambda_{I}/\lambda_{X})

From Eq. (10.11):

\lambda_{I}=0.693\,/\,t_{1/2I}=0.693\,/\,6.7=0.1034\,\mathrm{hr^{-1}}

\lambda_{X}=0.693\,/\,t_{1/2X}=0.693\,/\,9.2=0.0753\,\mathrm{hr^{-1}}

t={\frac{1}{0.1034-0.0753}}\ln(0.1034/0.0753)=11.3\,\mathrm{hr}