Question 10.12: Neptunium-238 has a thermal absorption cross section of 33 b......
Neptunium-238 has a thermal absorption cross section of 33 b, which we have neglected in deriving Eq. (10.30). In a reactor operating at a flux level of \phi = 5 × 10^{14} \mathrm{n/cm²/s} what fraction of the neptunium will capture a neutron instead of decaying to plutonium-239?
From Eq. (10.29) t_{1/2}^{39}=2.36\,\mathrm{d}\cdot(3600\cdot24\,\mathrm{s/d})=204\cdot10^{3}\,\mathrm{s} The capture rate in neptunium is \sigma_{a}^{39}\phi N^{39}(t) while the decay rate is \lambda^{39}N^{39}(t) hus the fraction captured is
{\frac{\sigma_{a}^{39}\phi N^{39}(t)}{\sigma_{a}^{39}\phi N^{39}(t)+\lambda^{39}N^{39}(t)}}={\frac{1}{1+\lambda^{39}\ /(\sigma_{a}^{39}\phi)}}
\lambda^{39}\,/(\sigma_{a}^{39}\phi)=0.693\,/(t_{1/2}^{39}\sigma_{a}^{39}\phi)=\frac{0.693}{204\cdot10^{3}\cdot33\cdot10^{-24}\cdot5\cdot10^{14}}=206
{\frac{1}{1+\lambda^{39}/(\sigma_{a}^{39}\phi)}}={\frac{1}{1+206}}=0.00483=0.483 \%