Question 7.1: Show that the most probable energy of a neutron emitted from......
Show that the most probable energy of a neutron emitted from the probability distribution shown in Figure 7.7 is 0.74 MeV
Since the neutron energy spectrum χ(E) = Ae^{-BE} sinh\left\{{\sqrt{CE}}\right\} describes the probability that a neutron will be emitted with an energy E, the most probable neutron energy will correspond to the peak of the curve shown in Figure 7.7. From college calculus, we know that this must correspond to the point on the curve where the slope is zero or where dχ(E)/dE = 0. Therefore, to find the energy that corresponds to the maximum of the curve, we must take the first derivative of χ(E) = Ae^{-BE} sinh\left\{{\sqrt{CE}}\right\} , set it to zero, and solve for E. Performing the required math, we find that
Multiplying through by e^{\sqrt{CE}}, we obtain (\sqrt{C/2B – 1})\left\{e^{2\sqrt{CE}} – 1\right\} = 0 or e^{2\sqrt{CE}} = 1/(\sqrt{C/2B – 1}) Finally, taking the natural log (ln) of both sides and solving for E, we obtain E = (1/2\sqrt{C})ln(1/(\sqrt{C/2B-1})) Upon substituting the appropriate values for B and C into this expression, we find that E = 0.74 MeV.