Direct Coupling of Energy Groups, Multigroup Neutron Diffusion

[NuclearVision]

This one comes from Duderstadt and Hamilton, Problem 7-3.

In multi-group diffusion theory What percentage of neutrons slowing down in hydrogen will tend to skip energy groups if the group structure is chosen such that $\frac{E_{g-1}}{E_{g}}$=100= 1/$\alpha_{approx}$.

I know that the probability of a neutron scattering to a lower energy in hydrogen is uniformly distributed as: $\frac{1}{E_{i}}$ (because $\alpha$ = 0 in this case).

My approach was to integrate the probability distribution from 0 to the bottom of the (approximated) energy group which should give me the probability that the final energy is less than $\alpha E_{i}$:

$\int^{E_{i} \alpha_{approx}}_{0} \frac{1}{E_{i}} dE_{f}$ which lead to a value of $\alpha_{approx}=\frac{1}{100}=1$%.

Am I doing this right?


It is worth noting that for neutron moderators with A> 1 (anything heavier than hydrogen) the s-wave downscattering PDF is given by:
$P(E_{i}\rightarrow E_{f})=\frac{1}{(1-\alpha)*E_{i}}$

where:
$\alpha=(\frac{A-1}{A+1})^{2}$

and $\alpha E_{i}$ is the minimum energy a neutron can scatter to from $E_{i}$ in a collision with a nucleus of mass A.

from this it is clear that if A=1 (as in hydrogen) α = 0 and thus the distribution goes as 1/E.


Thanks!

 

[Valkarie]

Yes, your approach is correct. Since the group structure is such that \frac{E_{g-1}}{E_{g}} = 1/\alpha_{approx}, then the probability that a neutron will skip energy groups is equal to \alpha_{approx} = 1/100 = 1%.