# I Critical radius for nuclear reactions

1. Mar 17, 2019 at 8:35 AM

### Decimal

Hello!

I am taking a course in nuclear physics using the book An Intro to the Physics of Nuclei & Particles by Dunlap. I am a little confused by an explanation related to the critical radius for nuclear reactions.

The author first defines a value $vq$ as the average number of neutrons produced by a free neutron in a sample of U-235. Here $v$ is the number of neutrons produced in a fission reaction and $q$ is a factor < 1 to account for the neutron loss in the sample. To obtain a controlled reaction in which the number of neutrons remains constant $vq$ should equal $1$. This I understand. The book derives that for a typical value $v = 2.5$ one should aim for $q = 0.4$.

My confusion starts when the author explains how one might obtain such a q value. The author derives the average path length before a neutron fission reaction occurs to be equal to $0.079$ m. Then he states that a sample sphere with this radius will be just the right size to keep the reaction controlled ($vq = 1$), also called the critical radius. However in my understanding this means that every neutron produced at the center will undergo on average 1 fission reaction, and thus $q \approx 1$. Then the reaction would obviously be unstable.

Thus I would think the critical radius would be some fraction of $0.079$ m, related to the q value one would aim for. However this is apparently not the case, so what am I missing? I feel like I am misunderstanding the probabilities involved, but some pointers would be greatly appreciated.

Thanks!

2. Mar 17, 2019 at 10:33 AM

### Staff: Mentor

That would mean no neutron can escape, that is not realistic. In addition neutrons are not only produced at the center. Fission processes elsewhere will on average have more neutrons escaping.

You'll need integration to get the average number of fission reactions from a fission reaction.

3. Mar 18, 2019 at 10:00 AM

### Decimal

Right, that makes a lot of sense! Thank you!