What Determines the Critical Radius for Fission in Nuclear Physics?

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Homework Statement


See the image below:
qncs.png


Homework Equations


In the previous part, it was proved that a particle moves through a distance x in a material without interacting with probability P given by

P=\exp(-\sigma n x)

Here σ represents the cross section of the reaction between the incoming particle and the material. n represents the target (material) particles per unit volume.

The Attempt at a Solution


The only thing I have figured out so far is that we need the probability of the neutron not escaping without a reaction, so there is a factor of

1- exp(- (\sigma_f + \sigma_c) nr )

in the equation.

I am not sure of where the fractional factor at the start of the equation comes from, or why the total cross section appears in the exponent.
 
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Bump! Does no one have any ideas, I haven't been able to solve this since the last post.
 
The sum in the exponent is simple: it is because cross-sections of different reactions are additive.

Thus (1 - exp) is the probability that the neutron will be captured or cause fission. You need to express from this the probability of fission.
 
Simple explanation! Looks like I wasn't thinking the right way, thanks.
 
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