Finding neutrino mean free path using cross section for interaction

[PhyStan7]

Homework Statement

A supernova can produce a neutron star with typical radius 10km. Assume the neutron star matter consists of iron nuclei (A=56), produced by the electron capture reaction:

$e^{-}$+Co$\rightarrow$Fe+$\nu_{e}$

The matter density is $\rho$=10$^{5}$ tonne mm$^{-3}$ and the neutrino cross section for interaction with this material is $\sigma$=3$\times$10$^{-46}$m$^{2}$.



Find the neutrino mean free path

Homework Equations

w$_{t}$=JσN (1)
w$_{t}$=Iσρ$_{T}$t$\frac{N_{A}}{A}$ (2)
I=JS (3)

Where w$_{t}$ is the rate of reaction
J is incoming flux
I is intensity of incoming beam
S is surface area of incoming beam
σ is the cross section of reaction
N$_{A}$ is Avogadro's number
A is number of protons in target material
ρ$_{T}$ is the density of the target material
t is the length the beam passes through

The Attempt at a Solution

Ok so in my course on nuclear and particle physics there are often questions like this but I am not sure how to approach them.

I figured that the target is essentially the neutron star, so t=20km (length through the star). I can see that I have most of the things required in equation (2), except w$_{t}$ and I. I figured that as w$_{t}$ is the rate of reaction it is essentially the number of reactions divided by the time. As we are only concerned with the reaction of 1 neutron in the star, I thought that in this case the time would be the mean free time (i.e the time before the neutrino reacts) which would be equal to the distance travelled/c (speed of neutrino). The distance traveled in this case would be the mean free path λ. This means that we can rewrite w$_{t}$ as:

w$_{t}$=c/λ


I come up with a problem though. As no information about the incoming beam is given (its flux J, intensity I or area S) I have no idea how the intensity I can be calculated so that (2) can be used.

I can see that, with the density and volume of the star, potentially the number of atoms in the target could be computed. However I cannot see how this would be useful to compute the quantity that I want.

The question is only 5 marks so I am guessing it is fairly simple but I can't see it. As I said, for my exam there are usually several similar questions concerning the cross section of interaction so any help would be much appreciated. The answer is λ=3km, I just want to find how it is done!

Thanks

 

[mark726]

for your question! To solve this problem, we can use equation (1) to find the rate of reaction (w_{t}), which is equal to the number of reactions per unit time. This can be rewritten as:

w_{t}=\frac{JσN}{S}

where J is the incoming flux, σ is the cross section of the reaction, N is the number of target particles (in this case, iron nuclei), and S is the surface area of the incoming beam. We can then use equation (3) to relate the intensity (I) of the incoming beam to the flux (J) and surface area (S):

I=JS

Substituting this into our expression for w_{t}, we get:

w_{t}=\frac{IσN}{S}

We can then use equation (2) to relate w_{t} to the mean free path (λ):

w_{t}=\frac{Iσρ_{T}t\frac{N_{A}}{A}}{λ}

Rearranging, we get:

λ=\frac{Iσρ_{T}t\frac{N_{A}}{A}}{w_{t}}

Substituting in the values given in the problem, we get:

λ=\frac{(3\times10^{-46}m^{2})(10^{5}tonne mm^{-3})(20km)(6.022\times10^{23}mol^{-1})(56)}{1/s}

Simplifying, we get:

λ=3km

Therefore, the neutrino mean free path is 3km.