Calculating Cross Section from Neutron Absorption in a Gold Foil

[Dovahkiin]

Homework Statement

In an experiment carried out with a beam of thermal neutrons it is found that on traversing a 2mm thick foil of 197Au, some 70% of the neutrons are removed. What is the total thermal neutron cross-section for this isotope of gold? (Density of gold: 19300 kg m-3)

Homework Equations

1. N = N0 exp(-xnσ) - x is thickness, n is number of nuclei per unit volume, σ is cross section
2. moles = mass/A
3. n0 = Na x moles - Na is avagadros number, n0 is number of nuclei

The Attempt at a Solution

I have attempted this question and appear to have used the same method as in the solution given by the professor, but i get a different answer... Am i right?

Firstly, to find n i substitute equation 2 into 3, giving n0 = (Na x m)/A

Then i divide this by the volume: n = n0/V = (Na x m)/(A x V) = (Na/A) x ρ (ρ = density)

Next i rearrange eqn 1: N/N0 = (70%)/(100%) = 0.7 = exp(-xnσ)

This gives: σ = (-ln(0.7))/(xn)

Now I am not sure if i should be using m in grams or kilograms?

My answer:
σ = 3.02 x 10^(-30) m^2 = 0.0302 barns (using m in grams)
σ = 3.02 x 10^(-27) m^2 = 30.2 barns (using m in kg)

Neither of these answers matches the answer given, 102 barns...
Even though my method is identical to the professors method (although he gives no intermediate results, so i don't know if/where i have gone wrong...)

Any help is much appreciated.

 

[Dovahkiin]

I don't seem to be able to edit my OP anymore, but I've realized i haven't fully defined all the variables (units in brackets):

. N is the number of particles remaining after the foil
. N0 is the starting number of particles
. x is thickness (m)
. n is number of nuclei per unit volume (m^-3)
. σ is cross section (m^2)
. m is the mass (grams or kg?)
. A is the mass number
. Na is avagadros number
. n0 is number of nuclei