Finding the scattering cross section

[Muthumanimaran]

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? Comment on the result of the cross-section measurement in the light of the fact that the radius of a gold nucleus is
6.5 x 10-15 m. (Density of gold: 19300 kg m-3)
Question is from this link #chap1http://physics-database.group.shef.ac.uk/phy303/303prob1.html#chap1

Homework Equations

$$\frac{R_s}{R_i}=\frac{N_{A}L{\rho}{\sigma}}{A{\times}10^{-3}}$$
$\frac{R_s}{R_i}$ is fraction scattered
A is mass number
L is thickness
$\rho$ is density

The Attempt at a Solution

I simply substituted the values in the above expression
$$0.7=\frac{6.02{\times}10^{23}{\times}(2{\times}10^{-3}){\times}(19300){\times}{\sigma}}{197{\times}10^{-3}}$$

And finally got $ \sigma = 59.32{\times}10^{-28} $ or 59 Barns but the actual answer turns out to be 102 Barns. where did made the mistake, I re did the problem but getting same answer for cross section

 

[Charles Link]

I also got your same answer on the first attempt, but then made the observation that the formula you used for the fraction absorbed only holds accurately when the fraction absorbed is small. The problem involves an exponential drop in intensity of the beam as it traverses the 2mm foil: $I(x)=I_o exp^{-n \sigma x}$ with $\frac{I(L)}{I_o}=.3$.(70% absorption means 30% survived). Solving this gives $n \sigma L=ln(10/3)=1.20$ where $n=(\rho N_A)/(A.W.)$.(The 1.20 basically replaces the .7 in the formula that you used to solve it in the way that you did. 1.20/.7=102/59) $\\$ Editing: Notice in this second method we used $-ln(1-.7)=1.2$ . Now $ln(1-x)=-x$ approximately for x<<1, but .7 was too large to give the result that $-ln(1-.7)=.7$, thereby the simplified formula of the first method gave an inaccurate result.

 

[Charles Link]

In the previous problem, it should be recognized that the intensity $I$ satisfies the differential equation $\frac{dI(x)}{dx}=-n \sigma I(x)$. What is happening in this case is scatterers/blockers of area $\sigma$ can each block the beam, but if $A$ is the area of the target with thickness $L$, the total scattering cross section $\sigma_{total}$ can not get any larger than the area $A$. For the first method of solving the problem, the total scattering cross section is computed to be $\sigma_{total}=n \sigma V =n \sigma A L$. This calculation does not account for possible overlap in the areas that are getting blocked, and ultimately the computed $\sigma_{total}$ will become larger than $A$ by this method as $L$ becomes large.