- #1
Muthumanimaran
- 81
- 2
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