Question 2.13: Equal volumes of graphite and iron are mixed together. Fifte......
Equal volumes of graphite and iron are mixed together. Fifteen percent of the volume of the resulting mixture is occupied by air pockets. Find the total macroscopic cross section given the following data σ_C = 4.75 b , σ_{Fe} = 10.9 b , ρ_c = 1.6 gm/cm³ , ρ_{Fe} = 7.7 gm/cm³ . Is it reasonable to neglect the cross section of air? Why?
The density of air is too small to contribute significantly to the cross section. Therefore
\frac{V_{F e}+V_{c}}{V}=0.85~~~\mathrm{but}~V_{F e}=V_{c}~~\mathrm{Thus}~~\frac{V_{F e}}{V}=\frac{V_{c}}{V}=0.425
\Sigma\;=\frac{V_{F e}}{V}\frac{\rho_{F e}N_{o}}{A_{F e}}\sigma_{F e}+\frac{V_{C}}{V}\frac{\rho_{C}N_{o}}{A_{C}}\sigma_{C}
\Sigma=.425\frac{7.87^*0.6023\cdot10^{24}}{55.85}\cdot10.9\cdot10^{-24}+.425\frac{1.6^{*}0.6023\cdot10^{24}}{12.01}\cdot4.75\cdot10^{-24}=.555~c m^{-1}