answers are 0.504/cm, 0.099, 4.9% 4.16 The total cross section for uranium dioxi

Question

answers are 0.504/cm, 0.099, 4.9%

4.16 The total cross section for uranium dioxide of density 10 g/cm is to be measured by a transmission method. To avoid multiple neutron scattering, which would introduce error into the results, the sample thickness is chosen to be much smaller than the mean free path of neutrons in the material. Using approximate cross sections for UO2 of o 15 barns and oa of 7.6 barns, find the total macroscopic cross section SI. Then find the thickness of target t such that 0.05. How much attenuation in neutron beam would that thickness give?

Explanation / Answer

Firstly we have to calculate

density * NA / M

( 10 g/cm^3 * 6.022 x 10^23 atoms/mole / 270 g/mol

6.022 x 10^24 / 270

= 2.23 x 10^22 atoms / cm^3

Now,

total macroscopic cross section = 2.23 x 10^22 atoms/cm^3 * (7.6 + 15 barns) * ( 1 x 10^-24 cm^2 / barn)

total macroscopic cross section = (2.23 * 22.6) x 10^-2

total macroscopic cross section = 50.4 x 10^-2 cm^-1

total macroscopic cross section = 0.504 cm^-1

Now,

lambda(l) = 1 / total macroscopic cross section

lambda(l) = 1 / 0.504 = 1.984 cm

So,

t = 0.05 * 1.984 cm

t = 0.0992 cm

Attenuation in neutron beam = (total macroscopic cross section * t ) * 100%

Attenuation in neutron beam = 0.504 * 0.0992 * 100%

Attenuation in neutron beam = 0.0492 * 100% = 4.92%

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