)The total cross section for uranium dioxide of density 09/cm3 s to be measured
ID: 3281187 • Letter: #
Question
)The total cross section for uranium dioxide of density 09/cm3 s to be measured by a transmission method. To avoid multiple neutron scattering, which would introduce error ineo the resuits, the sample thickness is chosen to be much smailer than the mean frea path of neut- rons in the material Using approximate cross sections for UO2 of Os=15 born s and Ja . = ,6 barns, fnd the total macroscopic cross section Then find "thenickness of the tacpesucntna/At 0.05. How Mucn attenuortion in neutron beam would thot thickness give?Explanation / Answer
Here we need to solve
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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