Question

In: Physics

Fission neutrons are thermalized in a pure Beryllium moderator at 427 oC. Determine the number of...

Fission neutrons are thermalized in a pure Beryllium moderator at 427 oC. Determine the number of collisions (scattering reactions) needed to thermalize these neutrons, the macroscopic slowing down power, and the moderating ratio. The density of Beryllium is 1.82 g/cm3

Expert Solution

Beryllium A= 7

microscopic cross sections _s = 7.63 b

_a = 0.0076 b

density = 1.82 g/cc

macroscopic cross section _s = * N_a/A *_s = 1.82 * 0.602 E+24 * 7.63E-24 /7

= 1.194 /cm --- scattering

_a = 1.82 * 0.602 E+24 * 0.0076E-24 /7

= 0.0012 /cm

slow down decrement = 1+ (/ 1- )ln where = (A-1)² /(A+1)² = 9/16

= 0.26

slow down power = *_s = 0.26 * 1.194 = 0.3104

Moderating ratio = *_s/_a = 0.26 * 1.194/0.0012 = 258.7

is the average logarthmic energy loss ratio per collision

n = 1/ ln(E_o/E_n ) is number of collisions required for slowing down the neutrons

E_o = 2 Mev - fast neutron from the reactor.

E_n - thermal energy of neutrons at 427^oC = 3/2 kT 3/2 * 8.617E-5 *(273+427) ev

= 0.009 ev

n = 1/0.26 * ln(2.0E+6/0.009) = 73.92

about 74 collisions are required to thermalise the fission neutrons to the moderator temp.

genius_generous answered 2 years ago

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