Question

Problem 4. Boral is the material composed of 50 % Al, 40 % B, and 10 % C by weight. Its density is 2.56 g/cm³. A beam of 0.025-eV neutrons with intensity of 5´10⁶neutrons/(cm²×s) strikes a target of 2.5 cm² in area made of the boral.

1. Calculate the atom densities of Al, B, and C in boral.
2. Calculate the macroscopic cross-sections S_a, S_s, and S_t for boral at the neutron energy 0.025 eV.
3. How thick must the boral target be in order to reduce the intensity of such neutron beam by a factor of 7 ?

Answer 1

Boral:

Avogadro number N_a = 6.02 e+23

Al mass number = 27

Atomic density in Boral = (2.56*0.5/27) *6.02e+23 = 4.73 e+21 /cc

Boran A= 10

Atomic density = (2.56*0.4/10) * 6.02e+23 = 1.024 e+22 /cc

carbon A =12

Atomic density = (2.56*0.1/10) * 6.02e+23 = 2.56 e+21 /cc

0.025 ev are thermal neutrons

thermal cross sections

Al :   =  0.233

B :   = 3409   = 2.25

C : = 3.0e-3   = 4.81

Macroscopic cross section of Boral

= 0.233e-24* 4.73*e+21 + 3409e-24 * 1.024e+22 + 3.0e-27 *2.56e+21

= 34.77 /cm

Cross section of B dominates and that of others is negligible

For scattering cross section again Al is negligible compared to B and C

= 2.25e-24 * 1.024e+22 + 4.81e-24* 2.56e+21

= 3.53 e-2 /cm

Incident intensity I₀ = 5.0e+6

I = I_o exp (- t)

I/I_o = 1/7 = exp(-3477 t)

thickness of the target t = 5.6 e-4 cm