Question

Both heavy water (D2O) and light water (H2O) are used as moderators for neutrons in nuclear reactors. Calculate the number of collision required to reduce the energy of a neutron from 2 MeV to thermal energy (0.025 eV) for each moderator. Assume that oxygen plays a minor role in the thermalization. Therefore, mass number of A = 2 and A = 1 can be used for calculations of number of collisions for heavy water (D2O) and light water (H2O), respectively.

Answer 1

Number of collisions for light water (H₂O)

E _high = 2 MeV

E _low = 0.025 eV

Calculated value of ξ for water = 2/(1 + 2/3) = 1.2

This approximation is relatively accurate for mass numbers (A) greater than 10, but not for some low values of A.

Convert 2 MeV to eV by multiplying with conversion factor 10^6

2 MeV = 2.0 x 10^6 eV

The number of collisions can be calculated by using formula

N = ln (E _high/E _low)/ ξ

N = ln (2.0 x 10^6/0.025)/ 1.2

N = 18.197/1.2 =15collisions

ξ is constant for each type of material . Average logarithmic energy decrement (ξ) for water = 0.948 (from table)

N = ln (2.0 x 10^6/0.025)/ 0.948

= 18.197/0.948 = 19.2 collisions

Number of collisions for heavy water:

Calculated value of ξ for D₂O = 2/(2 + 2/3) = 0.75

The number of collisions can be calculated by using formula

N = ln (E _high/E _low)/ ξ

N = ln (2.0 x 10^6/0.025)/ 0.75

= 18.197/0.75 = 24 collisions

Number of collisions = 24 collisions

Average logarithmic energy decrement (ξ) for heavy water = 0.510 (from table)

N = ln (2.0 x 10^6/0.025)/ 0.510

   = 18.197/0.510 = 35 collisions