Question

In: Chemistry

Both heavy water (D2O) and light water (H2O) are used as moderators for neutrons in nuclear...

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.

Expert Solution

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

queen_honey_blossom answered 2 years ago

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