A thermal nuclear reactor uses a moderator to lower the kinetic energy of fast-moving neutrons - AQA - A-Level Physics - Question 6 - 2019 - Paper 2

Starting with an initial kinetic energy of 2.0 MeV, after each collision, the neutron loses 63% of its kinetic energy. The remaining kinetic energy after one collision is:

$E_1 = 2.0 imes (1 - 0.63) = 2.0 imes 0.37 = 0.74 ext{ MeV}$

For subsequent collisions, we can express this as:

$E_n = E_{n-1} imes 0.37$

Continuing this process for five collisions:

1. After 1 collision: $0.74$ MeV
2. After 2 collisions: $0.74 imes 0.37 = 0.2738$ MeV
3. After 3 collisions: $0.2738 imes 0.37 \\approx 0.101m$ MeV
4. After 4 collisions: $0.101m imes 0.37 \\approx 0.0373$ MeV
5. After 5 collisions: $0.0373 imes 0.37 \\approx 0.0138$ MeV

Thus, after five collisions, the kinetic energy of the neutron is approximately 0.0138 MeV.

The number of collisions required depends on the nucleon number of the moderator atoms because heavier nuclei have a greater ability to slow down neutrons. The probability of a neutron being absorbed or scattered effectively is linked to the mass and density of the moderator. Moderators with higher nucleon numbers can reduce neutron energy more efficiently, requiring fewer collisions to achieve thermalization.

First, we need to calculate the mass defect by summing the masses of the reactants and subtracting the sum of the products' masses:

• Mass of $^{235}U$ = 235.044 u
• Mass of $^{1}n$ = 1.0087 u
• Mass of $^{142}Xe$ = 141.930 u
• Mass of $^{90}Sr$ = 89.908 u
• Mass of $^{4}n$ = 4.0026 u (approximately)

Mass defect = (mass of reactants) - (mass of products)

$= (235.044 + 1.0087) - (141.930 + 89.908 + 4.0026)$ $= 236.0527 - 235.8406 = 0.2121 ext{ u}$

Now convert this mass defect to energy using Einstein’s equation, $E = mc^2$. Knowing that 1 u = 931.5 MeV:

$E = 0.2121 imes 931.5 ext{ MeV/u} ≈ 197.77 ext{ MeV}$

Thus, the energy released in this fission process is approximately 197.77 MeV.

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