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

The neutron loses 63% of its kinetic energy in each collision. Its kinetic energy after the first collision is:

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

For subsequent collisions, we can apply the same reduction iteratively:

1. After the first collision: 0.74 MeV
2. After the second collision: $0.74 imes (1 - 0.63) = 0.2738$ MeV
3. After the third collision: $0.2738 imes (1 - 0.63) = 0.1014$ MeV
4. After the fourth collision: $0.1014 imes (1 - 0.63) = 0.0375$ MeV
5. After the fifth collision: $0.0375 imes (1 - 0.63) = 0.0138$ MeV

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

The number of collisions required is influenced by the nucleon number of the moderator atoms because heavier nuclei tend to scatter neutrons more effectively. A higher nucleon number means a greater mass, which can lead to more energy transfer during each collision. Consequently, with moderators that have a higher nucleon number, fewer collisions are needed to reduce the neutron's kinetic energy significantly.

To calculate the energy released during the fission process, we must first find the mass difference. The reaction involves:

• Mass of $^{235}U$: 235.044 u
• Mass of $^{142}Xe$: 141.930 u
• Mass of $^{90}Sr$: 89.908 u
• Mass of $^1n$: 1.0087 u

Now, we can calculate the mass difference:

$ext{Mass difference} = (235.044 + 1.0087) - (141.930 + 89.908 + 4 imes 1.0087)$

This simplifies to: $ext{Mass difference} = 236.0527 - (141.930 + 89.908 + 4.0348) = 236.0527 - 235.8728 = 0.1799 ext{ u}$

Then we convert the mass difference to energy using the conversion factor (1 u = 931.5 MeV):

$E = 0.1799 ext{ u} imes 931.5 ext{ MeV/u} = 167.65 ext{ MeV}$

Therefore, the energy released in this fission process is approximately 167.65 MeV.

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