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. After one collision, the kinetic energy is given by:

$E_1 = E_0(1 - 0.63) = 0.37 E_0$

For five collisions, the formula becomes:

$E_5 = E_0(0.37)^5$

Substituting $E_0 = 2.0 \text{ MeV}$:

$E_5 = 2.0 \times (0.37)^5 \approx 2.0 \times 0.00519 \approx 0.0104 \text{ MeV}$

The number of collisions is influenced by the mass of the moderator atoms. Heavier moderator nuclei are more effective at slowing down neutrons through elastic collisions. Thus, as the nucleon number increases, fewer collisions are required to reduce the neutron’s kinetic energy to the desired level, due to enhanced momentum transfer during interactions.

First, calculate the mass difference:

$\Delta m = (m_{\text{initial}} - m_{\text{final}})$

Initial mass:

$235.044 \, u + 1.0087 \, u = 236.0527 \, u$

Final mass:

$141.930 \, u + 89.908 \, u + 4 \times 1.0087 \, u = 141.930 + 89.908 + 4.0348 = 235.8728 \, u$

Mass difference:

$\Delta m = 236.0527 - 235.8728 = 0.1799 \, u$

Convert mass difference to energy:

Using the conversion $1 \, u \approx 931.5 \, MeV$:

$E = \Delta m \times 931.5 \, MeV/u = 0.1799 \times 931.5 \approx 167.4 \, MeV$

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