If a 2.0-MeV neutron released in a fission reaction loses half its kinetic energy in each moderator collision, how many collisions are needed to reduce its kinetic energy to $1 / 25 \mathrm { eV }$ ?

Question

Quizplus · Accepted Answer

The neutron starts with 2.0 MeV of energy and loses half its energy with each collision. To find the number of collisions needed to reduce its energy to $1 / 25 \mathrm { eV }$, we can use the formula for exponential decay: $E_f = E_i \times (\frac{1}{2})^n$, where $E_f$ is the final energy, $E_i$ is the initial energy, and $n$ is the number of collisions. Converting the initial and final energies to the same unit (eV) for ease of calculation, we have $2.0 \mathrm{MeV} = 2.0 \times 10^6 \mathrm{eV}$ and $1 / 25 \mathrm{eV} = 0.04 \mathrm{eV}$. Setting up the equation: $0.04 = 2.0 \times 10^6 \times (\frac{1}{2})^n$, solving for $n$ gives $n = 26$.

If a 20-MeV Neutron Released in a Fission Reaction Loses Half Its