Question

The average kinetic energy of thermal neutron is of the order of:
(Boltzmann’s constant ${k_B} =8 \times 10^{ - 5}{eV/K} $)
A. $0.03eV$
B. $3eV$
C. $3keV$
D. $3MeV$

Answer 1

Hint: First of all we know neutron energy is also called neutron detection temperature. It indicates a free neutron kinetic energy, usually given in electron volts. Neutron energy is distributed in different ranges; one of them is the thermal neutron. A thermal neutron is a free neutron with a kinetic energy of about 0.025eV.

Complete step by step solution:
The kinetic energy of thermal neutrons is calculated by the formula ($E = {k_B}T$), where ${k_B}$ the Boltzmann’s constant and T is the temperature in the Kelvin.
At $T = 20^\circ C = (293K)$
$E = {K_B}T = 8 \times 10^{ - 5}{eV/K} \times 293K$
$=0.03eV$

Additional information:
When we say a thermal neutron has energy ${k_B}T$, it means that there is a distribution of neutron energies owing to the temperature. This distribution is a Maxwell distribution and the most probable energy for a neutron in this distribution is ${k_B}T$. We choose this energy ${k_B}T$ as representative of a thermal neutron.

Note:
The atoms in a reactor will exhibit the same energy distribution as the neutrons because they will be at the same temperature. Thus, the most probable energy for the atoms in a reactor will be the same as that for the neutrons, 0.0253$eV$. Thermal neutrons have a different and sometimes much larger effective neutron absorption for a given nuclide than fast neutrons, and can therefore often be absorbed more easily by the atomic nucleus. This event is called neutron activation.