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THERMAL NEUTRONS
Maxwellian Distribution
The thermal neutron velocity/energy (
2 2
1
E K = mnv ) distribution is Maxwellian. The number of
neutrons of energy E per unit energy interval, N ( E ), and the number of neutrons of velocity v per unit
velocity interval, N ( v ), can be expressed in terms of the neutron energy:
E k T Ee kT
N
N E
dE
dN (^) /
3 / 2
0 0
− = =
or the neutron velocity:
mv k T e kT m
v N N v dv
dN (^) / 2
3 / 2
0
2 0
2
− = =
where k is Boltzmann’s constant; T is the absolute temperature of the medium; and N 0 is the total number of
neutrons per unit volume, that is,
∫ ∫
∞ ∞ = = 0 0
N (^) 0 N ( v ) dv N ( E ) dE (3)
0
0 1 2 3 4
Energy, kT, and Velocity, sqrt(2kT/m)
N(E), No/kT
0
N(v), No/sqrt(2kT/m)
N(E)
N(v)
Figure: Maxwellian neutron energy and velocity distributions with normalized units.
By setting the derivative of the N ( E ) and N ( v ) expressions equal to zero, the most probable energy
and velocity, respectively, can be solved for. We use the product rule to find the derivative of dN ( v )/ dv :
− −
−
k T
m v ve v e kT m
N
e kT m
v N
dv
d
dv
dN
mv kT mv kT
mv kT
/ 2 2 / 2 3 / 2
0
/ 2 3 / 2
0
2
2 2
2
By setting the above expression equal to zero, we find the most probable velocity
m
kT v
kT
mv v v dv
dN
p
2
This result agrees with the graphical plot of N ( v ). For a neutron at 20°C, the most probable velocity is then
2197 m/sec J
kg m /s
( 1. 675 10 kg)
2 ( 2 )( 1. 38 10 J/K)( 273 20 K)
2 2
27
23
×
× ° + °
−
−
m
kT v (6)
where the zero subscript implies thermal equilibrium at this reference (room) temperature. Because these
are low velocity neutrons, the energy at the most probable velocity may be found from the classical
expression for kinetic energy
k T m
kT E (^) T mvp m ⎟= ⎠
which is the thermal neutron energy. At 20° the thermal neutron energy is
( 8. 617 10 eV/K)( 273 20 K) 0. 0253 eV
5 0 =^ = × ° + ° =
− E k T (8)
The formulae for the most probable neutron energy, and its corresponding velocity, can be obtained in
similar fashion. Left as an exercise, the most probable neutron energy is
E (^) p k T 2
1 = (9)
which agrees with the earlier graph. Note the difference between the most probable energy of Eq. (7), and
the energy at the most probable velocity from Eq. (9).
The average energy and velocity can be found from
∫
∫
∫
∫
∞
∞
∞
∞
0
0
0
0
Nv dv
N v vdv
v
N E dE
N E EdE
E (^) avg avg (10)
We note that the denominator of both expressions above is equal to N 0. The average velocity can be found
using variable substitutions of
k T
m x v dx vdv a 2
such that
THERMAL NEUTRON FLUX
2200 meter-per-second Flux, φ 0
φ 0 = nv 0 (1)
Thermal Flux, φT
T E dE nvavg nv T
Beam vs. Reactor Flux
· relating the two different fluxes
T
T
E
E
v
v
nv
nv
T T T
T
0 0 0 0 0
where
293. 6 K
1. 38066 10 J/K
( 1. 67492 10 kg)( 2200 m/s)
23
27 2 2
(^21) 2 0
1
× °
×
−
−
k
mv T (4)
Reaction Rate
Fa = Σ a ( E 0 )φ 0 =Σ a , th φ T (5)
For 1/ v absorbers, the absorption rate is independent of the neutron energy distribution. Non-1/ v absorbers
include: U-233, U-235, U-238, Pu-239, Cd, In, Xe-135, Sm-149.
Thermal Absorption Cross Section
th
a
MK
a ms a th T T E
T 0. 0253 eV
293 K
, 2200 m/s
,
0 ,^2200 / , 0
σ σ σ
π σ =
Thermal Scattering Cross Section
σ (^) s , th = σ s , 2200 m/s (7)
