# Neutron Current Density

The neutron flux, φ, does not characterize the flow of neutrons. There may be no flow of neutrons, yet many interactions may occur (I = Σ.φ). The neutrons move in random directions and hence may not flow. Therefore the neutron flux φ is more closely related to densities. Neutrons will exhibit a net flow when there are spatial differences in their density. Hence we can have a flux of neutron flux!

This flux of neutron flux is called the neutron current density. We have to distinguish between the neutron flux and the neutron current density. Although both physical quantities have the same units, namely, neutrons.cm-2.s-1, their physical interpretations are different. In contrast to the neutron flux, the neutron current density is the number of neutrons crossing through some arbitrary cross-sectional unit area in a single direction per unit time (a surface is perpendicular to the direction of the beam). The neutron current density is a vector quantity.
The vector J is defined as the following integral:

## Physical Interpretation

The physical interpretation is similar to the fluxes of gases. The neutrons exhibit a net flow in the direction of least density. This is a natural consequence of greater collision densities at positions of greater neutron densities.

Consider neutrons passing through the plane at x=0 from left to right due to collisions to the left of the plane. Since the concentration of neutrons and the flux is larger for negative values of x, there are more collisions per cubic centimeter on the left. Therefore more neutrons are scattered from left to right, then the other way around. Thus the neutrons naturally diffuse toward the right.

Neutron Flux Density
In reactor physics, the neutron flux is often used, because it expresses the total path length covered by all neutrons. The total distance these neutrons can travel each second is determined by their velocity. Therefore the neutron flux density value is calculated as the neutron density (n) multiplied by neutron velocity (v).

Ф = n.v

where:
Ф – neutron flux (neutrons.cm-2.s-1)
n – neutron density (neutrons.cm-3)
v – neutron velocity (cm.s-1)

The neutron flux, the number of neutrons crossing through some arbitrary cross-sectional unit area in all directions per unit time, is a scalar quantity. Therefore it is also known as the scalar flux. The expression Ф(E).dE is the total distance traveled during one second by all neutrons with energies between E and dE located in 1 cm3.

The connection to the reaction rate, respectively, the reactor power is obvious. Knowledge of the neutron flux (the total path length of all the neutrons in a cubic centimeter in a second) and the macroscopic cross-sections (the probability of having an interaction per centimeter path length) allows us to compute the rate of interactions (e.g.,, rate of fission reactions). The reaction rate (the number of interactions taking place in that cubic centimeter in one second) is then given by multiplying them together:

where:
Ф – neutron flux (neutrons.cm-2.s-1)
σ – microscopic cross-section (cm2)
N – atomic number density (atoms.cm-3)

Example - Neutron flux in a typical thermal reactor core

A typical thermal reactor contains about 100 tons of uranium with an average enrichment of 2% (do not confuse it with the enrichment of the fresh fuel). If the reactor power is 3000MWth, determine the reaction rate and the average core thermal flux.

Solution:

Multiplying the reaction rate per unit volume (RR = Ф . Σ) by the total volume of the core (V) gives us the total number of reactions occurring in the reactor core per unit time. But we also know the amount of energy released per one fission reaction to be about 200 MeV/fission. Now, it is possible to determine the rate of energy release (power) due to the fission reaction. It is given by the following equation:

P = Ф . Σf . Er . V = Ф . NU235 . σf235 . Er . V

where:
P – reactor power (MeV.s-1)
Ф – neutron flux (neutrons.cm-2.s-1)
σ – microscopic cross section (cm2)
N – atomic number density (atoms.cm-3)
Er – the average recoverable energy per fission (MeV / fission)
V – total volume of the core (m3)

The amount of fissile 235U per the volume of the reactor core.

m235 [g/core] = 100 [metric tons] x 0.02 [g of 235U / g of U] . 106 [g/metric ton] = 2 x 106 grams of 235U per the volume of the reactor core

The atomic number density of 235U in the volume of the reactor core:

N235 . V = m235 . NA / M235
= 2 x 106 [g 235 / core] x 6.022 x 1023 [atoms/mol] / 235 [g/mol] = 5.13 x 1027 atoms / core
The microscopic fission cross-section of 235U (for thermal neutrons):

σf235 = 585 barns

The average recoverable energy per 235U fission:

Er = 200.7 MeV/fission

References:
Nuclear and Reactor Physics:
1. J. R. Lamarsh, Introduction to Nuclear Reactor Theory, 2nd ed., Addison-Wesley, Reading, MA (1983).
2. J. R. Lamarsh, A. J. Baratta, Introduction to Nuclear Engineering, 3d ed., Prentice-Hall, 2001, ISBN: 0-201-82498-1.
3. W. M. Stacey, Nuclear Reactor Physics, John Wiley & Sons, 2001, ISBN: 0- 471-39127-1.
4. Glasstone, Sesonske. Nuclear Reactor Engineering: Reactor Systems Engineering, Springer; 4th edition, 1994, ISBN: 978-0412985317
5. W.S.C. Williams. Nuclear and Particle Physics. Clarendon Press; 1 edition, 1991, ISBN: 978-0198520467
6. G.R.Keepin. Physics of Nuclear Kinetics. Addison-Wesley Pub. Co; 1st edition, 1965
7. Robert Reed Burn, Introduction to Nuclear Reactor Operation, 1988.
8. U.S. Department of Energy, Nuclear Physics and Reactor Theory. DOE Fundamentals Handbook, Volume 1 and 2. January 1993.