The reason I ask this is because I need to calculate the de Broglie wavelength of a neutron in thermal equilibrium with a nuclear reactor at temperature $500$K.

The only way I can think to do this is to use the internal energy equation of an ideal gas, namely $$E=\frac32 NK_B T\tag{1}$$ where $N$ is the number of particles, $K_B$ is the Boltzmann constant and $T$ is the thermodynamic temperature.

Since we are only considering $1$ neutron here; $N=1$ and $(1)$ reduces to $$E=\frac32 K_B T\tag{2}$$

The momentum $p$ is given by $$p=\sqrt{2m_n\,E}=\sqrt{3m_n\,K_B\,T}$$ where $m_n$ is the mass of a neutron, so $$\lambda=\frac{h}{p}=\frac{6.63\times 10^{-34}}{\sqrt{3\times 1.675\times10^{-27}\times 1.38\times 10^{-23}\times 500}}\approx1.13\times 10^{-10}\,\text{m}$$ This is actually the correct answer. What I would like to know is whether I can use equation $(1)$ to find the de Broglie wavelength.

In other words; is it plausible to think of radiation as a gas?


One answer suggests that it is okay to think of radiation as an ideal gas, but my main concern is that this equation $(1)$ is identical to the formula used in the theorem of Equipartition of energy.

So the question that remains is: Was the internal energy equation $(1)$ a simple coincidence in that it is equal to the formula used in the theorem of equipartition of energy?

Or put in another way; Which theorem is applicable to this situation?


The issue here is one of time scales. The neutrons come off fission events with large and non-thermal energy. They will also either capture or decay before to long.

So there are a range of possibilities between

Neutrons decay or are capture while their energy spectrum is still dominated by the conditions of their creation.

(in which case treating them with equipartition is completely erroneous) through

The neutrons thermalize fast enough and last long enough that the bulk of their existence as a separate species is spent in thermal equilibrium with their surroundings.

(under which conditions the approximation is entirely reasonable).

All of which means we need a way to estimate how long thermalization takes. To do that we ask how neutrons interact with their surroundings and how often that happens.

For the purposes of a BotE calculation we can treat neutron interaction with their surroundings as a contact interaction with nuclei, which gives us a cross-section on order of a few tens of millibarns. (Nucleon radius is about one femtometer and nuclear radius is up to a few times that, so the impact parameter for significant energy transfer is order of $2$–$5 \,\mathrm{fm}$, and cross-sections go by the square of the impact parameter...)

Mean free path is cross-section times number density. In a reactor environment the number density is roughly a few thousand moles per cubic meter. So mean-free path is of order $10^{-4}$–$10^{-3}\,\mathrm{m}$.

At nuclear decay and fission energies neutron speeds are non-relativistic, but still quite fast. Let's use $10^6\,\mathrm{m/s}$.

So the fast neutrons have an initial scattering rate around a billion times per second or a thousand times per microsecond. On average the neutron will lose about half it's energy on each scattering until it reaches energies comparable to thermal environment. That's about 25 scatterings for a $1 \,\mathrm{MeV}$ neutron.

Absorption times in a water environment are tens of microseconds and up, so we can safely treat them as thermal for most of their lifetime.

  • $\begingroup$ Hi, you mention "under which conditions the approximation is entirely reasonable" in brackets under the second quote. Does this mean that the $E=\dfrac32 K_B T$ is valid via equipartition of energy theorem or by the ideal gas equation? Many thanks for your answer. $\endgroup$ – BLAZE Oct 31 '16 at 3:20
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    $\begingroup$ The expression $ E = 3 n k_b T/2$ is proper for any equilibrium system with a mean of three (quadratic) modes per component; ideal gases have three modes per component. Given the nature of neutrons (their size and interaction behavior), at temperatures under, say, a few thousand Kelvin the act roughly as an ideal gas. The most significant deviation comes from the possibility of capture by various nuclei in the environment (chlorine, hydrogen, and carbon all contribute non-trivially and other nuclei present may also have a big effect). $\endgroup$ – dmckee --- ex-moderator kitten Oct 31 '16 at 3:52
  • $\begingroup$ Thanks for explaining that to me, I just needed to check that the formula was due to the ideal gas approximation and not the equipartition argument. Best regards. $\endgroup$ – BLAZE Oct 31 '16 at 4:57

Yes i think it is correct in thinking of the emitted neutrons as ideal gas molecules since you get $E = \frac{3}{2}NK_BT$ from kinetic theory of gases which has under its number of assumptions, the following two :

1) Molecules dont interact with each other (no attractive or repulsive force; they are neutral to other molecules' presence)

2)They are thought of hard, pointlike spheres(very negligible size) which dont deform on collision (which are perfectly elastic)

These can be clearly applied to a bunch of neutrons inside a reactor.

  • $\begingroup$ Again, molecules, atoms or other neutral stuff....all which are assumed to have no internal structure in kinetic theory are regarded in equal footing......so when you say that you give half KT to each translational degree of freedom for a molecule, you may just as well replace the word molecule with the word neutron $\endgroup$ – Prasad Mani Oct 30 '16 at 13:41
  • $\begingroup$ So which theory should be used? Or are you saying that either theory could be used? Regards. $\endgroup$ – BLAZE Oct 30 '16 at 14:04
  • $\begingroup$ In this case, doesnt matter what theory you use, the assumption for both of them are applicable to your problem.......for other questions, it depends....! I think you are getting confused with what is kinetic theory and what is equipartition theory. The internal energy that you have ...the $E$ = $KE$ + $E_{attr}$ + ....all of which are negligible except the $KE$..the kinetic energy.....which is calculated from equipartition theorem to have the value $\frac{3}{2}K_B T$ for each translational degree of freedom $\endgroup$ – Prasad Mani Oct 30 '16 at 14:11

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