# What is the spectrum of a nuclear bomb in a vacuum?

This question about 'nukes in space' mentions that the two forms of energy released from a nuclear bomb come from neutrons and photons (the latter about 104 times the former).

It's mentioned that the photons are in the form of X-rays, but what is the actual spectrum of the light emitted? How much of the light comes from

• the fission (assume pure-fission bomb for simplicity) where 239Pu is split into a mish-mash of lighter elements (is this energy quantized, i.e. has "peaks"?)
• black-body emission (the results being heated to 10x K; is this continuous or does strange stuff happen at very high temperatures?)
• extremely short-lived fission products
• The mix of all of these will depend on the particular design of the bomb. A lot of research went into "clean" nuclear weapons designs, which probably use fewer heavy elements in the material of the weapon than a more efficient but more "dirty" bomb would use. One would have to model the explosion in detail, using, at least, a shell model of the bomb's design, to make a useful prediction. Do you have a detailed construction plan for one of these? I have never seen one. – CuriousOne Jul 13 '16 at 9:12
• @CuriousOne for simplicity, I suppose say a 6.2 kg sphere of Pu-239 magically goes supercritical and loses 88 TJ/1 g of energy. I would imagine the casing is relatively inert anyways and would just transform fission photons into thermal? – Nick T Jul 13 '16 at 9:27
• Depends what's in the casing. In real nukes, from what little information is available, part of the magic is in the casing. In your example, however, if only a small fraction of the material actually fissions, there should be a non-black body spectrum (I doubt that x-ray photons can thermalize in the thinning expanding plasma, but I might be wrong about that) followed by a few relativistic electrons, neutrons and then a plasma of Pu ions and electrons, with slight admixtures of spallation products. – CuriousOne Jul 13 '16 at 9:54

One can get a rule of thumb understanding of things. The temperature in the immediate environment of a nuclear bomb is about $5\times 10^7$K. Using Wein's law $\nu_{max}~\simeq~(2.9/hc)kT$ for the frequency at the black body peak for this temperature $\nu~=~3.3\times 10^{18}$Hz and equivalently $\lambda~=~9\times 10^{-11}m$. We can also calculate the energy $E~=~h\nu$ that is $12$KeV. This in the X-ray range of energy. If you were to place a photon detector in space to measure a nuclear burst this would be about where the peak of the EM spectrum would be.
These photons are are secondaries after interacting with the materials of the bomb. The initial nuclear induced photons are at higher energy. The nuclear process does not primarily generate photons, which are generated by QED interactions. However, the motion of fission and fusion products induced by the nuclear interaction produces photons as these ions scatter off of each other by their electrostatic potentials. These photons are in the $100$KeV to $1$ MeV range of energy.
Neutrons are produced, and in the case of fusion they constitute $18$MeV of energy produced per fusion $D~+~T~\rightarrow~{}_2^4He~+~n$, which in turn produce photons as secondaries when they interact with matter. There is the neutron bomb that is a $T-T$ nuclear bomb meant to produce lots of neutrons. These have a magnetic moment that they interact with matter, and these are damaging to biological molecules. The neutron bomb is then largely an anti-personnel weapon and fashioned into a mini-hydrogen bomb.