In atomic bombs, nuclear reactions provide the energy of the explosion. In every reaction, a thermal neutron reaches a plutonium or a uranium nucleus, a fission reaction takes place, and two or three neutrons and $\gamma$ radiation are produced. I know that it happens in a very short time, and an extreme amount of energy is released which can be calculated from the mass difference between $m_\mathrm{starting}$ and $m_\mathrm{reaction\ products}$.

So my question is: Why exactly does it explode? What causes the shockwave and why is it so powerful? (Here I mean the pure shockwave which is not reflected from a surface yet) I understand the reactions which are taking place in nuclear bombs but I don't understand why exactly it leads to a powerful explosion instead of just a burst of ionising radiation.

  • $\begingroup$ This is illuminating though it's unclear whether this particular bomb is a fission bomb or a fusion bomb. $\endgroup$
    – davidbak
    Feb 23, 2018 at 21:20
  • $\begingroup$ Do you understand the concept of a chain reaction? $\endgroup$
    – MaxW
    Feb 23, 2018 at 22:48
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    $\begingroup$ Yes , I understand... the question is about the production of the shockwave. (As pointed out in the question...) $\endgroup$
    – L.Gyula
    Feb 24, 2018 at 0:17
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    $\begingroup$ And it is a fission bomb but I thought it was evident since I only mentioned $U$ and $Pu$ in my question.(Not deuterium..) $\endgroup$
    – L.Gyula
    Feb 24, 2018 at 0:36
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    $\begingroup$ +1 I've had the same question for a long time but forgotten to ask; thanks for asking. $\endgroup$
    – user541686
    Feb 24, 2018 at 6:26

4 Answers 4


I don't understand why exactly it leads to a powerful explosion instead of just a burst of ionising radiation.

This radiation, representing most of the initial energy output by a nuclear weapon, is swiftly absorbed by the surrounding matter. The latter in turn heats almost instantly to extremely high temperature, so you have the almost instantaneous creation of a ball of extremely high kinetic energy plasma. This in turn means a prodigious rise in pressure, and it is this pressure that gives rise the blast wave.

The same argument applies to the neutrons and other fission fragments / fusion products immediately produced by the reaction. But it is the initial burst of radiation that overwhelmingly creates the fireball in an atmospheric detonation, and the fireball that expands to produce most of the blast wave.

  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – rob
    Feb 26, 2018 at 15:23

In an implosion fission bomb, the bulk of the nuclear reactions (100 GJ) occur in the final microsecond--it is a tremendous amount of power--$10^{17}$ Watts. This power is dumped into the compressed Pu pit--which is a few kilogram with a density of 100 g/cm^3-it is small.

The nuclear material is heated to an extremely high temperature, tens of millions of K--it black body radiates (X-rays) to the nearest material: the "cold" bomb components (air quotes because they have just been involved in a conventional explosion). They are heated to a lesser temperature, which blackbody radiates isotropically, heating material further away from the bomb. Note that the atmosphere is opaque to these X-rays, so the process of radiative diffusion continues for several hundred of meters. This is called the fireball--it's all about photons and radiative diffusion. As it diffuses outward, at some point it becomes slower than a shockwave, so that a shockwave separates from the fireball, which is now on the order of a few hundred thousand K. (In photographs of the fireball in the 1st few milliseconds, when it's 100's of meters, the 'shadows' of the bomb components can still be seen).

The shockwave heats air as it passes, but it is not as hot as the fireball. It is at least 50,000K, maybe 100,000K--so it's at least as radiant as a lightning bolt--but it is not transient, and it could subtend a much larger solid angle--hence the phenomenal thermal damage. Nevertheless, it is significantly colder than the material behind it.

As the shockwave propagates, it weakens an eventually no longer heats air to luminance--at this point the traditional fireball has reached it's maximum size. (I say traditional, because its what we see in test footage, but it should be distinguished from the initial fireball which is radiatively diffusing photons with some plasma thrown in--and radiation is the dominant mode of energy transfer, even though the bomb casing can have hypersonic velocities.)

Now for an airburst the shockwave that is reflected from the ground move through heated air, and is thus faster than the direct shockwave: it catches up and the two combine to for a more powerful shock, called the Mach stem. This goes on to produce blast damage, as buoyant forces lift the fireball, producing the infamous mushroom cloud.

For tests like Starfish Prime, which occurred in space--the initial X-rays (very hard X-rays) from the bomb components aren't absorbed by air--they continue to the upper atmosphere where wide scale Compton scattering produces a huge and sudden current, leading to continental scale EMP.

A point of clarification: since the OP asked about shockwave formation--as others pointed out, the temperature rise leads to huge pressure, which leads to a shockwave, but it does not form in the fireball--the radiative diffusion is at first much faster--and it's only as the fireball petters out that the shockwave separates from it.

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    $\begingroup$ "It is at least 50K, maybe 100K" - are you still referring to degrees Kelvin here, and are several orders of magnitude out, or are you referring to another unit that I'm missing? $\endgroup$
    – ArtOfCode
    Feb 24, 2018 at 15:57
  • $\begingroup$ @ArtOfCode Kelvin (or Centigrade). But with a factor of 2 margin, I guess Fahrenheit isn't too bad either. $\endgroup$
    – JEB
    Feb 24, 2018 at 16:36
  • $\begingroup$ @ArtOfCode The real problem is I don't remember the numbers, and the book whence the narrative is from (bookdepository.com/Physics-Shock-Waves-Y-B-Zeldovich/…) is back in my office. $\endgroup$
    – JEB
    Feb 24, 2018 at 16:45
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    $\begingroup$ Sure, I just wanted to point out that 50K or 100K is extremely cold, not extremely hot as I think you meant. I'd expect the figure to be two or three orders of magnitude greater. $\endgroup$
    – ArtOfCode
    Feb 24, 2018 at 19:25
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    $\begingroup$ Adding to what @ArtOfCode already wrote, I think you meant fifty to a hundred kilokelvin, so kK, not K. 50-100 K is the temperature range around which most things we think of as gases (such as the oxygen and nitrogen that make up about 99% of the Earth's atmopshere...) turn to solids. $\endgroup$
    – user
    Feb 24, 2018 at 20:10

The easiest way to answer this is to direct you to an explanation of the products of nuclear fission and their energy components. You are correct in thinking that ionizing radiation is released. However, it is only a small part of the energy release. For U235, the energy released in a single fission is about 195 MeV. Of that, 170 MeV is in the kinetic energy of the two fission fragments (physics is nice in limiting things to two fragments most of the time). Another 12 MeV is in the kinetic energy of the released neutrons which can cause the self-perpetuating chain reaction. Only about 8 MeV is found in the released gamma radiation. The heat generated is then due to the transfer of kinetic energy in collisions with other matter. Since most of the initial energy release occurs within 10E-12 seconds of the absorption of the initial neutron, and the collisions begin shortly after that, the heat generation begins within fractions of a second. The plasma expansion, pressure rise, and blast wave follow.

  • $\begingroup$ I understand the nuclear reactions. The part I am really interested in is the production of the shockwave. $\endgroup$
    – L.Gyula
    Feb 24, 2018 at 0:45

The shock wave is created when the high temperatures of the explosion vaporise surrounding material resulting in it's rapid expansion.


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