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Given the heat and pressure, is there some reason why a relatively small amount of atoms of various elements in the atmosphere or soil or in the metals used to make to bomb itself do not fuse?

I do realize that a fission bomb is used to sort of ignite an H-bomb, but that is a very elaborate device. Maybe fusion is very hard to achieve without a very specific design -- or maybe indeed the energy release by an A-bomb inevitably forces atoms together.

And if this is true, could it be that the implosion devices tend to cause fusion more readily than the gun-type, or is the design of the A-bomb completely irrelevant?

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To ignite a fusion reaction requires the pressures and temperatures typical at the center of a fission explosion. Elements outside the fireball of a fission explosion are not heated enough to fuse.

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  • $\begingroup$ As a guess, I think the T/P conditions in the second stage (pre fusion) is more than a regular plutonium pit fission reaction. You have 80% of the fission reaction energy in X-rays around a Pu spark plug: nuclearweaponarchive.org/Library/Teller.html $\endgroup$
    – JEB
    Oct 13, 2022 at 13:32
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The fusion part relies on a particular mechanism that is not present outside the fuel, and thus fusion outside the fuel is almost impossible.

The main fusion fuel is arranged in a cylinder (in older designs anyway) that is compressed from the outside by the x-ray flux from the primary. This causes the secondary to be crushed down into something smaller than a pencil. At this density, the alpha particles released in the D-T reactions cannot travel very far, and slow down almost immediately, depositing their energy in the surrounding fuel. It is this energy that causes the reaction to continue, burning outward from the trigger in the center.

In the surrounding material, the density is simply too low to slow down the alpha enough, and any alphas that make it out of the secondary will deposit their energy along a much longer track and not heat up the material nearly enough. Additionally, the surrounding material is generally much higher mass and requires correspondingly higher energies (orders of magnitude) to undergo fusion.

It's a carefully balanced thing, unlike, say, a supernova where there's so much energy the fusion keeps going even when its an energy sink.

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Most of the nuclear fusion in the explosion of fusion bombs occurs inside the bomb within the hollow sphere (or a hollow cylindrical body) that constitutes the fission bomb. This is based on the fact that when the critical mass is exceeded, most of the inner energy in the core fission in the nuclear bomb, viewed segmentally, goes inwards and thus a higher pressure and a higher temperature build up inside, which are the conditions for nuclear fusion. The pressure and temperature on the outside of the bombs that we can produce are usually not sufficient for this.

At least that's the theory. This theory is taken up, for example, in the implosion design of fission bombs:

schematic simplified model structure of an implosion bomb

(The Original Image Source)

Note: the fusion core is usually a compressed enriched gas mixture of Deuterium $\operatorname{D} = \operatorname{_{1}^{2}H}$ and Tritium $\operatorname{T} = \operatorname{_{1}^{3}H}$ (Isotopes Of Hydrogen) and the fission explosive body is usually composed of uranium isotopes and or plutoniom isotopes.

Even if this is a common design, there are also variants in which an absolute maximum of nuclear fusion is attempted. In these experiments, attempts are also made to defend small parts of the shell during nuclear fusion. Attempts are also being made to use small parts of the hull in nuclear fusion.

To do this, you build the shells of the nuclear bomb out of lithium or lithium deuteride. Some of the Lithium Isotopes (Lithium-$6$ ($\operatorname{_{3}^{6}Li}$) and Lithium-$7$ ($\operatorname{_{3}^{7}Li}$)) can fuse with Deuterium $\operatorname{D}$ to form Helium-$4$ ($\operatorname{_{4}^{2}H}$) for this see see reaction Formulas $1$ and $2$. In addition, the Lithium-$6$ Deuteride $\operatorname{_{3}^{6}LiD}$ (This sometimes occurs, for example, in the Teller–Ulam-Design.) should also serve as a neutron source for the fission bomb (for this see Formula $3$).However, these Lithium Isotopes can also react directly with a neutron (for this see Formulas $4$ und $5$): $$ \begin{align*} \operatorname{_{3}^{6}Li} + \operatorname{D} = \operatorname{_{3}^{6}Li} + \operatorname{_{1}^{2}H} &\to 2 \cdot \operatorname{_{2}^{4}He} + 22.4 ~\mathrm{MeV} \tag{1}\\ \operatorname{_{3}^{7}Li} + \operatorname{D} = \operatorname{_{3}^{7}Li} + \operatorname{_{1}^{2}H} &\to 2 \cdot \operatorname{_{2}^{4}He} + n + 15.1 ~\mathrm{MeV} \tag{2}\\ \end{align*} $$ $$ \begin{align*} \operatorname{_{3}^{6}LiD} + n = \operatorname{_{3}^{6}Li_{1}^{2}H} + n &\to \operatorname{_{2}^{4}He} + \operatorname{_{1}^{3}H} + \operatorname{_{1}^{2}H} + 17.59 ~\mathrm{MeV} = \operatorname{_{2}^{4}He} + \operatorname{T} + \operatorname{D} + 17.59 ~\mathrm{MeV} \tag{3}\\ \end{align*} $$ $$ \begin{align*} \operatorname{_{3}^{6}Li} + n &\to \operatorname{_{2}^{4}He} + \operatorname{_{1}^{3}H} + 4.78 ~\mathrm{MeV} = \operatorname{_{2}^{4}He} + \operatorname{T} + 4.78 ~\mathrm{MeV} \tag{4}\\ \operatorname{_{3}^{7}Li} + n &\to \operatorname{_{2}^{4}He} + \operatorname{_{1}^{3}H} + n - 2,46 ~\mathrm{MeV} = \operatorname{_{2}^{4}He} + \operatorname{T} + n - 2,46 ~\mathrm{MeV} \tag{5}\\ \end{align*} $$

All of this is still happening inside the nuclear bomb, but it's not necessarily inside the fusion device.

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