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- Thread starter denni89627
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mathman

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The other materials you mentioned were used to generate neutrons to make sure the chain reaction started as soon as the bomb was triggered. They were not fission materials.

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Astronuc

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denni89627 said:

To help understand binding energy, fission and fusion, see -

http://hyperphysics.phy-astr.gsu.edu/hbase/nucene/nucbin.htmlNuclear Binding Energy

Nuclei are made up of protons and neutron, but the mass of a nucleus is always less than the sum of the individual masses of the protons and neutrons which constitute it. The difference is a measure of the nuclear binding energy which holds the nucleus together. This binding energy can be calculated from the Einstein relationship -

Current nuclear weapons are based on Pu-239, whereby spheres of Pu-239 are imploded. The technology is certainly classified. I believe all U-235 weapons have been deactivated and the U-235 recycled into commercial fuel stream.

Small weapons are primarily Pu-239, maybe with a DT booster. Thermonuclear weapons use a Pu-239 'trigger' to 'ignite' (initiate) the fusion reaction. That technology is also classified.

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E=mc^2 and the nuclear bomb...the energy/mass relationship and how it factors in when making nuclear weapons.

The simplest factor is the amount of total mass that is converted into energy:

[tex]E = mc^2[/tex]

1 kiloton of TNT = 4.184 * 10^12 joules (j)

1 Megaton of TNT = 4.184 * 10^15 joules (j)

[tex]m(1 kt) = \frac{E_1}{c^2} = \frac{4.184 \cdot 10^{12} \; \text{j}}{c^2} = 4.655 \cdot 10^{-5} \; \text{kg}[/tex]

[tex]m(1 kt) = 4.655 \cdot 10^{-5} \; \text{kg}[/tex]

[tex]m(1 Mt) = \frac{E_1}{c^2} = \frac{4.184 \cdot 10^{15} \; \text{j}}{c^2} = 4.655 \cdot 10^{-2} \; \text{kg}[/tex]

[tex]m(1 Mt) = 4.655 \cdot 10^{-2} \; \text{kg}[/tex]

Reference:

http://en.wikipedia.org/wiki/Megaton

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selfAdjoint

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Orion1 said:

The simplest factor is the amount of total mass that is converted into energy:

[tex]E = mc^2[/tex]

1 kiloton of TNT = 4.184 * 10^12 joules (j)

1 Megaton of TNT = 4.184 * 10^15 joules (j)

[tex]m(1 kt) = \frac{E_1}{c^2} = \frac{4.184 \cdot 10^{12} \; \text{j}}{c^2} = 4.655 \cdot 10^{-5} \; \text{kg}[/tex]

[tex]m(1 kt) = 4.655 \cdot 10^{-5} \; \text{kg}[/tex]

[tex]m(1 Mt) = \frac{E_1}{c^2} = \frac{4.184 \cdot 10^{15} \; \text{j}}{c^2} = 4.655 \cdot 10^{-2} \; \text{kg}[/tex]

[tex]m(1 Mt) = 4.655 \cdot 10^{-2} \; \text{kg}[/tex]

Reference:

http://en.wikipedia.org/wiki/Megaton

And a hundredth of a kilogram is ten grams, so the mass equivalent for a megaton explosion is 46.55 grams.

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This is not the case. Fission bombs do use very heavy elements (and yet not the heaviest). However fusion bombs use the lightest ones and have higher yields. The issue is not the mass of the elements used. More important is the difference in mass between the nuclei before and after the reaction and the efficiency with which one reaction causes other reactions.denni89627 said:Einstein's famous equation would have me believe the heaviest elements would yield much greater energy, fission or fusion.

By the way, Einstein's famous equation is not limited to certain types of mass or energy. It applies universally to all energy flows such as conventional explosions and even the flap of a butterfly's wings.

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The critical mass of plutonium is roughly 10 kg. As can be seen in Orion1's post #4, the amount of mass converted in a 1 kiloton nuclear explosion is on the order of one one-millionth of this, much less than 1%. The mass loss in the burning of petroleum or other chemical reactions is much smaller yet. It is so small that until Einstein published his famous equation, no one had noticed it. Had it been 1% as you suggest, the effect would have been easily measured and the famous equation would be somebody else's.dansydney said:I believe that when something like petroleum is burned that only something like 1% is converted to energy, the rest just changing state to carbon monoxide and water. I assume that in a nuclear explosion that most of the mass of the plutonium or uranium is converted to energy.

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so in conclusion,I don't think E=mc^2 realy helps here...but you can aply it to an antimatter bomb,to see what it does(don't forget to add the mass of the normal matter,too! )

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U-238 is much more unstable in a reactive environment because of its number of electron shells.

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Vanadium 50

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Modern fission based nuclear warheads use U-238, for your information, not Pu-239.

I find this very unlikely, as a) U-238 is difficult to fission, and b) it would allow the use on unenriched uranium, so why all the fuss about enrichment.

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vanesch

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U-238 is much more unstable in a reactive environment because of its number of electron shells.

That's totally erroneous. It is not possible to make a critical mass with U-238 and moreover the electrons have nothing to do with this.

What is correct, is that in a fission-fusion-fission weapon, the second "fission" part can be done with U-238 (which also serves as a tamper), but it is not a chain reaction. It just uses the fast neutrons created in the fusion stage to induce fission to boost the energetic yield (the neutrons themselves have 14 MeV, and if they induce a fission, they liberate 200 MeV, so there's in principle a factor of about 10 to be won with the second fission stage over the pure fusion although this is less of course as not every fusion-liberated neutron will give you a fission reaction).

The Tsar bomb test actually left out the second fission stage, which is why it "only" had 50 megatons, and not the 100 megatons or more.

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vanesch

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The critical mass of plutonium is roughly 10 kg. As can be seen in Orion1's post #4, the amount of mass converted in a 1 kiloton nuclear explosion is on the order of one one-millionth of this, much less than 1%. The mass loss in the burning of petroleum or other chemical reactions is much smaller yet. It is so small that until Einstein published his famous equation, no one had noticed it. Had it been 1% as you suggest, the effect would have been easily measured and the famous equation would be somebody else's.

Well, with a 10 kg plutonium core, you can have normally a few tens of kilotons of yield.

A 1 kiloton weapon is or a specially designed low-yield weapon, or a fizzle.

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...maybe they should work on an antimatter power plant instead. who's with me?

If we could compress some big object so much that it becomes a small black hole, we would have a source of anti-matter. As long as the Hawking temperature is higher than about

10^10 K, the black hole will emit positrons.

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