# E=mc^2 and the nuclear bomb

1. Jul 3, 2006

### denni89627

Just curious about the energy/mass relationship and how it factors in when making nuclear weapons. (or hopefully a more peaceful use for such a reaction in the future.) I know manhattan era nukes used plutonium and polonium/ beryllium for fission reaction. Einstein's famous equation would have me believe the heaviest elements would yield much greater energy, fission or fusion. Do current nukes use the heaviest elements? Maybe they are too rare or too hard to isotope? Just curious. maybe its classified anyway.

2. Jul 4, 2006

### mathman

The two principle materials for fission weapons are U235 and Pu239. They both fission and can be made in quantity. Moreover, they are relatively stable in a stored environment. Half life of Pu239 is several hundred thousand years, while U235 is in millions. Heavier elements are much harder to produce in quantity and have much shorter half-lives.

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.

3. Jul 4, 2006

### Astronuc

Staff Emeritus
To help understand binding energy, fission and fusion, see -
http://hyperphysics.phy-astr.gsu.edu/hbase/nucene/nucbin.html

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.

4. Jul 5, 2006

### Orion1

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

$$E = mc^2$$

1 kiloton of TNT = 4.184 * 10^12 joules (j)
1 Megaton of TNT = 4.184 * 10^15 joules (j)

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

$$m(1 kt) = 4.655 \cdot 10^{-5} \; \text{kg}$$

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

$$m(1 Mt) = 4.655 \cdot 10^{-2} \; \text{kg}$$

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

Last edited: Jul 5, 2006
5. Jul 5, 2006

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

6. Jul 6, 2006

### dansydney

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. Thats what a teacher told me when I asked a similiar question a while ago. Only a small amount of mass contains a huge amount of energy as the equation states so if the nuclear weapon is very efficient in converting all the mass to energy then you will have a big explosion.

7. Jul 7, 2006

### denni89627

there was actually a really good special on the National geographic channel about explosives a couple days ago. In the 50's scientists used a uranium or plutonium fission reaction as a trigger for hydrogen fusion. (as stated by astronuc also). The largest one ever detonated was "Tsar Bomba" by the russians. It yielded 38,000 times the energy of the atomic bomb dropped on Nagasaki. The video footage is unreal. I guess it's a bit rediculous to try and create something with a larger yield than that. But.... scientists are doing just that! Experts agree that antimatter will be the next weapon of choice. It's too costly and difficult to make and contain currently, but may not be in the future. In theory a mass of antimatter equivalent to the mass of a paper clip would be enough to destroy an entire city. maybe they should work on an antimatter power plant instead. who's with me?

8. Jul 7, 2006

### Jimmy Snyder

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.

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.

9. Jul 7, 2006

### Jimmy Snyder

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.

Last edited: Jul 7, 2006
10. Jan 8, 2009

### theallknower

actualy,nuclear bombs don't use the nuclar force...they rather fight it...all that destructive force comes from the electric energy! you see,protons are all positive,so the electric force makes them get away from each other! luckely,there is the nuclear force that keeps them togeter,but the nuclear force has a very very short range...if you take a very hevavy element and bombard it with a neutron,the protons get out of the range of the enslaving nuclear force,and the electric force can do it's thing...to get an ideea, electric force is 1 bilion bilion bilion bilion times stronger then gravity!
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! )

11. Jan 14, 2009

### tormund

Modern fission based nuclear warheads use U-238, for your information, not Pu-239.
U-238 is much more unstable in a reactive environment because of its number of electron shells.

12. Jan 15, 2009

Staff Emeritus
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.

13. Jan 15, 2009

### vanesch

Staff Emeritus
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.

14. Jan 15, 2009

### vanesch

Staff Emeritus
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.

15. Jan 15, 2009

### Count Iblis

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.