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badmana
01-16-2004, 06:07 PM
Most nuclear bombs contain large amounts of fusion material which I figure is required to create the chain reaction for a good explosion.

So what is the explosive rating of a signal atom (if we could split it without having to use special explosives that might mask the splitting of the atom)? I know the whole E=MC2 would the the energy released...so how does that translate to an explosive rating?

Would it be enough to blow up a car? a small building? a sky scraper?

Q.E.D.
01-16-2004, 06:22 PM
A single atom? It wouldn't be enough to blow up an amoeba. A single atom of uranium releases in the neighborhood of 200 MeV, which translates to roughly 0.00003204 joules.

Desmostylus
01-16-2004, 06:25 PM
Let's take a single carbon atom. It has a mass of 1.92 x 10-26 kg.

Assuming the whole thing gets turned into energy, the energy released is:

E = mc2 = 1.92 x 10-26 (3 x 108)2 = 1.73 x 10-9 J

That's enough to light a 1 W lightbulb for about a nanosecond.

Desmostylus
01-16-2004, 06:28 PM
A single atom? It wouldn't be enough to blow up an amoeba. A single atom of uranium releases in the neighborhood of 200 MeV, which translates to roughly 0.00003204 joules.No it doesn't. 200 MeV = 3.2 x 10-11 J.

badmana
01-16-2004, 06:34 PM
That's it? So this whole "splitting the atom" thing is just hype? :D


So then how much power is contained in a gram of matter? A kilogram?

Just how much uranium (or whatever modern nukes are made with) is in a nuclear bomb these days? I always figured it was just a few dozen Kg!

Q.E.D.
01-16-2004, 06:37 PM
Just how much uranium (or whatever modern nukes are made with) is in a nuclear bomb these days? I always figured it was just a few dozen Kg! It is only a few kg. But there's a LOT of atoms in one kg of uranium.

Q.E.D.
01-16-2004, 06:38 PM
No it doesn't. 200 MeV = 3.2 x 10-11 J. Quite right. For some reason I calculated 200 million MeV instead.

Squink
01-16-2004, 07:19 PM
200 MeV (7.6 x 10-14 kcal) is about 5 billion times as much energy as is stored in a single carbon-carbon bond (90 kcal/mol -> 1.49x 10-22 kcal/bond). That's not much energy in terms of lighting lightbulbs and such, but if its released in your liver, there's enough punch to blow quite a few molecules apart.

Teelo
01-16-2004, 07:33 PM
How does one go about splitting an Atom?

In Science 30, This is brought up a few times for nuclear fission or fusion...or something, and as far as I can tell my crappy text book never says how.
I have a Diploma exam coming up next monday, and if I need to answer a written response about Nuclear reactions, I would like to know.

Q.E.D.
01-16-2004, 07:39 PM
How does one go about splitting an Atom?

In Science 30, This is brought up a few times for nuclear fission or fusion...or something, and as far as I can tell my crappy text book never says how.
I have a Diploma exam coming up next monday, and if I need to answer a written response about Nuclear reactions, I would like to know.
Depends on the atom. In the case of certain isotopes of uranium or plutonium, for instance, you shoot a neutron at it.

Teelo
01-16-2004, 07:43 PM
How do you do that? :rolleyes:

Q.E.D.
01-16-2004, 07:48 PM
As it happens, those very same isotopes of uranium and plutonioum emit fast neutrons when they decay naturally. So, you just put a whole bunch of them in close proximity to one another, and the neutrons that fly out of the naturally-decaying nuclei hit intact atoms, causing them to decay, releasing more more neutrons, which, in turn, hit still more intact atoms...ad nauseum.

This is how an atomic bomb works. And when the reaction is controlled, nuclear reactors.

Teelo
01-16-2004, 07:51 PM
Thanks Q.E.D., but I still think I have studying to do though. :)

Q.E.D.
01-16-2004, 08:22 PM
This site (http://atomicarchive.com/index.shtml) is chock full of nuclear goodness.

Hermitian
01-16-2004, 09:48 PM
It sounds like you are underestimating how many atoms are in a kilogram. If I am correct (which sometimes happens), there are approximately 2.53x10^24 atoms in a kilogram of Uranium. Thats 2530000000000000000000000 atoms. Thats a lot. Even a little bit of energy multiplied that many times is a lot of energy.

(4.2 moles of U per kg, times avogadro's number was my calculation)

Tierce
01-16-2004, 10:43 PM
And for hardcore physics, this site (http://nuclearweaponarchive.org/) is pretty cool too -- scroll to NWFAQ about two-thirds of the way down.

_xiao_wenti_
01-17-2004, 06:09 AM
I am still getting used to the new forum software...
Avogadro's number is used measure a mole of atoms. Carbon was the element of standard. 12 grams of carbon has avogadro's number of carbon atoms 6.023e23.

The amount of fissile material to go critical (that is to have a decay event cause more than one decay event) depends on the amount of the fertile isotope and the shape. Uranium contains two isotopes 238 (nonfissile) and 235 (fissile). IIRC the 235 is on the order of 2.5% of bulk uranium and has to be separated from the bulk (this is the enriching process). Most uranium for fuel and bombs is enriched. For fuels the highest enrichment that I know is for submarines -- maybe 80% enrichment.

In fission the uranium or plutonium will break down into two or more other atoms. The mass difference between the U/Pu and the products is used for the e=mc^2.

If there are errors, I apolgize in advance. I took my Nuclear Engineering class 20-some years ago.

EvilGhandi
01-17-2004, 06:41 AM
Back to the OP. For those of us who cannot calculate the number of atoms per gram of fissile material.

If it were possible to split a single atom, say a plutonium isotope, would the reaction be visible? What if it occured in a cloud chamber?

What would we SEE? if anything?

swansont
01-17-2004, 07:38 AM
Back to the OP. For those of us who cannot calculate the number of atoms per gram of fissile material.

If it were possible to split a single atom, say a plutonium isotope, would the reaction be visible? What if it occured in a cloud chamber?

What would we SEE? if anything?

In a cloud chamber? Yes (http://lateralscience.co.uk/cloud/diff.html)

Also in certain rocks/crystals (http://allserv.ugent.be/~pvdhaute/fission.htm) such as mica.

SlickRoenick
01-17-2004, 08:52 AM
What is a MeV an abbreviation of?

rsa
01-17-2004, 09:41 AM
Megaelectronvolt, i.e. a million electron volts.

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

Skeptico
01-17-2004, 10:16 AM
To put Avogrado's number in perspective, if a baseball was blown up to the size of the Earth, each atom making up the baseball would be the size of a grape. Imagine the whole earth filled with nothing but grapes, and try counting them. Now, that's a LOT of grapes. Even at this zoom level, the nucleus would still be invisible. You'd need to put the grape under a microscope to see the nucleus.

(Hey Hermitian, are you a transpose conjugate? ;) )

Hermitian
01-17-2004, 11:13 AM
(Hey Hermitian, are you a transpose conjugate? ;) )

Yes, I am my own adjoint. :D

Ring
01-17-2004, 11:24 AM
Because of the consequent reduction in the Coulomb repulsion energy itís energetically favorable for a U235 nucleus to split into less massive nuclei. This both increases the binding energy and decreases the potential energy.

However this doesnít normally occur because as the nucleus attempts to split, its surface area and therefore its surface energy increases imposing a potential barrier of 6 MeV. Spontaneous fission via Barrier penetration can occur but it is very rare, and if this were the only method of decay the lifetime of the nucleus would be ~ 1016 years.

A much more important process is induced fission via the capture of a slow neutron. This converts the nucleus from a fermion to a boson and releases binding energy of almost exactly 6 MeV. This high excitation energy often goes into violent oscillations in which the nucleus becomes sufficiently elongated to fission.

When fission occurs potential energy is converted to kinetic and radiation energy and a local mass deficit obtains. In other words the local reduction in mass is the result of the energy loss not the cause. Mass cannot be converted to energy.

gluteus maximus
01-17-2004, 03:15 PM
I have the answer!



boom

Daylon
01-18-2004, 06:26 AM
That's it? So this whole "splitting the atom" thing is just hype? :D


So then how much power is contained in a gram of matter? A kilogram?

Just how much uranium (or whatever modern nukes are made with) is in a nuclear bomb these days? I always figured it was just a few dozen Kg!


I'm sure there is someone on here who is more of an expert, BUT..

There's a great deal of power in a kilogram....it's my understanding that the bomb dropped on Hiroshima had a kilogram of uranium in it.. they suspect that only 100 grams or so was converted to pure energy, and we all know the result. If memory serves that was referred to as a "10 Kiloton" blast, or 10 thousand tons of TNT.

That's basically 100 paperclips.. I'm sure we can explode things more efficiently these days...

We have weapons today that push 100 MEGAton yields.


I believe that most nuclear bombs that we use (the USA) are basically fusion bombs. Instead of high explosives compressing a uranium sample to the point where uncontrolled fission occurs, we use weapons that start a fission reaction (in place of high explosive) in order to compress matter to the point where fusion occurs.

Basically fusion makes a bigger boom that fission.

Just my two cents..

D.

Q.E.D.
01-18-2004, 09:16 AM
it's my understanding that the bomb dropped on Hiroshima had a kilogram of uranium in it.. they suspect that only 100 grams or so was converted to pure energy, and we all know the result. If memory serves that was referred to as a "10 Kiloton" blast, or 10 thousand tons of TNT.
1) The Hiroshima bomb contained approximately 60 kg of enriched uranium.

2) The blast was estimated to be about 20 kT, not 10.

3) In the uranium fusion reaction, about 0.1% is converted to mass. For the Hiroshima bomb, this amounts to about .06 kg. Of course, as you note, not all the uranium was consumed in the reaction. It's estimated that about 4% was actually consumed, or 2.4 kg, which means that only about 2.4 g of matter caused that 20 kT blast.

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