# What is the explosive power of 1 atom?

**URL:** <https://boards.straightdope.com/t/what-is-the-explosive-power-of-1-atom/224861>\
**Category:** Factual Questions\
**Created:** [January 17, 2004, 12:07am UTC](https://boards.straightdope.com/t/what-is-the-explosive-power-of-1-atom/224861 "2004-01-17T00:07:02Z")\
**Posts on this page:** 7\
**Page:** 2

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**Author:** ![rsa](https://avatars.discourse-cdn.com/v4/letter/r/839c29/32.png) [@rsa](https://boards.straightdope.com/u/rsa)\
**Post date:** [January 17, 2004, 3:41pm UTC](https://boards.straightdope.com/t/what-is-the-explosive-power-of-1-atom/224861/21 "2004-01-17T15:41:40Z")

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Megaelectronvolt, i.e. a million electron volts.

[http://en.wikipedia.org/wiki/Electron\_volt](http://en.wikipedia.org/wiki/Electron_volt)

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**Author:** ![Skeptico](https://avatars.discourse-cdn.com/v4/letter/s/e19b73/32.png) [@Skeptico](https://boards.straightdope.com/u/Skeptico)\
**Post date:** [January 17, 2004, 4:16pm UTC](https://boards.straightdope.com/t/what-is-the-explosive-power-of-1-atom/224861/22 "2004-01-17T16:16:45Z")

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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? 😉 )

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**Author:** ![Hermitian](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/hermitian/32/470_2.png) [@Hermitian](https://boards.straightdope.com/u/Hermitian)\
**Post date:** [January 17, 2004, 5:13pm UTC](https://boards.straightdope.com/t/what-is-the-explosive-power-of-1-atom/224861/23 "2004-01-17T17:13:30Z")

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> [@Skeptico](#):
>
> (Hey Hermitian, are you a transpose conjugate? 😉 )

Yes, I am my own adjoint. 😃

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**Author:** ![Ring](https://avatars.discourse-cdn.com/v4/letter/r/6a8cbe/32.png) [@Ring](https://boards.straightdope.com/u/Ring)\
**Post date:** [January 17, 2004, 5:24pm UTC](https://boards.straightdope.com/t/what-is-the-explosive-power-of-1-atom/224861/24 "2004-01-17T17:24:51Z")

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Because of the consequent reduction in the Coulomb repulsion energy it’s energetically favorable for a U[sup]235[/sup] nucleus to split into less massive nuclei. This both increases the binding energy and decreases the potential energy.

However this doesn’t 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 ~ 10[sup]16[/sup] 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.

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**Author:** ![gluteus\_maximus](https://avatars.discourse-cdn.com/v4/letter/g/f17d59/32.png) [@gluteus\_maximus](https://boards.straightdope.com/u/gluteus_maximus)\
**Post date:** [January 17, 2004, 9:15pm UTC](https://boards.straightdope.com/t/what-is-the-explosive-power-of-1-atom/224861/25 "2004-01-17T21:15:13Z")

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**I have the answer!**

boom

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**Author:** ![Daylon](https://avatars.discourse-cdn.com/v4/letter/d/45deac/32.png) [@Daylon](https://boards.straightdope.com/u/Daylon)\
**Post date:** [January 18, 2004, 12:26pm UTC](https://boards.straightdope.com/t/what-is-the-explosive-power-of-1-atom/224861/26 "2004-01-18T12:26:55Z")

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> [@badmana](#):
>
> That’s it? So this whole “splitting the atom” thing is just hype? 😃  
> 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.

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**Author:** ![Q.E.D](https://avatars.discourse-cdn.com/v4/letter/q/51bf81/32.png) [@Q.E.D](https://boards.straightdope.com/u/Q.E.D)\
**Post date:** [January 18, 2004, 3:16pm UTC](https://boards.straightdope.com/t/what-is-the-explosive-power-of-1-atom/224861/27 "2004-01-18T15:16:25Z")

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> [@Daylon](#):
>
> 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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