# Re: antihydrogen atoms

From: Michael M. Butler (butler@comp-lib.org)
Date: Sat Feb 23 2002 - 18:29:56 MST

> Roughly speaking, about how many kilotons per gram are we talking about
> here (mega, giga, peta?)

I seem to remember the figure of 21 Mt/kg, but I'll check.

New Mexico's WMD Preparedness site ( http://dem.state.nv.us/definitionsM.htm )
reminds me that a megaton is 4.18 * 10^15 Joules. Joules are kms units, so a
Joule is a Newton-meter; its units are kg*m^2/s^2.

Let's crank the handle for one kilogram:

e=mc^2; (1) * (2.998 * 10^8 m/s)^2 = 8.998 * 10^16 J ;

(8.998 * 10^16) / (4.18 * 10^15) = 21.50 Mt ; yup, just about spot on.

That's for total conversion of a mass totaling 1 kg--half of that is matter,
half antimatter.

Now, by "a gram", I assume you mean "a gram of antimatter", totally converted
by contact with matter (100% efficiency).

Divide by a thousand, then multiply by two (to account for the mass of the
_matter_ annnihilated); that gives us _43 kt per gram of antimatter_.

A big enough lump might be partly ejected by an initial blast, somewhat
mitigating the nominal 100% efficiency and thus the total local devastation.

```--
butler a t comp - lib . o r g
I am not here to have an argument. I am here as part of a civilization.
Sometimes I forget.
```

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