Re: How much energy comes from matter/antimatter annihilation, anyway?

Hi Isaac,

  For the first part of your question, I'm going to direct you to another 
post I made about matter/antimatter annihilation: http://
www.madsci.org/posts/archives/may2000/957365745.Ph.r.html

  In short, without looking at that page, it's Einsteins famous:

  E=m*c^2

  that lets you calculate how much energy is emitted upon annihilation.

  An electron/positron annihilation would emit about 2MeV (2 million 
eV..... 1eV is 1.6 x 10^19 Joules.)

  The second part of your question I'll answer here. It's not really joules 
that are harmful to humans. (Though there is some debate about this. For 
example, if a DNA molecule is destroyed by a photon (energy carrier), how 
many DNA destructions does it take before forming a cancer is likely? Some 
say just one would do, others say that the DNA has a sort of repair 
mechanism that makes it need more than one photon hit.)

If you're talking about something on a less sub-cellular level, like 
burning your hand, then instead of talking about Joules, we should talk 
about Power density. Power density is the amount of energy (Joules) 
delivered to a unit area (inch squared) in a unit time (seconds).

It also then depends on what energy the Power is at. For example, if you 
shone a laser, like those in CD players but more powerful, on a piece of 
glass, you wouldn't see much, because the glass wouldn't absorb much of the 
energy, it would mostly come out of the other side of the glass. So it 
would take a LOT of power to melt the glass. However, if you shone the 
laser on something that absorbed the wavelength (energy) of the photons in 
the light really well, then you wouldn't need the laser to be as powerful 
for the material to absorb the same amount of energy.

So to answer your question, the amount of Joules varies. I know that I have 
hurt myself by accidentally putting my finger in the beam of a laser that 
had power density of around 1000 Watts per square inch at a wavelength of 
roughly 350nm (3.5eV). But I didn't leave my finger in there for too long, 
I can tell you. But as I said, it would depend on the energy of the photon,
and also the power density.

Sorry I couldn't be more help.