Patrick D. Rockwell

2010-09-17 00:42:15 UTC

I've long been interested in probability. Some time ago, I read a

formula for the probability

of a chain reaction is given by

1.5-0.5*((12/k)-3)^0.5.

I got this at the following site, not to mention others.

http://www.newworldencyclopedia.org/entry/Nuclear_reaction

I'd like to know just how this formula is derived. From what I've

read, k

is the average number of neutrons which exit a fission, and in the

above

formula the allowable valuse for k are 1, 2, or 3. Couldn't k have

values outside

that range? Is there a way to make this formula more general? If not,

why? I should

think that it would be possible to make up a hypothetical problem in

probability where

the number of neutrons which exit the fission of the atom in question

is higher than

3.

Any info or insight on this is appreciated. :-)

formula for the probability

of a chain reaction is given by

1.5-0.5*((12/k)-3)^0.5.

I got this at the following site, not to mention others.

http://www.newworldencyclopedia.org/entry/Nuclear_reaction

I'd like to know just how this formula is derived. From what I've

read, k

is the average number of neutrons which exit a fission, and in the

above

formula the allowable valuse for k are 1, 2, or 3. Couldn't k have

values outside

that range? Is there a way to make this formula more general? If not,

why? I should

think that it would be possible to make up a hypothetical problem in

probability where

the number of neutrons which exit the fission of the atom in question

is higher than

3.

Any info or insight on this is appreciated. :-)