i m a newbie in nuclear field. i m just wondering how difficult plutonium is produced? is it that u just irradiate U-238 with a neutron and u will get a Pu-239? is there any constraint or technical difficulty for that?
Well, the difficulty is in that you have to do it inside a nuclear reactor. Then you have to separate out the Pu.
actually i was given a question that sound like this. using a portable neutron generator which generates 14 MeV neutrons at 1 * 10^12 neutrons/second via the D-T fusion reaction, how long does it take to produce 20kg of Pu-239? I am just wondering is that possible?
A major unknown in the problem you stated is how much U238 do you have and its geometry with respect to the neutron source. Also are there other materials present which could absorb some of the neutrons? To give you a start use Avogadros no. (approx 6x10^{23}) to get the number of atoms in 238 grams of U238.
thank you mathman. the quantity of U-238 is not given, i was only asked how long will it take to produce 20-kg of Pu-239. based on the equation U-238 + n -> U-239 -> Np-239 -> Pu-239 i interpreted it as one mole of U-238 when irradiated by 1 neutron will produce one mole of Pu-239. And so, to produce 20kg of Pu-239 (5.04*10^25) u will need 5.04*10^25 of U-238. Is that thought correct here? please help. thanks.
oops sorry, i forgot to mention just now. i was asked to approach this problem based on the assumption that all neutron is absorbed in the fertile material. so i guess there should be no other material which will absorb the neutron. of course i can be wrong. i m just a newbie.
I'm going to merge this thread with the identical one in homework help. nuclear, If you're going with the assumption that every neutron produced is involved in the reaction, then this problem is really just plug and chug. You know how much Plutonium you need to produce, so with Avogadro's number and the mass, you can get the total number of neutrons required. Once you have that number, you use the rate to determine the time it took to get them. Does that help any?