4 ms·
That much Californium in one place would probably go critical anyway. It might not explode, but it would glow and evaporate rather quickly.
by ars 7y ago
That much Californium in one place would probably go critical anyway. It might not explode, but it would glow and evaporate rather quickly.
- petschge 7y agoCritical mass of Cf-249 is 6kg, of Cf-251 5.46kg and of Cf-252 is 2.73kg. So no it would not go critical. 1.4 kg of Cf-252 would however emit something like 4e15 neutrons per second and 1.2kW (I first thought it was 700kW) of decay heat. Glowing and evaporation would be rather likely.
- LeifCarrotson 7y ago700 kW emitted by a grapefruit-sized object will be a white-hot, rapidly-expanding gas cloud. Maybe it's not a nuke, but I don't know that your decay heat math is correct...half a critical mass of uranium or plutonium is slightly above room temp, not something that violently disintegrates. Is Cf252 different?
- petschge 7y agoYou are right, the 700kW was wrong. The calculation I did was as follows (and I will point out the mistake I made): 1.4 kg is 1.4e9 micrograms and as per [1] "one microgram emits 170 million neutrons per minute" will emit 1.4e9 * 1.7e8 / 60 = 4e15 neutrons per second. Decays of Cf-252 have a high neutron multiplicity, about 3.7 per decay, so the neutron came from 1.1e15 decays per second. Most of those (> 95%) came from the decat Cf-252 -> Cm-248. The masses of those two isotopes is 252.081626 and 248.072349 atomic mass units respectively, as per [2] and [3]. An alpha nucleus of 4.001506179127 u is also produced. (If forgot about that one, vaastly overestimating the decay heat.) A mass loss of 7.77e-3 u is equivalent to the release of 7.77e-3 * 0.9314941 GeV / u = 7.2 MeV. Each MeV is 1.60e-13 J. So in total we get 1.1e15 decays per second * 7.2 MeV per decay * 1.60e-13 J / MeV = 1.2 J/s = 1.2 kW. If you forget that the alpha nucleus exists you over estimate the mass loss by a factor 500 and get a decay power of 700kW... [1] http://nuclearweaponarchive.org/Nwfaq/Nfaq6.html#nfaq6.2 http://nuclearweaponarchive.org/Nwfaq/Nfaq6.html#nfaq6.2 [2] https://en.wikipedia.org/wiki/Isotopes_of_californium https://en.wikipedia.org/wiki/Isotopes_of_californium [3] https://en.wikipedia.org/wiki/Isotopes_of_curium https://en.wikipedia.org/wiki/Isotopes_of_curium
- Jenda_ 7y agoI think you are underestimating the number of decays. You have 5.5 moles of the substance, that is, 5.5 * 6e23 atoms. The half-life of Cf-252 is 2.6 years. You therefore need at least 5.5 * 6e23/2.6/365/24/3600/2 = 20e15 decays per second to decay half of that (it will be actually more because it is exponential). (as a naive crosscheck, WolframAlpha for the query "activity of 1.4 kg Cf-252" thinks the same)
- petschge 7y agoIt is possible that the neutron emission number I found and used includes self shielding or is simply wrong. if the activity is 2.78e16 decays per second and not just 1.1e15/s then the decay heat would be about 30kW. But to be honest I am not sure if it is worthwhile to keep fixing bugs in this estimate. We know several good reason why there is no 1.4kg of Californium anywhere.
- russfink 7y agoI am awestruck at how smart people on this forum are. Many mod points for your calculations!