3 ms·
You 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] "
by petschge 7y ago
You 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!