3 ms·
I 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
by Jenda_ 7y ago
I 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!