also, uranium’s half life is 700 million years, so we expect (207/235)*7.5 (of lead) + 7.5 (uranium) ~ 14.106382978723405 lump.
also, a lot of the helium produced will remain trapped inside (most heavy metal lumps act as sponges for little gasses). but 700 mil years is also a large amount of time, so much of it would diffuse out. I could checkup diffusion statistics for he d pb-u but i would have to probably do a double integral (as pb-u combination is not fixed, and we can not simply do the error function calculation), so skipping that. but it is safe to say that we will have a lump of ~50% U, 44% pb, and 6% He (by mass), and a significant amount of he will remain in
also, uranium’s half life is 700 million years, so we expect (207/235)*7.5 (of lead) + 7.5 (uranium) ~ 14.106382978723405 lump.
also, a lot of the helium produced will remain trapped inside (most heavy metal lumps act as sponges for little gasses). but 700 mil years is also a large amount of time, so much of it would diffuse out. I could checkup diffusion statistics for he d pb-u but i would have to probably do a double integral (as pb-u combination is not fixed, and we can not simply do the error function calculation), so skipping that. but it is safe to say that we will have a lump of ~50% U, 44% pb, and 6% He (by mass), and a significant amount of he will remain in
So it would be more accurate to say that 13.2 lbs would be a minimum for the lump’s mass.