10.6 years
After twice the half-life, 1/4 of the original amount remains. After 3 times the half-life, 1/8 of the original amount remains. Etc.
The half life of Tritium is 12.32 years. it would therefore take 24.64 years for the amount to fall to a quarter of the original.
Carbon has a half life of 5730 years. That means that after 5730 years there'll only be half the amount left. After about 11500 there will be a quarter of the original. After 75000 years there will be about 0.0001 of the original sample which is too small to date accurately?
100 grams
Cobalt like iron and many other metals occurs mainly in ore form and has been used in compound form to colour glass and pottery for over 2,000 years. However Cobalt-60 a radioactive isotope is artificiality produced from the stable common isotope Cobalt 59
After twice the half-life, 1/4 of the original amount remains. After 3 times the half-life, 1/8 of the original amount remains. Etc.
This seems to refer to radioactive decay. The answer would depend on the isotope of cobalt used! For example, cobalt-59 is stable, so in this case, all of the original cobalt would remain.For more information, check the Wikipedia article entitled "Isotopes of cobalt".
The half life of Cobalt-60 is 5.27 years, so 32 years is almost exactly six half lives. (Which is probably why the question is put this way). Every half life the activity halves, so after six half lives it is reduced by a factor 26, which is 64. Therefore the activity after six half lives is 1/64 of the original level, or 1.56 percent.
Cobalt-60 has an 1/2 life of 5.24 years
1/32 of the original amount.
5.27 years.
Half-life of Co-60 = 5,27 year by beta- emission (0.315 MeV max.)
The half-life of 27Co60 is about 5.27 years. 15.8 years is 3 half-lives, so 0.53 or 0.125 of the original sample of 16 g will remain, that being 2 g.
The half life of Tritium is 12.32 years. it would therefore take 24.64 years for the amount to fall to a quarter of the original.
The equation for half-life is ... AT = A0 2 (-T/H) ... where A0 is the starting activity, AT is the activity at some time T, and H is the half-life, in units of T. 55134Cs has a half-life of 2.0652 years. Plugging in the known values, we get ... AT = 5.8 2 (-11.5/2.0652) AT = 5.8 2 -5.5685 AT = 0.12222
Of Course It Does as it is the hair one was born with - once this is shed the actual texture is reveal - the remnants of the original texture remain on some adults for years ( hairline, nape ) if they allow it to remain untouched
3 to 6 years