A material has a half-life of 3 years. there is initially 12 grams of this material, when is there 1 gram left

Answer 1

F = i (x) ^ t / T

F = Final amount
i = Initial amount
x = Half life
t = Total time for the entire process
T = Time for the half life

So, all we have to do is plug in what we know. Then, we can use logs to solve for "t"

1 = 12(0.5)^t/3
1/12 = 0.5^t/3
log (1/12) = log 0.5^t/3
log (1/12) = t/3 log 0.5
3log (1/12) = t log 0.5
[3log (1/12)] / log 0.5 = t
10.8 years = t

Answer 2

11

Answer 3

ok, if the a million/2 existence of a few thing is one thousand years, this implies that when one thousand years you will have one a million/2 of the quantity you began with, as a result the 4 hundred grams, in a single thousand years, would be 2 hundred grams. So, at one thousand years you have 2 hundred grams of the stuff, and in yet another one thousand years, one a million/2 of this would be long previous, as a result at 2000 years (one a million/2 existence + yet another) you will have a hundred grams of the stuff. And in case you think of approximately yet another a million/2 of this going away (this could be the 0.33 a million/2 existence) you will have 50 grams, as a bring about yet yet another a million/2 existence (fourth a million/2 existence) we can be right down to twenty-5 grams of the stuff. And the fourth a million/2 existence will happen at 4000 years (4 a million/2 lives at one thousand years consistent with a million/2 existence = 4000 years). wish this facilitates. in case you like extra advantageous than this, like the truthfully algebra or something, please repost. identity be chuffed to assist ya out!!

Answer 4

Use the formula for exponential decay.

Let
t = time in years

1 = 12(1/2)^(t/3)

12(1/2)^(t/3) = 1
(1/2)^(t/3) = 1/12
ln[(1/2)^(t/3)] = ln(1/12)
(t/3)ln(1/2) = -ln(12)
(t/3)[-ln(2)] = -ln(12)
t/3 = ln(12) / ln(2)
t = 3ln(12) / ln(2) ≈ 10.75 years

Answer 5

after five and a half years