The remainder is 2-p or 0.5p of the original amount.
Roughly a third.
1g (1/2)4 = 1/16 g
Age = no. of half lives x duration of each half life ratio of Parent nuclei : Daughter Nuclei 1 : 15 total of 16 parts, and for every 16 parts one part is a parent nuclei Mt = Mo x (0.5)^n Mt = amount remaining Mo = Initial amount n = number of 1/2 lives Mt = 1/16 (1/2)^n = 1/16 (1/2)^n = (1/2)^4 n=4 Age = 4 x duration of each half life Age = 4 x 300 Age = 1200 Therefore, the age of the specimen is 1200 years.
To calculate the remaining amount of cobalt-60 after 21.2 years, we can use the half-life formula. Since the half-life is 5.3 years, we find the number of half-lives in 21.2 years by dividing 21.2 by 5.3, which is approximately 4.0 half-lives. After 4 half-lives, the remaining amount can be calculated as (10.0 , \text{g} \times \left(\frac{1}{2}\right)^4 = 10.0 , \text{g} \times \frac{1}{16} = 0.625 , \text{g}). Thus, 0.625 g of cobalt-60 will remain after 21.2 years.
To determine the remaining amount of a 200 gram sample after 36 seconds with a half-life of 12 seconds, we first calculate how many half-lives fit into 36 seconds. There are three half-lives in 36 seconds (36 ÷ 12 = 3). Each half-life reduces the sample by half: after the first half-life, 100 grams remain; after the second, 50 grams; and after the third, 25 grams. Therefore, 25 grams of the sample would remain after 36 seconds.
Approx 1/8 will remain.
12.5%
Not sure what you mean by "had-lives". After 3 half lives, approx 1/8 would remain.
Massive nuclei are unable to remain bound aganst the repulsive force of their protons, which all have positive charge. This also occurs with elements #43 (technetium) and #61 (promethium) which because of their particular nuclear geometry are inherently unstable and radioactive, with the only natural isotopes having half-lives of just over 2 years.
1/8 of the original amount remains.
After seven half lives, approximately 0.78125% (1/2^7) of the original radioactive element will remain. This can be calculated by repeatedly halving the remaining amount after each half life.
12.5%
After three half-lives, only 1/8 (or 12.5%) of the original radioactive sample remains. This is because each half-life reduces the amount of radioactive material by half, so after three half-lives, you would have (1/2) * (1/2) * (1/2) = 1/8 of the original sample remaining.
That depends on the "half-life" of that particular radioactive element, which the question forgot to state. They're all different. Various radioactive elements have half-lives ranging from microseconds to millions of years.
After 5 half-lives, 3.125% (or 1/2^5) of a radioactive sample remains. Each half-life reduces the sample by half, so after 5 half-lives, there is only a small fraction of the original sample remaining.
Only 1/32 of the original radioactive material will remain. (½)5 = 1/32
If I take a radioactive sample of 400 moles of an unknown substance and let it decay to the point of three half-lives I would have 50 moles left of the sample. 1/2 of what is left will decay in the next half-life. At the end of that half-life I will have 25 moles left of the unknown substance or 4/25.