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.
After each half-life, the number of undecayed nuclei is halved. Starting with 600 nuclei, after one half-life, 300 would remain; after the second half-life, 150 would remain; and after the third half-life, 75 would remain. Thus, after three half-lives, 75 undecayed headsium nuclei would remain in the sample.
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