1/24 = 1/16
Assuming that "half" refers to "half-life", 360/60 = 6 so fraction left = 1/26 = 1/64
3/8
6 hours = 2 half lives, thus 25 % would remain. 0.25 x 2 mg = 0.5 mg.Done another way...fraction remaining = 0.5^n where n = number of half lives = 6hr/3hr = 2fraction remaining = 0.5^2 = 0.250.25 x 2 mg = 0.5 mg
1/2 (50%) of them.
There is no possible answer. For any given fraction, half that fraction is another fraction and it will be closer to 0. And then half of that will be closer still, and then half of that ... . Hope you get the idea.
After 5 half-lives, 31.25% of the original iodine remains, meaning 68.75% has decayed. Each half-life reduces the amount of the original substance by half, so after 5 half-lives, the remaining fraction is ( \left(\frac{1}{2}\right)^5 = \frac{1}{32} ). Therefore, the decay percentage is calculated as ( 1 - \frac{1}{32} = \frac{31}{32} ), which is approximately 96.875%.
If a radioisotope undergoes six half-lives, only (1/64) or (0.015625) of the original radioisotope remains, because half of the remaining material decays at each half-life.
half life is 8.1 days, so it takes 8.1 days for half the iodine sample to decay. It takes another 8.1 days for half of the remaining sample (ie. 1/4th of the original sample) to decay. So it takes 16.2 days for 3/4th of the sample to decay.
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.
After three half-lives, 12.5% of the radioactive isotope is remaining. This is because each half-life reduces the amount of radioactive material by half.
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.
Assuming that "half" refers to "half-life", 360/60 = 6 so fraction left = 1/26 = 1/64
Fraction remaining = 0.5^n where n = # of half lives that have elapsed60 yrs x 1 half life/12 yrs = 5 half lives have elapsed Fraction remaining = 0.5^5 = 0.03125 mass remaining = 0.03125 x 80.0 g = 2.5 g remaining
Carbon-14 has a half-life of about 5,730 years. After 18,000 years, which is approximately three half-lives (5,730 x 3 = 17,190), the remaining fraction of carbon-14 can be calculated using the formula ( \left(\frac{1}{2}\right)^n ), where ( n ) is the number of half-lives. Therefore, after three half-lives, the fraction of carbon-14 remaining is ( \left(\frac{1}{2}\right)^3 = \frac{1}{8} ). Thus, about 12.5% of the original carbon-14 would still remain.
After two half lives, 25% of the original carbon-14 would remain. This is because half of the remaining carbon-14 decays during each half life, leaving you with 50% after one half life and 25% after two half lives.
After 32 days, approximately 5 milligrams of the 80-milligram sample of Iodine-131 would be left. Iodine-131 has a half-life of about 8 days, so after each 8-day period, half of the remaining sample will decay.
91.16% of the daughter product has formed after 3.5 half lives.