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.
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.
To solve this problem, you can use the formula for radioactive decay: N(t) = N0 * (1/2)^(t/T), where N(t) is the amount of the isotope remaining at time t, N0 is the initial amount, t is the elapsed time, and T is the half-life. Plugging in the values: 1.50 g = 12.0 g * (1/2)^(t/8.07), solve for t to find the time it takes for the iodine-131 to decay to 1.50 g.
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.
After 60 years, there will be 2.5 grams of hydrogen-3 remaining from the original 80.0 grams sample. This is calculated by dividing the time elapsed by the half-life multiple times (60 years รท 12 years = 5 half-lives, 1/2^5 = 1/32). Multiplying 80.0 grams by 1/32 gives 2.5 grams remaining.
Assuming that "half" refers to "half-life", 360/60 = 6 so fraction left = 1/26 = 1/64
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.
To calculate the amount of thorium remaining after 2 half-lives, you use the formula: amount = initial amount * (1/2)^n, where n is the number of half-lives. If we assume the initial amount is 1 gram, after 2 half-lives, there would be 0.25 grams of thorium remaining.
After 2,000,000 years, only about 2.48% of chlorine-36 would remain undecayed. This is because chlorine-36 has a half-life of about 301,000 years, so after 2,000,000 years, multiple half-lives would have passed.