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 three half-lives, 12.5% of the original radioactive material will remain. Each half-life reduces the amount of material by half, so after three half-lives the remaining material will be 0.5^3 = 0.125 or 12.5%.
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.
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.
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.
Using the formula Nt = N0*(1/2)t/t1/2 where Nt is the amount of stuff remaining after an amount of time, t, and t1/2 is the half-life, you get Nt = .036N0. So about 3.6% of the radioactive stuff is left.
After 5 half-lives, 3.125% of the original carbon-14 would remain. Each half-life reduces the amount by half, so after 5 half-lives, it would be reduced by a factor of 2^5 = 32.
Cobalt-60 has a half-life of approximately 5.27 years, meaning that after this period, half of the original amount will have decayed. After 14 years, which is about 2.65 half-lives, the remaining amount can be calculated using the formula: remaining amount = original amount × (1/2)^(time/half-life). Therefore, after 14 years, approximately 1/6 of the original amount of cobalt-60 will remain.
After 2 half-lives, you would have 25% of the original amount remaining. Each half-life reduces the amount by half, so after two half-lives, you would have 25% left (50% reduced by half twice).
After two half-lives, only one-fourth (1/2 * 1/2 = 1/4) of the original material remains. Each half-life reduces the amount by half, so after two half-lives, the remaining material is one-fourth of the original amount.
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.
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 three half-lives, 12.5% of the original radioactive material will remain. Each half-life reduces the amount of material by half, so after three half-lives the remaining material will be 0.5^3 = 0.125 or 12.5%.
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.
After four half-lives, the amount of Carbon-14 remaining would be reduced to ( \frac{1}{16} ) of the original quantity, since each half-life halves the remaining amount. Given that the half-life of Carbon-14 is 5,700 years, four half-lives would total ( 4 \times 5,700 = 22,800 ) years. Thus, after 22,800 years, only a quarter of the original Carbon-14 remains.
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.
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.
1/8 of the original amount remains.