22.8 seconds
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.
The remainder is 2-p or 0.5p of the original amount.
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.
The answer depends on 3240 WHAT: seconds, days, years?
1/4
25
THE QUESTION IS INVALID. THE HALF-LIFE OF FLUORINE-20 is 11.07 SECONDS. There will be about 0.92 of of an isotope with a half-life of 114 seconds remaining after 14 seconds. The equation for half-life decay is AT = A0 2(-T/H) where T is time and H is half-life. 2(-14/114) is equal to about 0.92. In the case of fluorine-20, 2(-14/11.07) is about 0.42.
After 4 days, only 1/16 of the original amount of gold (198/16 = 12.375) will remain.
After 1 year, 50% of the original amount of cobalt-60 will remain. This means that 50% will decay and 50% will be left. After 4 years, 6.25% of the original amount (50% of 50%) of cobalt-60 will remain.
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.
5.4
The remainder is 2-p or 0.5p of the original amount.
After 2 half-lives, 25% of the original amount of thorium-234 will remain. This is because half of the substance decays in each half-life period.
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1/4 of the amount that was originally there will remain. Note that the rest doesn't simply evaporate, it will decay into something else.
After 14 years, 1/16th of the original amount of cobalt-60 will remain, because 14 years is equivalent to 2.64 half-lives of cobalt-60 (14 years / 5.3 years/half-life). Each half-life reduces the amount of cobalt-60 by half, so after 2.64 half-lives, the original amount will be reduced to 1/2^2.64 which is approximately 1/16th.
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