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Sagot :
The percentage of the original sample which would remain after 17.5 days according to the given half-life of the isotope is; 22.06%.
What percentage of the original activity in the sample remains after 17.5 days?
The given sample in discuss has a half life of; 8.02 days, this therefore indicates that after 17.5 days, the number of half-life periods completed is; 17.5/8.02 = 2.18 half-lifes.
Hence, the percentage of the original sample remaining after 17.5 days can be evaluated as follows;
N = 100% × (0.5)^2.18
N = 100 × 0.22
N = 22.06%.
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