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Answer :
Final answer:
The half-life of Au-198 is 2.7 days, determined by dividing the total time decayed (10.8 days) by the number of half-lives occurred (which is 4, as 100 grams decayed to 6.25 grams). (2^4 = 16)
Explanation:
The half-life of an isotope is defined as the time it takes for half of the atoms in a sample to decay. In this problem, the original mass of the Au-198 was 100 grams and it decayed to 6.25 grams in 10.8 days. Since 6.25 is 1/16th of 100, this means the Au-198 had to undergo 4 half-lives (since 2^4 equals 16). Therefore, the half-life of Au-198 can be calculated by dividing the total time elapsed (10.8 days) by the number of half-lives (4). The result is 2.7 days, which is the half-life of Au-198.
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Answer: 2.7 days
Explanation:
You'd divide 100 g by 2 until you reach 6.25 g:
100/2
=
50
;
50/2 (fraction
=
25
;
25/2(fraction
=
12.5
;
12.5/2 (fraction
=
6.25
Since there were four 2s, divide 10.8 days by 4:
10.8 days
4
=
2.7 days
100 g of Au-198 would have a half life of 2.7 days.