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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Rewritten by : Barada

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.