The radioactivity of a given sample of whisky | vedclass

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(D) The activity $A$ at time $t$ is given by $A = A_0 (1/2)^n$, where $n$ is the number of half-lives.Given that the activity is $3\%$ of the original

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$70 	ext{ years}$ back

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(D) The activity $A$ at time $t$ is given by $A = A_0 (1/2)^n$, where $n$ is the number of half-lives.Given that the activity is $3\%$ of the original, we have $0.03 = (1/2)^n$.Taking the logarithm, $\ln(0.03) = n \ln(0.5)$, which gives $n = \ln(0.03) / \ln(0.5) \approx 5.06$.Thus, the time elapsed for the tritium decay is $t = n 	imes T_{1/2} = 5.06 	imes 12.5 	ext{ years} \approx 63.25 	ext{ years}$.Since the reference bottle was already $7 	ext{ years}$ old, the total age of the sample is $63.25 + 7 \approx 70.25 	ext{ years}$.Therefore, the sample was prepared about $70 	ext{ years}$ back.

The radioactivity of a given sample of whisky due to tritium (half-life $12.5 \text{ years}$) was found to be only $3\%$ of that measured in a recently purchased bottle marked "$7 \text{ years}$ old". The sample must have been prepared about:

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