The radioactivity of a given sample of whisky | vedclass

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(D) The activity $A$ of a radioactive sample at time $t$ is given by $A = A_0 e^{-\lambda t}$ or $A = A_0 (1/2)^{n}$, where $n = t/T_{1/2}$ is the num

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$70$ years back

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(D) The activity $A$ of a radioactive sample at time $t$ is given by $A = A_0 e^{-\lambda t}$ or $A = A_0 (1/2)^{n}$, where $n = t/T_{1/2}$ is the number of half-lives.Given that the activity is $3\%$ of the reference sample, we have $3/100 = (1/2)^n$.Taking the logarithm on both sides: $\log_{10}(0.03) = n \log_{10}(0.5)$.$-1.5228 = n (-0.3010) \implies n \approx 5.06$.Since $n = t/T_{1/2}$, the time elapsed for the tritium decay is $t = n 	imes T_{1/2} \approx 5.06 	imes 12.3 \approx 62.2$ years.Adding the $7$ years age of the reference bottle, the total time since preparation is $62.2 + 7 \approx 69.2$ years, which is approximately $70$ years.

The radioactivity of a given sample of whisky due to tritium (half-life $12.3$ years) was found to be only $3\%$ of that measured in a recently purchased bottle marked "$7$ years old". The sample must have been prepared about:

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