Natural water contains a small amount of tritium ( ${}_{1}^{3}H$ ). This isotope beta-decays with a half-life of 12.5 years. A mountaineer while climbing towards a difficult peak finds debris of some earlier unsuccessful attempt. Among other things he finds a sealed bottled of whisky. On returning, he analyses the whisky and finds that it contains only 1.5 per cent of the ${}_{1}^{3}H$ radioactivity as compared to a recently purchased bottle marked '8 years old'. Estimate the time of that unsuccessful attempt.

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Solution

Given:
Half-life time of tritium, $T_{\frac{1}{2}}$ = 12.5 years
Disintegration constant, $λ = \frac{0.693}{12.5} per year$
Let A₀ be the activity, when the bottle was manufactured.
Activity after 8 years $(A)$ is given by
$A = A_{0} e^{\frac{- 0.693}{12.5} \times 8}$ ...(1)
Let us consider that the mountaineering had taken place t years ago.
Then, activity of the bottle $(A')$ on the mountain is given by
$A' = A_{0} e^{- λ t}$
Here, A' = (Activity of the bottle manufactured 8 years ago) × 1.5%
$A' = A_{0} e^{\frac{- 0.693}{12.5} \times 8} \times 0.015 . . . (2)$
Comparing (1) and (2)
$\frac{0.693}{12.5} t = \frac{- 0.6931 \times 8}{12.5} + In [0.015] \Rightarrow \frac{- 0.693}{12.5} t = \frac{- 0.693}{12.5} \times 8 - 4.1997 \Rightarrow 0.693 t = 58.040 \Rightarrow t = 83.75 years$

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Natural water contains a small amont of tritium $(_{1}^{3} H)$ . This isotope beta-decays with a half-life of 12.5 years. A mountaineer while climbing towards a difficult peak finds debris of some earlier unsuccessful attempt. Among other things he finds a sealed bottle of whisky. On return he analyses the whisky and finds that it contains only 1.5 per cent of the $(_{1}^{3} H)$ . radioactivity as compared to a recently purchased bottle marked '8 years old'. Estimate the time of that unsuccessful attempt.