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Question
Natural water contains a small amount of tritium (`""_1^3H`). 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 `""_1^3H` radioactivity as compared to a recently purchased bottle marked '8 years old'. Estimate the time of that unsuccessful attempt.
Solution
Given:
Half-life time of tritium, `T_"1/2"` = 12.5 years
Disintegration constant, `lambda = 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_0e^((-0.693)/(12.5) xx 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_0e^(-lambdat)`
Here, A' = (Activity of the bottle manufactured 8 years ago) × 1.5 %
`A' = A_0e^((-0.693)/(12.5) xx 8) xx 0.015` ...(2)
Comparing (1) and (2)
`(-0.693)/12.5 t = (-0.6931 xx 8)/12.5 + "In" [0.015]`
⇒ `(-0.693)/12.5 t = (-0.693)/12.5 xx 8 - 4.1997`
⇒ `0.639 t = 58.040`
⇒ t = 83.75 years
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