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* **A*_{0} 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