Question

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.

Answer 1

The activity when the bottle was manufcatured = $A_0$

Activity after 8 years

= $A_0{e}^{\frac{-0.653}{12.5}\times 8}$

Let the time of the mountaineering = t years from the present

A = $A_{0}e\frac{-0.653}{12.5}\times 8$ ...(1)

A = Acitivity of the bottle found on the mountain

$A_0$ = (Activity of the bottle manufactured 8 years before) $\times$ 1.5 %

$A=A_0{e}^{\frac{0.6931}{12.5}\times 8}\times \frac{1.5}{100}$

$=A_0{e}^{\frac{0.693}{12.5}\times 8}\times 0.015$ ...(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

$\Rightarrow$ 0.05544 t = 0.44352 + 4.1997

$\Rightarrow$ t = 83.75 years.