Question

Natural water contains a small amount of tritium ${(}_1{\text{H}}^3).$ This isotope $\beta -$decays with a half-life of $12.5\text{years}$. 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 $1.5\%$ of the ${}_1{\text{H}}^3$ radioactivity as compared to a recently purchased bottle marked $8\text{years old}$, estimate the time of that unsuccessful attempt. $(\ln 0.015=-4.1997)$

• A. $83.75\text{years}$
• B. $62.5\text{years}$
• C. $45\text{years}$
• D. $35.2\text{years}$

Answer 1

The correct option is A $83.75\text{years}$

The activity when the bottle was manufactured $=A_0$

By the law of radioactivity,

$A=A_0{e}^{-\lambda t}$
Activity after $8$ years, $A_8=A_0{e}^{\frac{-0.693}{12.5}\times 8}$

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

$A_t=A_0{e}^{\frac{-0.693}{12.5}\times t}$

Given that,

$A_t=1.5\%\times A_8$

$\Rightarrow A_0{e}^{\frac{-0.693}{12.5}\times t}=A_0{e}^{\frac{-0.693}{12.5}\times 8}\times 0.015$

Taking $\ln$ both side,

$\Rightarrow \frac{-0.693}{12.5}t=\frac{-0.693\times 8}{12.5}+\ln \left[0.015\right]$

$\Rightarrow 0.05544t=0.44352+4.1997$

$\Rightarrow t=83.75\text{years.}$

Hence, option $\left(A\right)$ is correct.

Why this question? To give an idea that how to solve real-life numericals.