NEET-XII-Physics
46: The Nucleus
- #34Natural water contains a small amount of tritium (
H13). 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
H13radioactivity as compared to a recently purchased bottle marked ‘8 years old’. Estimate the time of that unsuccessful attempt.Ans : Given:
Half-life time of tritium, `` {T}_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}`` = 12.5 years
Disintegration constant, `` \lambda =\frac{0.693}{12.5}\,\mathrm{\,per\,}\,\mathrm{\,year\,}``
Let A0 be the activity, when the bottle was manufactured.
Activity after 8 years `` \left(A\right)`` 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 `` \left(A\text{'}\right)`` on the mountain is given by
`` A\text{'}={A}_{0}{e}^{-\lambda t}``
Here, A' = (Activity of the bottle manufactured 8 years ago) × 1.5%
`` A\text{'}={A}_{0}{e}^{\frac{-0.693}{12.5}\times 8}\times 0.015...\left(2\right)``
`` ``
Comparing (1) and (2)
`` \frac{0.693}{12.5}t=\frac{-0.6931\times 8}{12.5}+\,\mathrm{\,In\,}[0.015]``
`` \Rightarrow \frac{-0.693}{12.5}t=\frac{-0.693}{12.5}\times 8-4.1997``
`` \Rightarrow 0.693t=58.040``
`` \Rightarrow t=83.75\,\mathrm{\,years\,}``
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