Chemical Kinetics - Result Question 73
72. The nucleidic ratio, ${ } _1^{3} H$ to ${ } _1^{1} H$ in a sample of water is $8.0 \times 10^{-18}: 1$. Tritium undergoes decay with a half-life period of $12.3$ $ yr$. How many tritium atoms would $10.0$ $ g$ of such a sample contain $40 $ $yr$ after the original sample is collected.
(1992, 4M)
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Answer:
Correct Answer: 72. $(5.6 \times 10^{5})$
Solution:
- Initially :
$ \begin{aligned} & N\left({ } _1 H^{3}+{ } _1 H^{1}\right) =\frac{10}{8} \times 2 \times 6 \times 10^{23}=\frac{20}{3} \times 10^{23} \\ \Rightarrow \quad & 1+\frac{N\left({ } _1 H^{1}\right)}{N\left({ } _1 H^{3}\right)} =\frac{20 \times 10^{23}}{3 N\left({ } _1 H^{3}\right)} \\ \Rightarrow \quad & 1+\frac{1}{8 \times 10^{-18}}= \frac{20 \times 10^{23}}{3 N\left({ } _1 H^{3}\right)} \approx 1.25 \times 10^{17} \\ \Rightarrow \quad & N\left({ } _1 H^{3}\right) =\frac{20 \times 10^{23}}{3 \times 1.25 \times 10^{17}}=5.33 \times 10^{6} \\ \Rightarrow \quad & k t =\ln \frac{N _0}{N} \Rightarrow \frac{\ln 2}{12.3} \times 40=\ln \frac{5.33 \times 10^{6}}{N} \\ \Rightarrow \quad & N =5.6 \times 10^{5} \end{aligned} $
BETA
