An Oceanographic Tracer. Nuclear weapons tests in the 1950 s and 1960 s released significant amounts of radioactive tritium (${}_{\mathbf{1}}{}^{\mathbf{3}}\mathit{H}$, half-life 12.3 years) into the atmosphere. The tritium atoms were quickly bound into water molecules and rained out of the air, most of them ending up in the ocean. For any of this tritium-tagged water that sinks below the surface, the amount of time during which it has been isolated from the surface can be calculated by measuring the ratio of the decay product, ${}_{\mathbf{2}}{}^{\mathbf{3}}\mathit{H}\mathit{e}$, to the remaining tritium in the water. For example, if the ratio of ${}_{\mathbf{2}}{}^{\mathbf{3}}\mathit{H}\mathit{e}$to ${}_{\mathbf{1}}{}^{\mathbf{3}}\mathit{H}$in a sample of water is 1:1, the water has been below the surface for one half-life, or approximately 12 years. This method has provided oceanographers with a convenient way to trace the movements of subsurface currents in parts of the ocean. Suppose that in a particular sample of water, the ratio of${}_{\mathbf{2}}{}^{\mathbf{3}}\mathit{H}\mathit{e}$to${}_{\mathbf{1}}{}^{\mathbf{3}}\mathit{H}$is 4.3 to 1.0. How many years ago did this water sink below the surface?

Step by step solution

01

Formula used to solve the question

Relation between half-life and decay constant

$\mathit{\lambda }\mathbf{=}\frac{\mathbf{In}\mathbf{}\mathbf{2}}{{\mathbf{T}}_{\mathbf{1}\mathbf{/}\mathbf{2}}}$.............(i)

Number of nuclei at time t in sample of radioactive element:

$\mathit{N}\mathbf{=}{\mathit{N}}_{\mathbf{0}}{\mathit{e}}^{\mathbf{-}\mathbf{\lambda t}}$ ............(ii)

02

Determine the time

Let ${\mathit{N}}_{\mathbf{0}}$be the initial number of ${}_{\mathbf{1}}{}^{\mathbf{3}}\mathit{H}$ atoms, N is the number at time t and ${\mathit{N}}_{\mathbf{H}\mathbf{e}}$is the number of ${}_{\mathbf{2}}{}^{\mathbf{3}}\mathit{H}\mathit{e}$ atoms at time t. In the beginning, there was no ${}_{\mathbf{2}}{}^{\mathbf{3}}\mathit{H}\mathit{e}$; But, as ${}_{\mathbf{1}}{}^{\mathbf{3}}\mathit{H}$decays, the number of ${}_{\mathbf{1}}{}^{\mathbf{3}}\mathit{H}$atoms increases, where ${\mathit{N}}_{\mathbf{0}}\mathbf{-}\mathit{N}\mathbf{=}{\mathit{N}}_{\mathbf{H}\mathbf{e}}$. ${}_{\mathbf{1}}{}^{\mathbf{3}}\mathit{H}$is a radioactive element, so use equation (ii) to calculate the number of its atoms at any time.

Now, the ratio of He to H at any time as follows:

$$\begin{gathered} \mathit{r}\mathbf{=}\frac{{\mathbf{N}}_{\mathbf{He}}}{\mathbf{N}} \\ \mathbf{}\mathbf{}\mathbf{=}\frac{{\mathbf{N}}_{\mathbf{0}}\mathbf{-}\mathbf{N}}{\mathbf{N}} \\ \mathbf{}\mathbf{}\mathbf{=}\frac{{\mathbf{N}}_{\mathbf{0}}}{\mathbf{N}}\mathbf{-}\mathbf{1} \end{gathered}$$

From equation (ii),

$\mathit{r}\mathbf{=}{\mathit{e}}^{\mathbf{\lambda }\mathbf{t}}\mathbf{-}\mathbf{1}$

Solve for t,

$\mathit{t}\mathbf{=}\frac{\mathbf{1}}{\mathbf{\lambda }}\mathit{I}\mathit{n}\mathbf{(}\mathit{r}\mathbf{+}\mathbf{1}\mathbf{)}$

Substitute,

$\mathit{t}\mathbf{=}\frac{{\mathbf{T}}_{\mathbf{1}\mathbf{/}\mathbf{2}}}{\mathbf{In}\mathbf{}\mathbf{2}}\mathit{I}\mathit{n}\mathbf{(}\mathit{r}\mathbf{+}\mathbf{1}\mathbf{)}$

Plug the given,

$$\begin{gathered} \mathit{t}\mathbf{=}\frac{\mathbf{12}\mathbf{.}\mathbf{33}}{\mathbf{I}\mathbf{n}\mathbf{}\mathbf{2}}\mathit{I}\mathit{n}\mathbf{(}\mathbf{4}\mathbf{.}\mathbf{3}\mathbf{+}\mathbf{1}\mathbf{)} \\ \mathbf{}\mathbf{=}\mathbf{29}\mathbf{.}\mathbf{6}\mathbf{}\mathit{y}\mathit{r} \end{gathered}$$

Thus, the water sink below the surface before 29.6 yr.

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