91Ó°ÊÓ

Relation between half-life and decay constant

$\mathit{λ}\mathbf{=}\frac{\mathbf{In}\mathbf{}\mathbf{2}}{{\mathbf{T}}_{\mathbf{1}\mathbf{/}\mathbf{2}}}$

Let the number of ${}^{\mathbf{87}}\mathit{R}\mathit{b}$nuclei now is N, the number of ${}^{\mathbf{85}}\mathit{R}\mathit{b}$is N, which is constant and the number of ${}^{\mathbf{87}}\mathit{R}\mathit{b}$at the beginning is ${\mathit{N}}_{\mathbf{0}}$.

So, the total number of rubidium atoms, now, is $\mathit{N}\mathbf{+}{\mathit{N}}_{\mathbf{S}}$, and its number at the beginning is ${\mathit{N}}_{\mathbf{0}}\mathbf{+}{\mathit{N}}_{\mathbf{S}}$.

Currently, only 27.83 % of all rubidium atoms on the earth are the radioactive ${}^{\mathbf{87}}\mathit{R}\mathit{b}$isotope.

Thus,

$\frac{\mathbf{N}}{\mathbf{N}\mathbf{+}{\mathbf{N}}_{\mathbf{s}}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{2783}$

Solve for N,

$$\begin{gathered} \mathit{N}\mathbf{=}\frac{{\mathbf{N}}_{\mathbf{s}}}{\mathbf{2}\mathbf{.}\mathbf{593}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{3856}\mathbf{}{\mathit{N}}_{\mathbf{s}} \end{gathered}$$ .............(i)

Now, the number of rubidium atoms as a percentage when the solar system was formed is,

$\mathit{f}\mathbf{=}\frac{{\mathbf{N}}_{\mathbf{0}}}{{\mathbf{N}}_{\mathbf{0}}\mathbf{+}{\mathbf{N}}_{\mathbf{s}}}$ ................(ii)

The number of remaining nuclei is given by,

$\mathit{N}\mathbf{=}{\mathit{N}}_{\mathbf{0}}{\mathit{e}}^{\mathbf{-}\mathbf{λ}\mathbf{t}}$

Solve for ${\mathit{N}}_{\mathbf{0}}$substitute it into equation (ii),

$\mathit{f}\mathbf{=}\frac{\mathbf{N}{\mathbf{e}}^{\mathbf{λ}\mathbf{t}}}{\mathbf{N}{\mathbf{e}}^{\mathbf{λ}\mathbf{t}}\mathbf{+}{\mathbf{N}}_{\mathbf{s}}}$

Plug the values,

$$\begin{gathered} \mathit{f}\mathbf{=}\frac{\mathbf{0}\mathbf{.}\mathbf{3856}\mathbf{}{\mathbf{N}}_{\mathbf{s}}{\mathbf{e}}^{\mathbf{λ}\mathbf{t}}}{\mathbf{0}\mathbf{.}\mathbf{3856}\mathbf{}{\mathbf{N}}_{\mathbf{s}}{\mathbf{e}}^{\mathbf{λ}\mathbf{t}}\mathbf{+}{\mathbf{N}}_{\mathbf{s}}} \\ \mathbf{}\mathbf{}\mathbf{=}\frac{\mathbf{0}\mathbf{.}\mathbf{3856}\mathbf{}{\mathbf{e}}^{\mathbf{λ}\mathbf{t}}}{\mathbf{0}\mathbf{.}\mathbf{3856}\mathbf{}{\mathbf{e}}^{\mathbf{λ}\mathbf{t}}\mathbf{+}\mathbf{1}} \end{gathered}$$ ..................(iii)

Here,

$$\begin{gathered} \mathit{λ}\mathbf{=}\frac{\mathbf{I}\mathbf{n}\mathbf{}\mathbf{2}}{\mathbf{4}\mathbf{.}\mathbf{75}\mathbf{×}{\mathbf{10}}^{\mathbf{10}}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{}\mathbf{1}\mathbf{.}\mathbf{46}\mathbf{×}{\mathbf{10}}^{\mathbf{-}\mathbf{11}}\mathbf{}\mathit{y}{\mathit{r}}^{\mathbf{-}\mathbf{1}} \end{gathered}$$

And

${\mathit{e}}^{\mathbf{λ}\mathbf{t}}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{069}$

Now, plug the values in equation (iii),

$$\begin{gathered} \mathit{f}\mathbf{=}\frac{\mathbf{0}\mathbf{.}\mathbf{3856}\mathbf{(}\mathbf{1}\mathbf{.}\mathbf{069}\mathbf{)}}{\mathbf{0}\mathbf{.}\mathbf{3856}\mathbf{(}\mathbf{1}\mathbf{.}\mathbf{069}\mathbf{)}\mathbf{+}\mathbf{1}} \\ \mathbf{}\mathbf{}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{290} \end{gathered}$$

Thus, the percentage of rubidium atoms ${}^{\mathbf{87}}\mathit{R}\mathit{b}$when our solar system was formed $\mathbf{4}\mathbf{.}\mathbf{6}\mathbf{×}{\mathbf{10}}^{\mathbf{9}}$ y ago is 29.20 %.