Q61P... - University Physics with Modern Physics

Question

Measurements indicate that 27.83 % of all rubidium atoms currently on the earth are the radioactive ${}^{\mathbf{87}}\mathit{R}\mathit{b}$ isotope. The rest are the stable ${}^{\mathbf{85}}\mathit{R}\mathit{b}$ isotope. The half-life of ${}^{\mathbf{87}}\mathit{R}\mathit{b}$ is $\mathbf{4}\mathbf{.}\mathbf{75}\mathbf{\times }{\mathbf{10}}^{\mathbf{10}}$ y. Assuming that no rubidium atoms have been formed since, what percentage of rubidium atoms were ${}^{\mathbf{87}}\mathit{R}\mathit{b}$ when our solar system was formed $\mathbf{4}\mathbf{.}\mathbf{6}\mathbf{\times }{\mathbf{10}}^{\mathbf{9}}$y ago?

Step-by-Step Solution

Answer

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

Step 1: Formulas used to solve the question

Relation between half-life and decay constant

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

Step 2: Determine the number of rubidium nuclei

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)

Step 3: Determine the percentage

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,

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

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

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

Plug the values,

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

Here,

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

And

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

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 $\mathbf{4}\mathbf{.}\mathbf{6}\mathbf{\times }{\mathbf{10}}^{\mathbf{9}}$ y ago is 29.20 %.