Q62 dgP - University Physics with Modern Physics

Question

The nucleus ${}_{\mathbf{8}}{}^{\mathbf{15}}\mathit{O}$ has a half-life of 122.2 s; ${}_{\mathbf{8}}{}^{\mathbf{19}}\mathit{O}$ has a half-life of 26.9s. If at some time a sample contains equal amounts of ${}_{\mathbf{8}}{}^{\mathbf{15}}\mathit{O}$ and ${}_{\mathbf{8}}{}^{\mathbf{19}}\mathit{O}$, what is the ratio of ${}_{\mathbf{8}}{}^{\mathbf{15}}\mathit{O}$ to ${}_{\mathbf{8}}{}^{\mathbf{19}}\mathit{O}$ (a) after 3.0 min and (b) after 12.0 min?

Step-by-Step Solution

Answer

a. The ratio of ${}_{\mathbf{8}}{}^{\mathbf{15}}\mathit{O}$ to ${}_{\mathbf{8}}{}^{\mathbf{19}}\mathit{O}$ after 3.0 min is 37.23.

b. The ratio of ${}_{\mathbf{8}}{}^{\mathbf{15}}\mathit{O}$ to ${}_{\mathbf{8}}{}^{\mathbf{19}}\mathit{O}$ after 12.0 min is $\mathbf{1}\mathbf{.}\mathbf{92}\mathbf{*}{\mathbf{10}}^{\mathbf{9}}$.

Step 1: Formula used to solve the question

Relation between half-life and decay constant

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

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

                                             $\mathit{N}\mathbf{=}{\mathit{N}}_{\mathbf{0}}{\mathit{e}}^{\mathbf{-}\mathbf{\lambda t}}$                                       ( 2 )

Step 2: Determine the ratio

Let ${\mathit{N}}_{\mathbf{01}}$ be the initial number of ${}^{\mathbf{15}}\mathit{O}$ and ${\mathit{N}}_{\mathbf{02}}$ be the initial number of ${}^{\mathbf{19}}\mathit{O}$.

From equation (2),

                                                     $$\begin{gathered} {\mathit{N}}_{\mathbf{1}}\mathbf{=}{\mathit{N}}_{\mathbf{01}}{\mathit{e}}^{\mathbf{-}{\mathbf{\lambda }}_{\mathbf{1}}\mathbf{t}} \\ {\mathit{N}}_{\mathbf{2}}\mathbf{=}{\mathit{N}}_{\mathbf{02}}{\mathit{e}}^{\mathbf{-}{\mathbf{\lambda }}_{\mathbf{2}}\mathbf{t}} \end{gathered}$$ 

The ratio will be:

                                                    $\frac{{\mathit{N}}_{\mathbf{1}}}{{\mathit{N}}_{\mathbf{2}}}\mathbf{=}\frac{{\mathit{N}}_{\mathbf{01}}{\mathit{e}}^{\mathbf{-}{\mathit{\lambda }}_{\mathbf{1}}\mathit{t}}}{{\mathit{N}}_{\mathbf{02}}{\mathit{e}}^{\mathbf{-}{\mathit{\lambda }}_{\mathbf{2}}\mathit{t}}}$ 

But at: $\mathit{t}\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{ }\mathbf{ }{\mathit{N}}_{\mathbf{01}}\mathbf{=}{\mathit{N}}_{\mathbf{02}}$ 

                                        $$\begin{gathered} \frac{{\mathit{N}}_{\mathbf{1}}}{{\mathit{N}}_{\mathbf{2}}}\mathbf{=}\frac{{\mathit{e}}^{\mathbf{-}{\mathit{\lambda }}_{\mathbf{1}}\mathit{t}}}{{\mathit{e}}^{\mathbf{-}{\mathit{\lambda }}_{\mathbf{2}}\mathit{t}}}\mathbf{=}{\mathit{e}}^{\mathbf{(}{\mathit{\lambda }}_{\mathbf{1}}\mathbf{-}{\mathit{\lambda }}_{\mathbf{2}}\mathbf{)}\mathit{t}} \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ } \end{gathered}$$                               (3)

Now, get the decay constant using equation ( 1 ),

                       ${\mathit{\lambda }}_{\mathbf{1}}\mathbf{=}\frac{\mathbf{In}\mathbf{ }\mathbf{2}}{\mathbf{122}\mathbf{.}\mathbf{2}}\mathbf{=}\mathbf{5}\mathbf{.}\mathbf{67}\mathbf{*}{\mathbf{10}}^{\mathbf{3}}\mathbf{ }{\mathit{s}}^{\mathbf{-}\mathbf{1}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{340}\mathbf{ }{\mathbf{min}}^{\mathbf{-}\mathbf{1}}$ 

And similarly,

                                          ${\mathit{\lambda }}_{\mathbf{2}}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{55}\mathbf{ }\mathit{m}\mathit{i}{\mathit{n}}^{\mathbf{-}\mathbf{1}}$     

Now, ${\mathit{\lambda }}_{\mathbf{2}}\mathbf{-}{\mathit{\lambda }}_{\mathbf{1}}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{55}\mathbf{-}\mathbf{0}\mathbf{.}\mathbf{340}\mathbf{ }\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{21}\mathbf{ }{\mathbf{min}}^{\mathbf{-}\mathbf{1}}$ 

Substitute in equation (3),

                                         $\frac{{\mathbf{N}}_{\mathbf{1}}}{{\mathbf{N}}_{\mathbf{2}}}\mathbf{=}{\mathit{e}}^{\mathbf{1}\mathbf{.}\mathbf{21}\mathbf{t}}$                                                      (4)

Now plug the values for t,

                     $$\begin{gathered} \frac{{\mathit{N}}_{\mathbf{1}}}{{\mathit{N}}_{\mathbf{2}}}{\mathbf{|}}_{\mathbf{3}\mathbf{.}\mathbf{0}\mathbf{m}\mathbf{i}\mathbf{n}}\mathbf{=}{\mathit{e}}^{\mathbf{(}\mathbf{1}\mathbf{.}\mathbf{21}\mathbf{)}\mathbf{(}\mathbf{3}\mathbf{.}\mathbf{0}\mathbf{)}}\mathbf{ }\mathbf{=}\mathbf{37}\mathbf{.}\mathbf{23} \\ \frac{{\mathit{N}}_{\mathbf{1}}}{{\mathit{N}}_{\mathbf{2}}}{\mathbf{|}}_{\mathbf{12}\mathbf{.}\mathbf{0}\mathbf{m}\mathbf{i}\mathbf{n}}\mathbf{=}{\mathit{e}}^{\mathbf{(}\mathbf{1}\mathbf{.}\mathbf{21}\mathbf{)}\mathbf{(}\mathbf{12}\mathbf{.}\mathbf{0}\mathbf{)}}\mathbf{ }\mathbf{ }\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{92}\mathbf{*}{\mathbf{10}}^{\mathbf{9}} \end{gathered}$$

Thus, the ratio of ${}_{\mathbf{8}}{}^{\mathbf{15}}\mathit{O}$ to ${}_{\mathbf{8}}{}^{\mathbf{19}}\mathit{O}$ after 3.0 min is 37.23. The ratio of ${}_{\mathbf{8}}{}^{\mathbf{15}}\mathit{O}$to ${}_{\mathbf{8}}{}^{\mathbf{19}}\mathit{O}$ after 12.0 min is $\mathbf{1}\mathbf{.}\mathbf{92}\mathbf{*}{\mathbf{10}}^{\mathbf{9}}$.