91Ó°ÊÓ

Relation between the activity of a sample and the number of radioactive nuclei

$\mathit{R}\mathbf{=}\mathit{λ}\mathit{N}$ .................(i)

Where

$\mathit{N}\mathbf{=}{\mathit{N}}_{\mathbf{0}}{\mathit{e}}^{\mathbf{-}\mathbf{λ³Ù}}$ ...................(ii)

Relation between the half-life and decay constant:

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

Multiply equation (ii) by $\mathit{λ}$, and substitute equation (i),

$\mathit{R}\mathbf{=}{\mathit{R}}_{\mathbf{0}}{\mathit{e}}^{\mathbf{-}\mathbf{λ}\mathbf{t}}$ ..............(iv)

Given: The half time is $\mathbf{28}\mathbf{}\mathit{y}\mathit{r}$.

(a) Let the daughter nucleus is ${}_{\mathbf{Z}}{}^{\mathbf{A}}\mathit{X}$, thus, the equation that describe the decay of ${}_{\mathbf{38}}{}^{\mathbf{90}}\mathit{S}\mathit{r}$by beta emission is:

${}_{\mathbf{38}}{}^{\mathbf{90}}\mathit{S}\mathit{r}\mathbf{→}{}_{\mathbf{Z}}{}^{\mathbf{A}}\mathit{X}\mathbf{+}{}_{\mathbf{-}\mathbf{1}}{}^{\mathbf{0}}\mathit{e}$

From the conservation of the atomic number and the number of nucleons,

localid="1667635842227" $$\begin{gathered} \mathbf{}\mathbf{38}\mathbf{=}\mathit{Z}\mathbf{-}\mathbf{1}\mathbf{}\mathbf{⇒}\mathbf{}\mathit{Z}\mathbf{=}\mathbf{39} \\ \mathbf{90}\mathbf{=}\mathit{A}\mathbf{+}\mathbf{0}\mathbf{}\mathbf{⇒}\mathit{A}\mathbf{=}\mathbf{90} \end{gathered}$$

From the periodic table, the element with $\mathit{Z}\mathbf{=}\mathbf{39}$is ${}_{\mathbf{39}}\mathit{Y}$;

Therefore, the daughter nucleus is localid="1663999463800" ${}_{\mathbf{39}}{}^{\mathbf{90}}\mathit{Y}$.

After $\mathbf{56}\mathbf{}\mathit{y}\mathit{r}$, two half-lives have passed; And after one half-life, the radioactivity is reduced to its half.

Thus, the percent of the activity after two half-lives is $\mathbf{24}\mathbf{}\mathbf{\%}$.

When the activity is reduced to 6.25%, we have:

localid="1667635798372" $\frac{\mathbf{R}}{{\mathbf{R}}_{\mathbf{0}}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{0625}\mathbf{⇒}\frac{\mathbf{R}}{{\mathbf{R}}_{\mathbf{0}}}\mathbf{=}\mathbf{16}$

Now, solve equation (iv) for t ,

localid="1667635805534" $\frac{{\mathbf{R}}_{\mathbf{0}}}{\mathbf{R}}\mathbf{=}{\mathit{e}}^{\mathbf{λ}\mathbf{t}}\mathbf{}\mathbf{⇒}\mathbf{}\mathit{t}\mathbf{=}\frac{\mathbf{1}}{\mathbf{λ}}\mathit{I}\mathit{n}\mathbf{\left(}\frac{{\mathbf{R}}_{\mathbf{0}}}{\mathbf{R}}\mathbf{\right)}$

Substitute for $\mathrm{λ}$from equation (iii),

localid="1667635811451" $\mathit{t}\mathbf{=}\frac{{\mathbf{T}}_{\mathbf{1}\mathbf{/}\mathbf{2}}}{\mathbf{I}\mathbf{n}\mathbf{}\mathbf{2}}\mathit{I}\mathit{n}\mathbf{\left(}\frac{{\mathbf{R}}_{\mathbf{0}}}{\mathbf{R}}\mathbf{\right)}$

Plug the values

localid="1667635827559" $$\begin{gathered} \mathit{t}\mathbf{=}\frac{\mathbf{28}}{\mathbf{I}\mathbf{n}\mathbf{}\mathbf{2}}\mathit{I}\mathit{n}\mathbf{}\mathbf{16} \\ \mathbf{}\mathbf{}\mathbf{=}\mathbf{}\mathbf{112}\mathbf{}\mathit{y}\mathit{r} \end{gathered}$$,

Thus, the daughter nucleus is ${}_{\mathbf{39}}{}^{\mathbf{90}}\mathit{Y}$. The percentage of the original level of ${}^{90}$left after ${}^{56}$years is ${}^{25\%}$. The waited time is ${}^{112yr}$.