Question

Isotopes such as${}^{26}\mathrm{A}\mathrm{l}$(half-life:$7.2\times {10}^5$years) are believed to have been present in our solar system as it formed, but have since decayed and are now called extinct nuclides.

(a)${}^{26}\mathrm{A}\mathrm{l}$decays by${\beta }^{+}$emission or electron capture. Write the equations for these two nuclear transformations.

(b) The earth was formed about$4.7\times {10}^9$(4.7 billion) years ago. How old was the earth when$99.999999\mathrm{\%}$of the${}^{26}\mathrm{A}\mathrm{l}$originally present had decayed?

Short answer

The equations for these two nuclear transformations.

${}_{13}^{26}\mathrm{A}\mathrm{l}{+}_{-1}^{0}e{\to }_{12}^{26}\mathrm{M}\mathrm{g}$.

The earth was $1.67\cdot {10}^7$years old when of the ${}^{26}\mathrm{A}\mathrm{l}$ originally present had decayed.

Step-by-step solution

Definition of extinct radionuclide

Radioactivity generated by nucleosynthesis before the formation of the Solar System is known as an extinct radionuclide.

Write the equations for these two nuclear transformations.

Let us solve the given problem.

If we have a positron emission, the reaction is:

${}_{13}^{26}\mathrm{A}\mathrm{l}{\to }_{12}^{26}\mathrm{M}\mathrm{g}{+}_{+1}^{0}e$

If we have an electron capture, the reaction is:

${}_{13}^{26}\mathrm{A}\mathrm{l}{+}_{-1}^{0}e{\to }_{12}^{26}\mathrm{M}\mathrm{g}$

Therefore, ${}_{13}^{26}\mathrm{A}\mathrm{l}{+}_{-1}^{0}e{\to }_{12}^{26}\mathrm{M}\mathrm{g}$.

Calculate the age of the substance

Consider the given problem.

If we want to calculate the age of the substance, there is a simple formula for that:

$t=\frac{1}{\lambda }\ln \frac{n_0}{n_t}$

Where$t$is time,$\lambda$is the decay constant for the radioactive decay,$n_0$is the initial amount of the radioactive substance, and$n_t$is the amount of the radioactive substance over time$t$.

The half-life of the$\mathrm{A}\mathrm{l}-26$is$7.2\cdot {10}^5\mathrm{y}$. From the half-life, we can get$\lambda$:

$\begin{array}{l}\lambda =\frac{\ln (2)}{t_{1/2}}\\ \lambda =\frac{\ln (2)}{7.2\cdot {10}^5\mathrm{y}}\\ \lambda =9.627\cdot {10}^{-7}{\mathrm{y}}^{-1}\end{array}$

We can assume that V is$1$. And$n_t$is:

$\begin{array}{l}99.999999:100\\ =0.999999\\ =9\cdot {10}^{-8}\end{array}$

Now we can calculate the age:

$\begin{array}{c}t=\frac{1}{9.627\cdot {10}^{-7}{\mathrm{y}}^{-1}}\ln \frac{1}{9\cdot {10}^{-8}}t\\ =1.67\cdot {10}^7\mathrm{y}\end{array}$

Which means the earth was$1.67\cdot {10}^7$years old.

Therefore, the earth was $1.67\cdot {10}^7$years old.