Measurements indicate that\(27.83\)% of all rubidium atoms currently on the earth are the radioactive\({}^{87}\)Rb isotope. The rest are the stable\({}^{85}\)Rb isotope. The half-life of\({}^{87}\)Rb is\(4.75*{10^{10}}\)y. Assuming that no rubidium atoms have been formed since, what percentage of rubidium atoms were\({}^{87}\)Rb when our solar system was formed\(4.6*{10^9}\)y ago?

Relation between half-life and decay constant

\(\lambda = \frac{{\ln 2}}{{{T_{1/2}}}}\)

Let the number of \({}^{87}\)Rb nuclei now is N, the number of \({}^{85}\)Rb is N, which is constant and the number of \({}^{87}\)Rb at the beginning is \({N_0}\).

So, the total number of rubidium atoms, now, is N + N\(_S\), and its number at the beginning is N\(_0\) + N\(_S\).

Currently, only\(27.83\)% of all rubidium atoms on the earth are the radioactive\({}^{87}\)Rb isotope.

Thus,

\(\frac{N}{{N + {N_S}}} = 0.2783\)

Solve for N,

\(N = \frac{{{N_S}}}{{2.593}} = 0.3856{N_S}\)

(1)

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

\(f = \frac{{{N_0}}}{{{N_0} + {N_S}}}\)

(2)

The number of remaining nuclei is given by,

\(N = {N_0}{e^{ - \lambda t}}\)

Solve for \({N_0}\) substitute it into equation (2),

\(f = \frac{{N{e^{\lambda t}}}}{{N{e^{\lambda t}} + {N_S}}}\)

Plug the values,

\(f = \frac{{0.3856{N_S}{e^{\lambda t}}}}{{0.3856{N_S}{e^{\lambda t}} + {N_S}}} = \frac{{0.3856{e^{\lambda t}}}}{{0.3856{e^{\lambda t}} + 1}}\) (3)

Here,

\(\lambda = \frac{{\ln 2}}{{4.75*{{10}^{10}}}} = 1.46*{10^{ - 11}}y{r^{ - 1}}\)

And

\({e^{\lambda t}} = 1.069\)

Now, plug the values in equation (3),

\(f = \frac{{0.3856(1.069)}}{{0.3856(1.069) + 1}} = 0.2920\)

Thus, the percentage of rubidium atoms \({}^{87}\)Rb when our solar system was formed \(4.6*{10^9}\)y ago is \(29.20\% \).