Question

The ages of rocks that contain fossils can be determined using the isotope 87Rb. This isotope of rubidium undergoes beta decay with a half‑life of 4.75×1010y . Ancient samples contain a ratio of 87Sr to Rb87 of 0.0125. Given that 87Sr is a stable product of the beta decay of 87Rb, and assuming there was originally no 87Sr present in the rocks, calculate the age of the rock sample. Assume that the decay rate is constant over the relatively short lifetime of the rock compared to the half-life of 87Rb.

Answer 1

Answer:

The age of the rock is $8.513 \ 10^{8}$ years

Explanation:

We can write the formula for the quantity of material in an exponential decay as:

$N(t) = N_0 \ e ^{ - \frac{t}{\tau}}$

where $N_0$ is the initial quantity of the material, and $\tau$ is the mean lifetime of the material.

This means that the total material that has decayed is:

$N_0 - N(t)$

The mean lifetime can be obtained from the half-life ( $t_{\frac{1}{2}}$) by the relationship

$\tau = \frac{t_{\frac{1}{2}}}{ln(2)}$

For our problem, as there was no Sr-87 present in the rock initially, and its a product of the decay,  its abundance will be given by $N_0 - N(t)$, this means, the total quantity of material that has decayed.

So, our equation for the ratio will be

$\frac{ N_0 - N(t) }{ N(t) } =  0.0125$

$\frac{ N_0}{N(t) } - 1 =  0.0125$

Working it a little

$\frac{ N_0}{N_0 \ e ^{ - \frac{t}{\tau}} } = 1+ 0.0125$

$e ^{ + \frac{t}{\tau}}  = 1.0125$

$ln (e ^{ + \frac{t}{\tau}} )  = ln( 1.0125)$

$\frac{t}{\tau}}   = ln( 1.0125)$

$t = \tau ln( 1.0125)$

$t = \frac{ t_{ \frac{1}{2} } }{ln(2)} ln( 1.0125)$

The half-life of the problem is

$t_{ \frac{1}{2}} = 4.75 \ 10^{10} y$

So, the age of the rock will be

$t = 8.513 \ 10^{8} y$