Roman lead and modern physics

22.2 years after the Roman lead has been mined, half of the $^{210} P_b$ atoms in it have decayed. The next 22.2 year, half of the remaining atoms has decayed and so on.

The ratio of the amount of $^{210} P_b$ atoms in the Roman lead now vs. when it was mined is therefore given by:

$\left(\frac{1}{2}\right)^{1713/22.2} \approx 2^{-77.61} \approx \boxed{5.91\times 10^{-24}}$

If every 22.2 years the amount of Mined lead is cut in half, we can model the situation mathematically as

A(t) = $A_{0} \times (\frac{1}{2})^{ t/h}$

Where A(t) is the amount of Mined lead as a function of time, $A_{0}$ is the amount of Mined lead you have at the beginning, and h is the half life.

If you were to divide A(t) by $A_{0}$ , you would get the ratio of the amount of Mined lead now to Mined lead at the beginning. Let's do it with math.

$\frac{A(t)}{A_{0}} = \frac{A_{0} \times (\frac{1}{2})^{ t/h}}{A_{0}} = (\frac{1}{2})^{ t/h}$

Since 1713 years have passed, we plug in 1713 for t and 22.2 for h , the half life.

This gives you the answer, which is about $\boxed{5.9139 \times 10^{-24}}$

Voilà!

Let the number of half-lives that has passed be $n$ . Then, the ratio is $\left( \dfrac {1}{2} \right)^n$ . The number of half-lives since the mining is the number of years divided by the half-life time. The number of years is $1712$ , so $n = \dfrac {1713}{22.2},$ and our answer is $\left( \dfrac {1}{2} \right)^{1713/22.2} \approx \boxed {5.91 \times 10^{-24}}.$

radioactive decay follow first order reaction. $\frac{dN}{dt} = -kN$ *note: negative is just for convention, indicating that it's actually a decay

by solving the differential equation with $N(0) = N_0$ , you would get an equation:

$N_t = N_0 e^{-\lambda t}$

by definition, half-life is the time taken for a radioactive material to decay into half of it's initial value. So by plugging $N_{T_{1/2}} = \frac{1}{2}N_0$ and doing simplification, you will get an equation: $N_t = N_0 e^{\frac{t}{T_{1/2}}}$

Plugging the number: $T_{1/2} = 22.2$ and $t = 1713$ into the calculator and you would get the ratio: $\boxed{\frac{N_t}{N_0} = 5.913909291 \times 10^{-24}}$

Half-life can be modeled by the first-order differential equation $\frac{dm}{dt} = rm$ , where m is mass (or amount) of material, since decay rate is proportional to mass. Rearranging and using separation of variables, we find the solution to this equation to be $m = Ce^{rt}$ , where t is time.

By setting an initial condition of 1 for the mass (ie original amount) at time t=0, we find that C is equal to one. We then solve for r, using the condition where m = 1/2 (This is half-life!), and t = 22.2 years:

$\frac{1}{2} = 1e^{22.2r}$

Taking the natural log of both sides, we find r to be -0.03122. We can now solve the problem: $m = e^{-0.03122 \times 1713} = 5.193 \times 10^-24$ years.