Half-life time computation, and percentage of isotope remaining after X years.

Question

The half-life of a certain radioactive substance is 1400 years. What is the percentage of radioactive isotopes still present after 700 years? The reference solution of my book is: 80.8%

Thank you very much for considering my request.

Answer 1

Of course this is dead wrong. If we have $80.8\%$ after $700$ years, then we have $(0.808)^2$ of the initial amount after $1400$ years, and that is definitely not $\frac{1}{2}$ of the initial amount.

In $1400$ years, the amount shrinks to $\frac{1}{2}$, so the amount at time $t\ge 0$ is $A\times 2^{-t/1400}$, where $A$ is the initial amount.

So the amount after $700$ years is $A\times 2^{-700/1400}$, which is about $70.7\%$ of the initial amount.

Answer 2

If you start out with an amount of radioactive material with half-life $T$denoted by $N_0$, then the amount remaining at time $t$, denoted by $N(t)$, is given by:

$$N(t) = N_0\cdot (\frac{1}{2})^{(\frac{t}{T})}$$

The percentage of radioactive material left at a given time is given by:

$$\frac{N(t)}{N_0} \times 100\%$$

Can you proceed from there?

By the way, I don't agree with your book's reference solution.