Question

Cobalt‑60 is subject to radioactive decay, and each year the amount present is reduced by 12.3%.

The amount of cobalt‑60 present is an exponential function of time in years. What is the half‑life of cobalt‑60?

Answer 1

The half-life of a radioactive substance is the time it takes for half of the substance to decay. Let's denote the half-life as T_half. We know that the remaining amount after one year is reduced by 12.3%, which means that it retains 100 - 12.3 = 87.7% of the initial amount. So, after one year, the amount remaining is 0.877 times the initial amount.

In general, the remaining amount after t half-lives is given by the formula:

Amount = Initial_amount * (1/2)^t

In our case, after one half-life (t = 1), we have:

Amount = Initial_amount * (0.877)^1

Since we want to find the time when only half of the initial amount remains, we need to find when:

0.5 * Initial_amount = Initial_amount * (0.877)^T_half

Dividing both sides by the initial amount:

0.5 = 0.877^T_half

To find T_half, take the log of both sides and solve for T_half:

log(0.5) = T_half * log(0.877)

T_half = log(0.5) / log(0.877) ≈ 5.27 years

So, the half-life of cobalt-60 is approximately 5.27 years.