# An isotope of lead, 201Pb, has a half-life of 8.4 hours. How many hours ago was there 40% more of the substance?

Nov 9, 2015

$t = 4.078$ hours
$\therefore t \approx 4$ hours

#### Explanation:

The half-life of a radioactive substance is the time in which half of any given sample of the substance undergoes radioactive decay. The mathematical basis of the phenomenon of radioactive decay is given by the half-life equation:

$N \left(t\right) = {N}_{0} {e}^{-} \left(\lambda \cdot t\right)$ ...(1)

where $N \left(t\right)$ is the amount of substance present at time $t$, ${N}_{0}$ is the initial amount of substance, $\lambda$ is the decay constant and $t$ is the time elapsed.

Now, the half-life ${t}_{\frac{1}{2}}$ is the time when the amount of the substance is halved, i.e. $N \left(t\right) = \frac{1}{2} {N}_{0}$

Using this in (1) to find ${t}_{\frac{1}{2}}$, we get

${t}_{\frac{1}{2}} = \ln \frac{2}{\lambda} = \frac{0.693}{\lambda}$ (refer here for the derivation) ...(2)

Now, in the question, we are given ${t}_{\frac{1}{2}}$ and the ratio of $\frac{N \left(t\right)}{N} _ 0$, and we can solve the half-life equation for $t$ if we know $\lambda$

$\lambda$ can be obtained from (2)

$\lambda = \frac{0.693}{{t}_{\frac{1}{2}}} = \frac{0.693}{8.4} = 0.0825$

The ratio of $\frac{N \left(t\right)}{N} _ 0$ is $1.4$ since we need to calculate the time when the substance was 40% more than it is right now, that is at $t = 0$.

Thus, equation (1) yields,

${e}^{-} \left(0.0825 \cdot t\right) = 1.4$

$\therefore {e}^{0.0825 \cdot t} = \frac{1}{1.4} = 0.7142$

On taking the natural log on both sides, we get

$0.0825 \cdot t = \ln \left(0.7142\right) = - 0.3365$

$\therefore t = - \frac{0.3365}{0.0825} = - 4.078$

The answer is a negative value of time because we were asked to calculate how many hours "ago" the substance was 40% more than it is now.