Struggling with this question here:
"One percent of a substance disintegrates in $100$ years. What is its half life?"
I'm not understanding how to apply the formula $T=\dfrac {\ln 2}k$ to this. Thanks
$\begingroup$
$\endgroup$
3
-
$\begingroup$ Welcome to MSE. Please read this text about how to ask a good question. $\endgroup$– José Carlos SantosCommented Feb 12, 2023 at 13:44
-
$\begingroup$ Before leaping directly to the formula you've provided, it may be worth considering the behaviour of the substance as it's decaying (i.e. exponential), and what the 'half-life' actually means. This will hopefully help you create a model of the situation, which will then let you extract necessary information to understand the application of the formula. It may also be a useful exercise to derive that equation in the first place as well, which would tie in to the question you've asked. $\endgroup$– Frankie S. PalmerCommented Feb 12, 2023 at 13:47
-
1$\begingroup$ You might consult the 70 (or 72) rule of doubling. So a growth rate of 5% leads to doubling in 70/5=14 years. Here you get 70/1 units of 100 years. The exact result will of course have more digits. $\endgroup$– Lutz LehmannCommented Feb 12, 2023 at 13:50
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
You need to find $k$ before applying the given formula.
If $N(t) = N_0e^{-kt}$ denotes the quantity of the considered substance at time $t$, where $N_0 = N(0)$ is the initial quantity, then we have the following equation : $$ N(t=100) = N_0e^{-100k} = 1\% \cdot N_0 \quad\Longrightarrow\quad k = -\frac{\ln(1\%)}{100} = \frac{\ln10}{50} $$ hence $$ T = \frac{\ln2}{k} = \frac{\ln2}{(\ln10)/50} = 50\log_{10}(2) \approx \underline{\underline{15.05}} \,[\mathrm{years}] $$
