# How do you find the associated exponential decay model given Q = 100 when t = 0; Half-life = 6?

##### 1 Answer
Jun 20, 2016

It is $Q \left(t\right) = 100 {e}^{- 0.115 t}$.

#### Explanation:

We try here to calculate the general law for decay.

We have an initial population ${Q}_{i}$ and, after a time interval $\setminus \Delta t$ we have a new population, ${Q}_{f}$, that was the initial population times the constant of decay (that we can call $\lambda$).

Mathematically we can say that

${Q}_{f} - {Q}_{i} = - \lambda {Q}_{i} \setminus \Delta t$.

The negative sign is to take into account that the populations at the end is smaller than the population at the beginning.

We can write our equations as

$\setminus \Delta Q = - \lambda {Q}_{i} \setminus \Delta t$.

The important part of this equation is that it is valid no matter how big is the time interval. If it is one hour, everything scale correctly with $- \lambda {Q}_{i}$ and the final result will be correct. If the time is a nanosecond it still work. Then we can consider the instantaneous change in population writing

$\mathrm{dQ} = - \lambda Q \setminus \mathrm{dt}$

Here I have to use $Q$ instead of ${Q}_{i}$ because also the initial distribution changes continuously with the time interval. Every infinitesimal time I consider I have to plug in $Q$ the new initial distribution that will be changed infinitesimally from the previous.
Then I can write the differential equation

$\frac{\mathrm{dQ}}{\mathrm{dt}} = - \lambda Q$

where the $Q$ is function of time.

This is a differential equation that has the solution

$Q \left(t\right) = {Q}_{0} {e}^{- \setminus \lambda t}$.

Where ${Q}_{0}$ is the population at the beginning of time (when $t = 0$).

Now we are ready to substitute our numbers. In your case ${Q}_{0} = 100$

Then the equation is $Q \left(t\right) = 100 {e}^{- \lambda t}$ and we also know that when $t = 6$ the population is the half (this is the definition of half life). So we have

$50 = 100 {e}^{- \lambda 6}$

$\frac{1}{2} = {e}^{- \lambda 6}$

$\ln \left(\frac{1}{2}\right) = - \lambda 6$

$\lambda = - \ln \frac{\frac{1}{2}}{6} \setminus \approx 0.115$

Then the law is

$Q \left(t\right) = 100 {e}^{- 0.115 t}$ that is the searched decay law.