# Question f25a9

Sep 6, 2017

$0.693 \cdot \frac{N}{n}$

#### Explanation:

As you know, a radioactive decay is essentially a first-order reaction

$\text{A " -> " products}$

which means that you can express the rate of the reaction, i.e. the rate of disintegration, in differential form like this

- (d["A"])/(dt) = k * ["A"]#

Here

• $k$ is the rate constant
• $\left[\text{A}\right]$ is the concentration of the radioactive element

In your case, you know that this radioactive element emits $n$ alpha particles per second. This is equivalent to saying that the rate of change of its concentration per unit of time is equal to

$\frac{d \left[\text{A}\right]}{\mathrm{dt}} = n$

Moreover, you know that your sample contains $N$ atoms of this radioactive element, so

$\left[\text{A}\right] = N$

Plug this into the first equation to get

$- n = k \cdot N$

It's important to realize that the minus sign is there to show that the concentration of $\text{A}$ is decreasing as the reaction proceeds, which means that you don't really need it for your purposes.

$n = k \cdot N$

This will get you

$k - \frac{n}{N}$

Now, the half-life of a first-order reaction is given by

${t}_{\text{1/2}} = \ln \frac{2}{k} \approx \frac{0.692}{k}$

This means that you have

${t}_{\text{1/2}} = \frac{0.693}{\frac{n}{N}} = 0.693 \cdot \frac{N}{n}$