Question

Question #f25a9

Answer 1

`0.693 * N/n`

Explanation:

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

`"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
• `["A"]` 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

`(d["A"])/(dt) = n`

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

`["A"] = N`

Plug this into the first equation to get

`-n = k* N`

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

`n = k * N`

This will get you

`k - n/N`

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

`t_"1/2" = ln(2)/k ~~ 0.692/k`

This means that you have

`t_"1/2" = 0.693/(n/N) = 0.693 * N/n`