# Question #f25a9

##### 1 Answer

#### 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 therate constant#["A"]# is the concentration of the radioactive element

In your case, you know that this radioactive element **emits** **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

#["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 **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#