Question #f25a9

1 Answer
Sep 6, 2017

Answer:

#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#