Determine Half-Lives & Initial Activities of 2 Isotopes - Q&A

[Soilwork]

Q. A sample of a certain element with two naturally occurring isotopes becomes
activated by neutron capture. After 1 hour in the reactor, it is placed in a
counting room, in which the total number of decays in 1 hour is recorded at
daily intervals. A summary of the recorded data appears below.
From the data, determine the (i) half-lives and (ii) initial activities of the 2
components. (iii) What is the element?

The thing with this question is that you're given a table with the time in one column and the total number of counts in the other column.
I know that if you take the natural log of the exponential decay function that you can find lambda and therefore the half-life.
But you have to find the half-life of two isotopes and I don't know how you can do that.
Any hint that can point me in the right direction would be greatly appreciated :)

 

[jtbell]

Make a graph of the log of the number of counts, versus time. What does it look like? (after mentally smoothing out any random wiggles, of course)

 

[Soilwork]

Yeah I get kind of two different straight lines.
The times given are as follows:
0,1,2,3,4,5,6,7,8,9,10,20,40,60,80,100,120,140,160,180,200.
So the first gradient is about 0.23 and is from the time scale of 0-10.
The second gradient is 0.0116 and is from 40 - 200.
This will give the two half lives of the isotopes, but I don't know why.
So the element is Antimony.

 

[jtbell]

Yeah I get kind of two different straight lines.

That's what I suspected. The hypothetical sample started out with the number of decays being dominated by the short-lived isotope, so the initial slope reflects the shorter half-life. After most of the short-lived isotope has decayed, the decays observed are mostly from the long-lived isotope, so the second slope reflects the longer half-life.

 

[Soilwork]

Ahhh k thanks for that :)
So now I can just use the LOGEST function on Excel over the relevant points to find the intial activities of the two isotopes.
Thank you so much for explaining this.