Calculating Exponential Decay of Thorium Isotope from Supernova Explosion

[Pomico]

[SOLVED] Exponential decay

Homework Statement

A certain amount of the radioactive isotope of thorium ^{232}Th was produced during a supernova explosion 2 billion years ago. This isotope decays according to the exponential law N(t) = Noe^{-t/to}, where No and N are the initial number of atoms and the number of atoms after time t respectively, and to = 2x10^{10} years. Calculate the fraction of initial atoms that have not decayed since the explosion.
What time is needed for one half of the initial atoms of thorium to decay?

Homework Equations

N(t) = Noe^{-t/to}

The Attempt at a Solution

I have an answer, I'd just like to check it.
Using t = 2x10^{9} years,
N = Noe^{-(2x10^{9})/(2x10^{10})} = Noe^{-1/10} years

For the second part,
0.5 = e^{-t/(2x10^{10})}
ln0.5 = \frac{-t}{2x10^{10}}, t = 1.386 x 10^{10}

 

[Defennder]

Well I think your first part needs to be expressed in terms of fraction. You shouldn't have No as an answer. And even without No, it still looks as though your answer is the fraction decayed, rather than not decayed.

Your answer for part 2 looks odd, but it seems that your numerical answer for it is correct.

 

[Pomico]

How do I express the first part then?
That's mostly what I was checking... Do I substitute 232 in for No?

 

[blochwave]

Well ask yourself, why would that be right? 232 has to do with the number of nucleons in thorium, not the number of atoms. You're never given a number of atoms, but it doesn't ask any questions involving that.

It asked you "what fraction..." and you gave just an expression for N. Also be careful about the difference between fraction decayed and fraction not decayed, like the other guy said

Your work for part 2 looks correct

The fact that you got part 2 correct means you know how to answer your question about part 1, if that helps

 

[Pomico]

Yeah I didn't think using 232 would be right, hence not doing it :p

So would the fraction decayed be \frac{No}{N} e^{\frac{-1}{10}}?

And the fraction not decayed would be 1 - \frac{No}{N} e^{\frac{-1}{10}}?

 

[Pomico]

The fact that you got part 2 correct means you know how to answer your question about part 1, if that helps

Oh, I'll have another look!

 

[blochwave]

In part 2 you were asked to find the time it took half the atoms to decay(the half-life, to is often seen as lambda, the decay constant, which is why in Half-Life and Half-Life 2 the game's kinda perpetual symbol is lambda! Fun fact)

So, you were asked blah blah "the fraction decayed = .5", and then you went and put .5 equal to the right thing. The fraction decayed. *nudges you really hard*

And yah, subtract it from 1 for the fraction not decayed

 

[Pomico]

Yeah, I got that from your previous message thanks :P
Ok I got 0.095.
Sound alright?

(And I know about the half-life game's symbol being lambda. Love it!)

 

[blochwave]

I was just about to say yes, BUT...

so 9.5% of the thorium from 2 billion years ago is left. As for whether that's reasonable, I googled the half life of thorium, which we got right(roughly 14 billion years)

So it takes 14 billion years for half of a sample of thorium to decay, yet 2 billion years from the supernova most of it's gone?

It looks like you just have a minor mathematical error in getting to e^(-1/10)

I believe the kinda "oh neat" point of the problem is that MOST of the thorium will still be there, owing to its huge half life.

In fact, to hammer home that point, the estimated age of the universe is just a little less than 14 billion years! So if you had a sample of Thorium AT the big bang, it would almost be down to 50%, with a few hundred million years to go ^_^

 

[Pomico]

Good point! So do we in fact not minus it from 1?
I re-read the question and it says N is the number of atoms after time t. Not decayed after time t...
So the answer would be 0.905?

 

[blochwave]

Yah, N/No should be the fraction left, I believe that was the problem. For some reason I thought you messed up dividing but you didn't, that's the phantom math mistake I was referring to

 

[Pomico]

Ah I see.
Thanks for the help!