Easyish radioactivity question

[Pezzoni]

Due to me being a bit dense, I'm having trouble with a question from a physics paper, and was wondering if anyone would mind lending a hand?

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λ for carbon-14 is 4.0x10^-12 (for this question, anyway)

A sample containing 1g of carbon was taken from a bog body. 65 atoms of carbon-14 were found to decay in 600s.

Calculate the number of carbon-14 atoms present in the sample

Any help would be greatly apreciated

[drachefly]

Well, it depends somewhat on the conventions. Here, the lambda looks like it should have units of inverse time, which isn't what I was expecting.

In any case, exponential decays proceed such that the amount of the material that decays in a constant period of time is proportional to the amount remaining.

That is,

decays per second = - d/dt[ amount remaining]= lambda * ln(a) * a^(-lambda* t)

amount remaining = a ^ (- lambda * t)

where a is either 2 or e, depending on the conventions of lambda.

Does that help? I don't want to do your homework for you, but that might get you started.

[Pezzoni]

I tried somthing similar the first time round I think, with:

N = No e(-λt)

(Where No is the number at 0 seconds)

Rearranged to:

No = N / e-λt

So

No = 65 / e(-4*10^-12 * 600)

Although that got be a (obviously incorrect) answer of 65.


I tried the formula you suggested as well, but got 600, which I'm also guessing was incorrect?

I also just tried

Change in N / Change in t = -λN

So

N = Change in N / Change in t * -λ

Which came out to 2.7x10^10, which looks vaugely sensible?

Thanks for the help

[Wyrm]

Although that got be a (obviously incorrect) answer of 65.

I tried the formula you suggested as well, but got 600, which I'm also guessing was incorrect?

Your instincts serve you well. drachefly's answer, although a correct derivation, doesn't answer the question at hand: what is the current number of C-14 atoms in the sample?

I also just tried

Change in N / Change in t = -λN

So

N = Change in N / Change in t * -λ

Which came out to 2.7x10^10, which looks vaugely sensible?

Here, what you did is almost right, though from your statements, you probably don't know why. If A is the current activity, then it is equal to -dN/dt (because dN/dt is the rate at which C-14 is decreasing). We know dN/dt = -λN, so A = λN! From here, it's just algebra.

A caution, the figure you're given is 65 decays over 600s, not 1s, so you have to work out the average decay over one second.

[sketerpot]

Due to me being a bit dense, I'm having trouble with a question from a physics paper, and was wondering if anyone would mind lending a hand?

You might get better results in general by asking someone else who's taking the class, or one of the TAs.

By the way, I haven't studied whatever you're studying, but wouldn't it be possible to simply find the number of atoms in 1 gram of Carbon-14 using the molar mass of C14 (14.003241988 g/mol) and Avogadro's number (6.022e23)? Or is there something I'm missing?

[Lancer]

Due to me being a bit dense, I'm having trouble with a question from a physics paper, and was wondering if anyone would mind lending a hand?

You might get better results in general by asking someone else who's taking the class, or one of the TAs.

By the way, I haven't studied whatever you're studying, but wouldn't it be possible to simply find the number of atoms in 1 gram of Carbon-14 using the molar mass of C14 (14.003241988 g/mol) and Avogadro's number (6.022e23)? Or is there something I'm missing?

you can't assume that the carbon sample is entirely C-14.

[sketerpot]

By the way, I haven't studied whatever you're studying, but wouldn't it be possible to simply find the number of atoms in 1 gram of Carbon-14 using the molar mass of C14 (14.003241988 g/mol) and Avogadro's number (6.022e23)? Or is there something I'm missing?

you can't assume that the carbon sample is entirely C-14.

Oh.

I thought it might be something like that, thanks.

[Wyrm]

By the way, I haven't studied whatever you're studying, but wouldn't it be possible to simply find the number of atoms in 1 gram of Carbon-14 using the molar mass of C14 (14.003241988 g/mol) and Avogadro's number (6.022e23)? Or is there something I'm missing?

That "1g" figure is a distractor. The number of C-14 atoms is proportional to the activity rate. That's the meaning of dN/dt = -λN.