# What has radio got to do with this activity?

1. Aug 10, 2010

### SpY]

what has "radio" got to do with this activity?

Probably sounds silly, but what has "radio" got to do with this activity?

I don't know if I'm interpreting this guy's formula right (http://www.hps.org/publicinformation/ate/q8270.html) so can someone confirm this expression for half life:

$$A= \frac{C_R}{CD}$$

$$=kN$$

$$= \Bigg( \frac{ln(2)}{T_\frac{1}{2}} \Bigg) \Bigg( \frac{m_0}{A_r} N_A \Bigg)$$

Where $$C_R$$ is the count rate, $$C_D$$ the counts per disintegration, k the decay constant, N the number of radioactive atoms, $$T_\frac{1}{2}$$ the half life, $$m_0$$ the original mass of the pure substance $$A_R$$the atomic weight $$N_A$$ Avogadro's number. (bad latex sorry)

If not, can you help me find a way to find the half life of a long lived radionuclide?

Last edited: Aug 10, 2010
2. Aug 10, 2010

### fatra2

Hi there,

This equation seems right. Is your problem lies in understanding the meaning of it. If so, I can try to walk you through it.

$$C_R$$ is the number of alpha particles detected per minute.
$$C_D$$ is the efficiency of your detector -- the number of disintegration compared to the number particles detected.

By dividing these two numbers, we obtain the number of disintegration in the last time interval, or the activity of the sample.

Coming back to the basic equation of radioactive disintegration:

$$dN = \lambda N dt$$

where dN is the number of disintegration per time interval of a sample (activity): dN = A, $$\lambda$$ is the decay rate: $$\lambda = \frac{ln 2}{T_{1/2}}$$, N is the number of unstable nuclei in the sample. A bit more about this last parameter.

We have a sample with $$4.41 \times 10^{-3}g$$ of U238. We need to find how many nuclei this mass represents. And for this we need to find the amount of moles in the sample:

$$\frac{m}{238.03g/mol}$$

With this number, we can find out the number of atoms in the system:

$$\frac{m}{238.03g/mol} \cdot N_A$$

You then need to isolate the half life variable in the equation, and it is as simple as that.

Hope this helps. Cheers

3. Aug 10, 2010

### SpY]

ye thanks it does help, was unsure about my formula and what Cr was

But one thing is how they actually determine BILLIONS of years as half-life: how long would they have to monitor the decay of a sample just to plot the first few points of this curve? Or would that matter? Cos I was guessing the more points (count values), the more accurate exponential fit, the more accurate half-life extrapolation is

4. Aug 10, 2010

### fatra2

Hi there,

I don't think I understand your question very well. If you would put this sample on this detector, it would count 1014 alpha particles per minute. The only thing is that if you want to measure some kind of "decrease" in the activity (or count rate), you would have to wait a very long time.

Cheers

5. Aug 10, 2010

### SpY]

Yes, that's what I'm asking, if you have a half life of billions of years, how long would you have to wait for a decrease? And then measure enough decreases to get an exponential curve (to predict the HL)

6. Aug 10, 2010

Staff Emeritus

You don't need to see the decrease - you know how fast it's decaying by seeing how much it emits.

7. Aug 11, 2010

### fatra2

Hi there,

I see your point. Well knowing that every measurement, even over many many hours, will have a certain uncertainty. From what I remember, for alpha measurement uncertainty in environmental radiation is in the order of a few %. Therefore, to measure an effective decrease of counts per minutes (of 25%), you would have to wait a couple of millions of years.

Cheers

8. Aug 13, 2010