# Calculating half life

1. Dec 4, 2014

### sam400

1. The problem statement, all variables and given/known data

Basically, the problem asked for the half life of Thorium-232, given the initial mass and the rate of decay.

\frac{dN}{dt} = 4100 \frac{ \alpha }{s}

initial mass = 1.0 g

2. Relevant equations

$$\frac{dN}{dt} = \lambda N$$
$$t_h = \frac{ \ln 2}{\lambda}$$

3. The attempt at a solution

Started off converting the mass into number of particles:

$$1.0 g \frac{1.0 mol}{232 g} 6.022*10^{23} molecule/mol$$

That gave me N, which was $2.59*10^{21}$ molecules. From that, I guess I could use the first equation to find the decay constant, so I got $1.58*10^{-18}$

From that, the second equation can be used to obtain a value for the half life, which then gave me the value, but I had to make sure to convert to years next. After using the proper conversion factor, I got a value of $8.1* 10^{11}$ years, which is different from the literature value of 1.4*10^{10}. I have a feeling I shouldn't have done what I did to find the decay constant, but I'm not entirely sure what other approach I can take, since I'm still new to nuclear. Thanks in advance.

Edit; I just realized that the method in my original post does give me the value found in literature. However, if I try to do it with a slightly different method, ie. convert the decay rate into grams/second, I seem to be getting the wrong answer.

$$4100 \frac{\alpha}{s} \frac{ 6.64E-27 kg}{ 1 \alpha} \frac{1 g}{10^{-3} kg} = 2.72E-20 g/s$$

This gives me a different decay constant value entirely, which then gives me the value I originally stated, so I think my original method is still wrong, or I made a conversion error.

Last edited: Dec 4, 2014
2. Dec 4, 2014

### Bystander

is in units of what?

is in units of what?

3. Dec 4, 2014

### sam400

dN/dt is in alpha particles per second, and the second is grams per second

4. Dec 4, 2014

### Bystander

... and, g/s would be how many alpha/s?

5. Dec 4, 2014

### sam400

to convert g/s to alpha/s, I would have to at first take 1 mol/alpha mass in g then multiply by avogadro. This would mean that the conversion I did in the second part was incorrect. However, doing the method described here also gives me a different decay constant value.

6. Dec 4, 2014

### Bystander

... and, how many alphas/s per how many thorium atoms?

7. Dec 4, 2014

### sam400

when I used that method, I do get the right answer, but for the alternate method, which is converting alpha/s to g/s, and leaving the thorium to just 1.0 g, I get the wrong answer. The conversion I used for alpha/s to g/s is after the edit on my original post. Since one method works and the other doesn't, I am guessing either I made a conversion error on the second method, or the process itself was incorrect and I got lucky in the first method.

8. Dec 4, 2014

### Bystander

When you're working with half-life, you're counting particles, not their masses. If half-life is discussed as N1/2/N0, and people say, "Gee after one half-life, there's only half the material left," it's natural to think that half the mass has disappeared. Half the original parent isotope mass has disappeared, but that has been replaced in the sample with whatever "daughter" product is the result of the decay --- for an alpha decay process, this is a measureable mass loss, 232 to 228, but the product is still present in the sample.

9. Dec 4, 2014

### sam400

so converting it into mass as I did it the second time would be like saying the sample is gaining or losing mass, when in reality the mass just stays constant, it's the particles that's change, which would mean everything has to be converted to number of particles, and the other way would just give an incorrect answer?

10. Dec 4, 2014

### Bystander

Yes, with the qualification that some wise guy is going to throw masses of this or that at you on tests, homework, quizzes, or give you a "mass half-life" and ask for a "population half-life," and all permutations/perversions and combinations of the above. Radioactive half-lives are all in terms of populations.