# How do you solve this?

1. Dec 19, 2006

### boredooom

I got an exam tomorrow, and since I missed the last week, I don't really know how to solve questions like these:

2. Relevant equations

1 gramm of Ra-226 radiates 3,7*10^10 alpha-particles in a second. Find out the half-life period of Ra-226.

We produce a radioactive Cu-64 and seal it. After 20 days, the activity is 20 Bq. What was the mass in gramm of the Cu-64 when we produced it?

Answers and explanations would be greatly appreciated,

Thanks,

Pear

2. Dec 19, 2006

### daveb

1) You have the mass of the sample, so from this you can find the number of atoms in the sample. This, with activity, can be used to find half-life.
2)Reverse that for the copper (though you'll need to look up half-life for Cu-64).

3. Dec 19, 2006

### boredooom

i don't think i understand.. can you explain more please?

4. Dec 19, 2006

### daveb

What don't you understand? If you don't know the formula for finding the number of atoms in a sample given its gram atomic weight and mass, you should be able to find it in your book. The same goes for half life if given the activity and number of atoms.

5. Dec 19, 2006

### ShawnD

The problem is difficult because you are given useless information. Try doing it with this information:

Half-life means you have 0.5g left

6. Dec 19, 2006

### OlderDan

What part of the given information is useless?

7. Dec 19, 2006

### ShawnD

You don't need to know which element or isotope is decaying.

8. Dec 19, 2006

### OlderDan

You might want to rethink that. The half-life depends on the fraction of the available nuclei that decay each second.

9. Dec 20, 2006

### ShawnD

I understand where you are coming from, but doing it that way would require doing an integral, which seems a few grades higher than the source of this question.

your way would be something like this (the E means 10 to the power of the next number):
(6.02E23)/226 = 2.66E21 atoms; the first number is the AMU equivalent of 1g
(3.7E10)/(2.66E21) = 1.389e-11 decaying per atom per second (each decay has a mass of 4)

I never really got the hang of integrals but this is the best I can do (which is wrong).
$$\int^{???}_{0} (4)(1.389E^{-11})(2.66E^{21})((1) - (1.389E^{-11}))^{t} dt$$
= (0.5)(6.02E23)
It's confusing as hell. How do I subtract the amount of decayed atoms from the nondecayed?

Last edited: Dec 20, 2006
10. Dec 20, 2006

### OlderDan

The underlying principle is that the probability of any one atom decaying in a small time interval dt is constant, so the rate of decay or activity A(t) is proportional to the number of atoms available for decay N(t)

A(t) = -dN(t)/dt = λN(t)

The minus is because N(t) is decreasing with time.

The function N(t) that satisfies this differential relationship is the familiar decaying exponential. It could be found by separating variables and integrating, but the result is well known, so you don't need to repeat that part of the derivation.

11. Dec 20, 2006

### tim_lou

I think you do need all the information give, since you have to find lambda to find half life (or the mean-life of the decay).

12. Dec 21, 2006

### daveb

All you need is the relationship between activity, number of atoms (implying you need the mass of the sample) and half life.
Activity = N*lambda = N*ln2/Half-life, so half-life = N*ln2/A. N = Avogardo's*mass/gram atomic weight, so
Half-life = Avogadro's number*mass*ln2/Gram Atomic Weight. Mass and atomic weight are given, so you don't need to use an integral, and you do need the isotope.