# Homework Help: Radioactive Decay Differential Equation Question

1. Nov 9, 2008

### TFM

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

Unobtainium is a radioactive (and fictional) element, with its rate of decay being proportional to the amount of Unobtainium, $$u' = −k u$$. The half-life of Unobtainium, in other words the time it takes for half the initial amount to decay, is $$2.3 × 10^9$$yr.

(a) Find the value of k.

(b) How many decay events will there be per second in a 1 kg block of Unobtainium, given
that it has an atomic mass of 415?

2. Relevant equations

N/A

3. The attempt at a solution

I think I am doing the right thing, but I can't seem to work out k. I have a feeling I need limits to me integrals. so far I have:

$$\frac{du}{dt} = -ku$$

$$\frac{1}{u} dt = -k dt$$

$$ln u = -kt + C$$

Where C is a constant of integration

taking exponentials of both sides gives:

$$u = e^{-kt} + C$$

If we had a $$u_{0}$$ somewhere, we know the half life, and thus could say that $$u = \frac{1}{2} u_o$$, stick some values for u and u0 in, and thus could probably work out k.

Any ideas where I am going wrong?

TFM

2. Nov 9, 2008

### tiny-tim

Hi TFM!

Nooo … u = Ce-kt

3. Nov 9, 2008

### TFM

That was rather silly of me, thanks.

I still need a $$u_0$$ in there somewhere though to calculate k, don't I...?

I also need to work out the constant C as well.

TFM

4. Nov 9, 2008

### tiny-tim

Are you tired?

5. Nov 9, 2008

### TFM

How did you get C to equal $$u_0$$

???

TFM

6. Nov 9, 2008

### tiny-tim

If u = Ce-kt, what is u when t = 0?

7. Nov 9, 2008

### TFM

I see, so:

$$u = Ce^{-k*0}$$

$$u = Ce^0$$

and:

$$e^0 = 1$$

Thus:

$$u = C$$ at t = 0

thus C = $$u_0$$

Thanks,

So now:

$$u = u_0e^{-kt}$$

Insert values for u and u_0 that follow half-life, eg u_0 = 10, u = 5:

$$5 = 10e^{-k*2.3 * 10^9}$$

Divide both sides by 10:

$$\frac{5}{10} = e^{-k*2.3 * 10^9}$$

Take logs of both sides:

$$ln(\frac{1}{2}) = -k*2.3 * 10^9$$

divide by half life:

$$\frac{ln\frac{1}{2}}{2.3 * 10^9} = -k$$

thus

$$k = -\frac{ln\frac{1}{2}}{2.3 * 10^9}$$

Does this look right?

TFM

8. Nov 9, 2008

### tiny-tim

Yes (but a bit long-winded ).

Now simplify -ln(1/2).

9. Nov 9, 2008

### TFM

So:

$$k = -\frac{ln\frac{1}{2}}{2.3 * 10^9}$$

$$-ln\frac{1}{2}$$

Thi simplifies to:

$$ln(1) - ln(2)$$

ln(1) = 0

thus:

$$k = -\frac{-ln2}{2.3 * 10^9} = \frac{ln2}{2.3 * 10^9}$$

TFM

10. Nov 9, 2008

### tiny-tim

Yup!

11. Nov 9, 2008

### TFM

Excellent. Thanks.

So for part b)

How many decay events will there be per second in a 1 kg block of Unobtainium, given
that it has an atomic mass of 415?

We know now that:

$$u = u_0e^{\frac{ln\frac{1}{2}}{2.3 * 10^9}t}$$

Do we insert t = 1 second, and u_0 = 1?

TFM

12. Nov 9, 2008

### tiny-tim

t = 1, and u0 is the number of atoms of Unobtainium in 1kg

13. Nov 9, 2008

### TFM

The question states that it has an atomic mass of 415. This means, if I remember correctly, it has 415 atoms per kilogram, thus 1 kg = 415...?

TFM

14. Nov 9, 2008

### tiny-tim

oooh, that's chemistry … i've no idea!

though that does see a rather heavy atom …

(and, come to think of it , doesn't hydrogen have an atomic mass of 1?)

15. Nov 9, 2008

### TFM

Well on wikipedia:

and gives an equation:

$$m_{\rm{u}}={m_{\rm{grams}} \over N_{A}}$$

Is this useful?

TFM

16. Nov 10, 2008

### tiny-tim

If you can use it, it is useful.

If it is useful, use it.

17. Nov 10, 2008

### Vuldoraq

Hey,

It should be $$415 g*mol^{-1}$$. To find the number of mols divide the total mass by the molar mass. To find the number of atoms multiply the number of mols by Avogadro's number. Then put the number of atoms as your value of u to find the number of atoms decaying per second.

18. Nov 10, 2008

### TFM

Thanks, so its

$$415 grams m^{-1}$$

we have 1000 grams

so:

$$\frac{1000}{415} = 2.4 mols$$

So now we times this number by Avagaro's Constant, $$6.022 * 10^{23}$$

This gives the number of particles to be:

$$1.45 * 10^{24}$$

So the equation is:

$$u = u_0e^{\frac{ln\frac{1}{2}}{2.3 * 10^9}t}$$

t = 1
$$u_0 = 1.45 * 10^{24}$$

$$u = (1.45 * 10^{24})e^{\frac{ln\frac{1}{2}}{2.3 * 10^9}1}$$

This gives the answer to be

U = 1.45e^24

But thats because the answer is so small.

we need to find the number of decay events, so I assume we take away u from u_o,

so this gives:

$$4.27*10^{14}$$ events per second. Seem right?

TFM

19. Nov 10, 2008

### Vuldoraq

Your amount for $$u_{0}$$ is correct.

The rest is incorrect. $$u \&\ u_{0}$$ are simply the number of atoms in your sample and as such have no units. You want the number of atoms decaying per second in a 1 kg sample. You have the total number of atoms in 1kg, $$u_{0}$$. You have already found the decay constant, in this case k. You need only to multiply these two values to find your answer. Can you see why? (Consider the units of k.)

Also don't forget to change k from per year to per second.

20. Nov 10, 2008

### TFM

Well the decay constant is 1/mean lifetime

So the k at the moment is:

$$k = -\frac{ln\frac{1}{2}}{2.3 * 10^9}$$

Converting into seconds gives a value of $$-9.4923*10^{-3}$$

So multiplying u0 with k(secs) gives: $$1.376*10^22$$

???

TFM