Radioactive Decay Differential Equation Question

[TFM]

Homework Statement

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?

Homework Equations

N/A

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

 

[tiny-tim]

$$ln u = -kt + C$$

Where C is a constant of integration

taking exponentials of both sides gives:

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

Hi TFM!

Nooo … u = Ce^-kt

 

[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

 

[tiny-tim]

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.

Are you tired?

How about u₀ = C?

 

[TFM]

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

?

TFM

 

[tiny-tim]

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

?

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

 

[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

 

[tiny-tim]

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

Yes (but a bit long-winded ).

Now simplify -ln(1/2).

 

[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

 

[tiny-tim]

Yup!

 

[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

 

[tiny-tim]

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

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

 

[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

 

[tiny-tim]

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

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?)

 

[TFM]

Well on wikipedia:

The atomic mass of an 56Fe isotope is 55.935 u and one mole of 56Fe will in theory weigh 55.935g,

and gives an equation:

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

Is this useful?

TFM

 

[tiny-tim]

If you can use it, it is useful.

If it is useful, use it. ​

 

[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.

 

[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 that's 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

 

[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.

 

[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

 

[Vuldoraq]

To convert k to per seconds you should times your k value by 1 year and divide by the number of seconds in one year, 60*60*24*365=3.1536*10^7.

 

[TFM]

this is probably going to sound rather silly but...How do you times it by 1 year? Wouldn't it just be times by 1?

?

TFM

 

[Vuldoraq]

K is the decay constant in this question, so the mean liftime is 1/k.

 

[Vuldoraq]

Yeah you just times it by one. I find it helps if you always consider the units you are using, so that per year times year removes the unit per year from your equation.

 

[TFM]

Okay, that does actually make a lot of sense.

So:

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

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

$$\frac{-\frac{ln\frac{1}{2}}{2.3 * 10^9}}{3.1536*10^7}$$

?

TFM

 

[Vuldoraq]

Looks good to me. Then just times that by the number of atoms and that should give you the number of atoms decaying per second.

 

[TFM]

So:

$$\frac{-\frac{ln\frac{1}{2}}{2.3 * 10^9}}{3.1536*10^7}$$

$$\frac{-\frac{ln\frac{1}{2}}{2.3 * 10^9}}{3.1536*10^7}* 1.45 * 10^{24}$$

I'm sure that this can be canceled down?

TFM

 

[Vuldoraq]

Err, just plug the numbers into a calculator to get the answer. That figure you have there is somewhat intimidating,

 

[TFM]

Now why didn't I think of that...?

I think its because usually they say to keep the logs as logs

So I get a value of $$1.39 * 10^7$$

Look about right?

TFM

 

[Vuldoraq]

I suppose you could collect termsto get,

$$\frac{\ln(2)}{5.002*10^{8}}$$

 

[Vuldoraq]

Looks good to me.

 

[TFM]

Execellent.

Thanks for your assitance, Vuldoraq and Tiny-Tim

TFM