What Was the Activity of Am-241 on June 16, 2004?

[Lissajoux]

Homework Statement

{}^{241}_{95}Am produces \alpha particles, which interact with {}_{4}^{9} Be to produce neutrons.

The Am sample emits \alpha particles at a rate of 70.0s^{-1} on June 16, 1996.

What was the activity on June 16, 2004?

Homework Equations

The half life for {}^{241}Am :

T_{1/2}=432.2y

Also know that:

A=-\frac{dN}{dt}=\lambda N

The Attempt at a Solution

I know this calculation isn't particularly tricky, but I just can't figure it out at the moment

I've calculated the wavelength:

\lambda = \frac{ln(2)}{T_{1/2}} = \frac{ln(2)}{1.363\times 10^{10}}} = 5.086\times 10^{-11}m = 0.05nm

Then guessing just need to calculate what the value of N is, and can multiply this with the value of \lambda to get A.

Now N, or rather N(t), is just the number of nuclei remaining in the sample after a time t.

I'm not sure, but is this just.. 241?! :shy:

I'm getting a bit confused with this, really shouldn't be I know , but a bit of guidance would be great.

 

[nickjer]

First note: \lambda isn't wavelength in this problem, it is called the decay constant. So check your units.

You don't know your initial or final N. But you know the initial dN/dt. Knowing the initial dN/dT you can plug this into your 2nd equation to get the initial N. Then you are right... Solve for the final N and get the final dN/dt.

A simpler way would be to take the ratio of both dN/dt's, like so:

\frac{\frac{dN_f}{dt}}{\frac{dN_i}{dt}} = \frac{N_f}{N_i} = \frac{N_0 e^{-\lambda t_f}}{N_0 e^{-\lambda t_i}} = e^{-\lambda (t_f-t_i)}

 

[Lissajoux]

Ah yes of course it's the decay constant not the wavelength, silly me!

Right so something like this:

\frac{dN}{dt}=70.0s^{-1}

Have calculated \lambda as: 5.086\times 10^{-11} s^{-1}

So can rearrange this equation:

A=-\frac{dN}{dt}=\lambda N \implies N = \frac{\frac{dN}{dt}}{\lambda}

Therefore:

N = 1.376 \times 10^{12}

.. which is the initial N.

Correct?

Then I need to calculate final N, which I can do from:

N(t)=N_{0}e^{-\lambda t}

Using the calculated value of \lambda and t value of .. 8? Or 252.288\times 10^{6}. I.e. I assume it's the latter and that I need to convert this too to seconds?

Once I've found this value of N, can use A=-\lambda N to get the activity at this time, right?

Hopefully I'm getting there with this question somewhat.

 

[Rajini]

Hello nickjer,
is it not correct to take 70 per sec as initial value?
just curious to know.

 

[nickjer]

Remember your t is in units of years. So convert them to seconds. You are right about plugging the time back into get the final N. Then you get the activity from this.

Or you can just use the final formula I gave in the last post. Just solve for the decay constant in units of years (very easy to do). Then plug it in, and keep your t in units of years. Since you know the initial dN/dt, the final dN/dt just comes out from it.

Edit: It is correct to take 70 per sec as the initial rate, dN/dt. Just make sure all the units work out correctly.

 

[Lissajoux]

So if I stick in t=252.288\times 10^{6} into N(t)=N_{0}e^{-\lambda t} I can find the final value of N, or rather the value of N at the later time stated in the question. Then can use this to find A at the time required but using A=-\lambda N

 

[nickjer]

Yes, that is correct. Just make the solution positive in the end.

 

[Rajini]

Okay,
then better use this:
<br /> <br /> N(t)=N_0\exp\left(\frac{-t\ln 2}{T_{1/2}}\right),<br /> <br />.
take N₀ as 70 s^-1.
t is 8 yrs and T_1/2 is 432.2 yrs

 

[nickjer]

That formula can be confusing with your notation. The same formula can be seen in my first post at the end.

 

[Lissajoux]

The minus sign simply denotes that the activity is decreasing, so I can ignore it in my actualy calculations because the value I calculate must be positive, right. If that makes sense.

 

[nickjer]

The minus sign just means the number of particles is decreasing. You can ignore it because they just want magnitude of the rate.

 

[Lissajoux]

Right so I've got that N(t)\le N_{0} , so that N(t)\approx 0.987 N_{0}

Then from this value, that A at time t=8y is 69.1.

Do I have to round this to an integer value i.e. 69? I assume not, because A before was 70.0, and this is average decay rate so doesn't have to be whole number of nuclei right? So I'd have that A=69.1 s^{-1}

.. correct?

 

[Rajini]

Yes, if you use the formula that i posted you get 69.1 per sec.
Also you no need to round off. But you mentioned that the solution is 241 ??[no unit!]
Do you understand the formula. ?

 

[Rajini]

First note: \lambda isn't wavelength in this problem, it is called the decay constant. So check your units.

You don't know your initial or final N. But you know the initial dN/dt. Knowing the initial dN/dT you can plug this into your 2nd equation to get the initial N. Then you are right... Solve for the final N and get the final dN/dt.

A simpler way would be to take the ratio of both dN/dt's, like so:

\frac{\frac{dN_f}{dt}}{\frac{dN_i}{dt}} = \frac{N_f}{N_i} = \frac{N_0 e^{-\lambda t_f}}{N_0 e^{-\lambda t_i}} = e^{-\lambda (t_f-t_i)}

Here please note that this formula is something different. Here 't' is not same for both initial and final states. So i guess this formula is something different and may not be applicable for you.
EDIT: Since you have 3 unknowns factors, t_f,t_i,N_f.

 

[nickjer]

Ummmm... You do know them:

t_f = June 16, 2004
t_i = June 16, 1996

So you are left with:

t_f - t_i = \Delta t = 2004 - 1996 = 8 yrs

It is just the change in time.

 

[Lissajoux]

Sorry Rajini you've now confused me.

Is this correct what I have calculated:

N(t) \approx 0.987 N_{0}

Then from this value:

A at time t=8y is A=69.1 s^{-1}

 

[Rajini]

Sorry!
Nickjer, I overlooked some thing.
soln. is correct 69.1 per sec. (I did not confuse you). Offcourse you can use the formula which i wrote or Nickjer's one.

 

[Lissajoux]

I also need to find out how many Am nuclei were in the same at each date. How do I go about this? can't figure it out.

 

[nickjer]

Use the 2nd equation from your first post. The equation with A.