Nuclear chemistry doubts

[ronthepon]

Now I have been told that the nuclear decay is a first order reaction, i.e. the rate of nuclear decay is proportional to the number of atoms present in a sample.

 

Why?

 

Why don't they all decay after the same amount of time?

 

Why do they take different amounts of time for decaying?

 

And if there is just one atom present, then when will it decay?

 

Is there any well defined, fully believed theory behind this phenomena?

[Jay-qu]

Think it out, though it is only a probability that the atom will decay at any moment we can calculate half-lives with a fair accuracy.

 

Im sure you know what a half life is (no not a video game) the time taken for half the sample to decay, half is only defined by how much you start with. Then from there you can work out the rate in bequerals, disintergrations per second.

 

As for the disintergration of individual atoms, I think it is inherently spontaneous (from our level of observation) but we can calculate probabilities of when it should decay. Im pretty sure if you co-related many results of this you would end up with a normal curve, ie most decays will happen after a specific time and as you move further from this time, either +ve or -ve it becomes less and less likely to have disintergrations at that time. Sorry if that doesnt make to much sense..

[ronthepon]

No, it makes good sense. You are saying that perhaps the probability of decay for an atom first increases, then reaches a maximum at some point of time, then reduces again?

 

Well, I do not fully agree. Nuclear decay is a first order process.

This means:

 

Suppose there are 2^64 atoms at time = 0. Probability of atom decay in half life time period(t) = 50%. (This is something I am supposing, and we'll see if it gives theoretical predictions which agree with observations.)

So 50% of atoms decay between time = 0 and time = t.

 

At time = t, 50% of atoms remain. Between time = t and time = 2t, again there is a 50% chance of decay for each atom. Thus 50% of the atoms decay.

 

And so on...

 

Is my reasoning correct? I'm not too sure of it.

 

 

 

So it eventually boils down to the probability game. Right?

 

The answer to Q#3 is 'nobody knows'.

 

The answer to Q#1,2,4 remain unanswered. Note that I'm asking why it happens... not what happens.

[Jay-qu]

Do you understand the concept of a normal distribution, the bell shaped curve? because I think you misunderstood what I was trying to say..

[hallenrm]

I think it has something to do with energy distribution and activation energy. As you will definetly be aware of not all molecules of a compound undergo transformation all at once, that is so because only some of them have sufficient energy to undergo the transformation. Same is the case with radioactive atoms; only the atoms that have energy in exccess of the energy neccessary for the nuclear decay do undergo the transformation at a particular moment. :)

[ronthepon]

Jay-Qu:

Well... I admit that I do not know about the concept fully, (i.e. mild case of -> :confused:)

 

But I have seen some references. Wanna help me a little?

 

Have a look at the equations I've been looking at.

 

Suppose:

 

[math]n[/math] - number of nuclei present at an instantaneous moment.

 

[math]-{\frac{dn}{dt}}[/math] - rate of change of number of nuclei.

 

Then,

[math] -{\frac{dn}{dt}} = kn[/math] This, I am 100% sure of. This is hard truth. (And I don't know how to make a sign of proportionality in latex, so I've put a k)

 

Uhh... now...:shrug:

 

Integerate... I think...

 

[math] {{Log}_e}{n} = {{Log}_e}{n_{o}} {\,} -k{{\Delta}t} [/math]

 

Yeah. That's perfect.:)

 

Note that [math]n_o[/math] refers to initial number of nuclei, and you'll need boundary condition evaluation to get the term it is in.

 

Now re-write the equation.

 

[math] n = {n_{o}{e^{-k{{\Delta}t}}} [/math]

 

From what I can guess, this does not form a normal curve. Right...? Uh... I'm not sure what the curve formed is called, but it does not seem like the ones on wiki.

 

Wanna help me out, Jay?

 

PS: I had begun to complie this post around half an hour ago.

[ronthepon]

I've just read your post hallenrm...

 

So if nuclei need a particular minimum energy to begin decay, then we can infer from the observation that they decay in a fashion: 'one after the another', that they are gaining energy from some source.

 

Is it so?

[hallenrm]

They gain/lose energy from one another. Energy is somewhat like money, someone's gain is another one's loss.

[ronthepon]

In which case... there must be only a limited amount of decays before all the energy free to do this is completely dissipated out of the system or used up.

[Jay-qu]

I dont think energy plays as big of a part in nuclear reaction as it does in chemical reactions.. but I must admit I have never heard anything contrary to the fact.

 

Ron, you math is fine, what you have shown is the number of remaining nuclei will exponentially decrease. What I am talking about is if you consider each nuclei on its own and count how long it takes to disintergrate, then plot a graph of seconds vs no. of desintergrations that second, you will see that it will be normally distributed around the half life.

 

I think I will try and draw this, just a min..

[Jay-qu]

ok here it is :lol:

 

 

This graph is approximately what I think it should look like, puts into words what I have been trying to say. Note the lines will have an asymptope of becquerals = 0, in maths the graph is continuous, that means it never really touches but gets really close, while physically if you graphed it, it would cross hit the line at some point because we are working with discreet atoms.

[ronthepon]

(how'd you make the graph so smooth?!)

 

You mean that maximum activity is shown at a particular time for a group of radioactive nuclei?

 

... and half life is that time?

[Jay-qu]

Yup, but at any one time (mathematically) from the start of the experiment there is a small amount of activity.

 

I just used MSpaint there is a little curved line tool on it, took a few goes but I got it :lol:

[ronthepon]

Aha! I think I got it!

 

You are plotting [math]\frac{dn}{dt}[/math] with t, right?

 

So let me try to get the equations.

 

[math]\frac{dn}{dt} = -kn [/math]

 

and

 

[math]n = {n_o}{e^{-kt}} [/math],

 

we get

 

[math]\frac{dn}{dt} = -k{n_o}{e^{-kt}} [/math]

 

...but this is another exponential one.

[Jay-qu]

no quite.. I will look into normal curve equations, they are fairly complex