# Question On Half Lives

### #1

Posted 18 October 2009 - 12:56 AM

"...you take a known mass of some radioactive isotope (i.e., a known number of atoms to start with), and count how many decays there are in a given period of time (i.e., how many atoms out of the original total decay). Because there are such a huge number of atoms in a macroscopically sized sample (on the order of 10^24 of them in a kilogram of U-238), even a very short period of time compared to the half-life is enough to get a large number of decay counts and, with an accurate enough counter, to show a change in the count rate (for example, the half-life of U-238 is about 10^17 seconds, so in one second we would expect about 10^7 counts from a kilogram of U-238; after a few minutes we would have more than 10^9 counts, and the number of counts per second would have decreased by about 10^2, enough to detect the change in rate). Then you can do a curve fit and show (1) that the decay curve is exponential (within the limits of measurement accuracy), and (2) what the time constant (half-life) is."

I'm assuming 10^24 is an exponent of 10--correct? This is where my question lies. He says that the "counts per second would have decreased by about 10^2, enough to detect the change in rate)."

Now the half life of U238 is 4.46 billion years.

1. So if it the count is slowing down by 100 every few minutes, has it always slowed down at the same rate? How would we know?

2.How would we know what we stated with for a coefficient in this calculation--unless you had an already assumed starting point in time. He states "(2) what the time constant (half-life) is..." as one of the coefficients for a calculation. SO he is in effect figuring the assumed half life to figure the half life. Am I right or wrong?

2. "Natural uranium is about 99.27 percent U-238, 0.72 percent U-235, and 0.0055 percent U-234." Wikipedia Why do we find natural uranium almost 100% U238--there should be detectable amounts of some isotopes that would show the decay process all the way to lead206 if it is truly been decaying.

3. If the half life of U238 is truly 4.46 BA then shouldn't we see natural uranium 50/50 with a U234 mixture?

### #2

Posted 18 October 2009 - 04:46 AM

This is my first reply in this forum, so I hope IÃ‚Â´m doing everything right.

first question: 10^24 is the normal scientific notation for exponent 24 to the base 10, else written: 1,000,000,000,000,000,000,000,000 and that would be impractical.

to 1.: the count isnÃ‚Â´t slowing down by 100 for every few minutes. It is slowing down by 100 counts for a given amount of atoms in a fixed period of time. If there are fewer atoms left, the "slowing down" is slower lets say by 90 counts then by 80 counts and so on. If it were always 100 counts per second it wouldnÃ‚Â´t be exponential but linear.

The radioactive decay is a reaction of "zeroth order" that means it is independent of the amount of atoms. These reactions always follow an exponential curve. If we measure a tiny bit of the curve, we know how the whole curve looks like.

to 2.: donÃ‚Â´t know what you mean by "coefficient". Exponential decay curves always have a half life and by measuring the beta decay counts you could measure the half life. I donÃ‚Â´t see your problem, sorry.

to 2 (the second 2 ): thats the natural distribution of uranium to the most abundand isotopes. We mostly find U-238 because it is easier to be produced by solar fusion processes and it has a really long half life (is really stable). We do find other isotopes in naturally occuring uranium, all of the decay chain to lead-206. Take for example radium. ItÃ‚Â´s a very strong alpha radiator (hence the name) it was first isolated by Marie Curie from Uranium minerals.

to 3.: If I understand you correctly, you mean: If the half life is correct and the age of the earth is really 4.5 BA half of all U-238 should have decayed to U-234 and we should have roughly a 1:1 distribution between U-238 and U-234?

But the half life of U-234 is much shorter (by the factor 2x 10^4) than of U-238. It gets 2x 10^4 "half lifes" in the time U-238 gets one. So theres a lot of it already decaying before it gets "refilled" by other decayed U-238.

I hope I could help you. Feel free to ask more

greetings

### #3

Posted 18 October 2009 - 12:27 PM

Thanks skeptic, it's what I thought--an exponent for the lack of ability to type it like shown in text books. I'm not specialized in math or physics so I wanted to make sure it wasn't some other notation I wasn't familiar with.first question: 10^24 is the normal scientific notation for exponent 24 to the base 10, else written: 1,000,000,000,000,000,000,000,000 and that would be impractical.

I understand what you are saying about proportionality. I was talking within the measured quantity he gave me (i.e. kg of U238) but I failed to mention it. But the point of my question still is valid--the count slows down in a few minutes.Hi,to 1.: the count isnÃ‚Â´t slowing down by 100 for every few minutes. It is slowing down by 100 counts for a given amount of atoms in a fixed period of time. If there are fewer atoms left, the "slowing down" is slowerÃ‚Â Ã‚Â lets say by 90 counts then by 80 counts and so on. If it were always 100 counts per second it wouldnÃ‚Â´t be exponential but linear.

I understand the concept and I'm not confusing the two. If it seemed that I did it was my fault I did not express it. I was putting it in question form to see if I was indeed understanding the concept. Thanks.The radioactive decay is a reaction of "zeroth order" that means it is independent of the amount of atoms. These reactions always follow an exponential curve. If we measure a tiny bit of the curve, we know how the whole curve looks like.

By coefficient I mean absolute known value such as 2, as opposed to an unknown variable such as x.to 2.: donÃ‚Â´t know what you mean by "coefficient". Exponential decay curves always have a half life and by measuring the beta decay counts you could measure the half life. I donÃ‚Â´t see your problem, sorry.

We are trying to find half life here, not an age. Now if I'm trying to find x and then I assign x an estimation (please bear with me I understand it is a curve and therefore you say you know the whole curve) --by so doing I am changing the unknown variable I am trying to solve into a known coefficient. I then use that estimated coefficient to solve the unknown variable.

It is like taking a middle of the river measurement in calculating the HL. By that I mean you measure the amount of water that passes through a 4 ft long by 400 foot width of river and estimate how much water has passed through the river--when you don't know how long the river has been flowing in the first place.

As well you must assume that the rate today is the rate in the distant past. SO you in effect assign absolute value on the rate of 100 years ago when you really don't know it, and you assign absolute value on the water flowing through the 4 x 400 section of the river by estimating how long the water has been flowing through the river. Otherwise you can't solve the equation.

By a 4 x 400 ft section of river--this an analogy of the present day measurements (coordinates) and trigonometrical curves of U238 decay counts. U238 has been allegedly decaying for 4.6 years, when the "statistically probable" half life is 4. 46 BA. Therefore you have to have an absolute value on the HL to calculate the HL.

This is circular and therefore invalid.

to 3.: If I understand you correctly, you mean: If the half life is correct and the age of the earth is really 4.5 BA half of all U-238 should have decayed to U-234 and we should have roughly a 1:1 distribution between U-238 and U-234?

But the half life of U-234 is much shorter (by the factor 2x 10^4) than of U-238. It gets 2x 10^4 "half lifes" in the time U-238 gets one. So theres a lot of it already decaying before it gets "refilled" by other decayed U-238.

Yes, I was wrong to say 50/50 or 1:1 because U238 is in magnitude of 10^9 and U234 is 10^5. But still they are half lives so 99% plus purity in the natural uranium is very high. It seems there should also be lead 206 mixed in and there should be more than under a percentage point. And the lead should be mixed homogeneously within the confines of the natural uranium--IF the the HL is correct and the age of the earth.

It's just not lining up.

Thank you skeptic. I hope we can continue this discussion.

### #4

Posted 18 October 2009 - 01:50 PM

Thanks skeptic,Ã‚Â it's what I thought--an exponent for the lack of ability to type it like shown in text books.Ã‚Â I'm not specialized in math or physics so I wanted to make sure it wasn't some other notation I wasn't familiar with.

I understand what you are saying about proportionality.Ã‚Â Ã‚Â I was talking within the measured quantity he gave me (i.e. kg of U238) but I failed to mention it.Ã‚Â But the point of my question still is valid--the count slows down in a few minutes.

I understand the concept and I'm not confusing the two.Ã‚Â If it seemed that I did it was my fault I did not express it.Ã‚Â I was putting it in question form to see if I was indeed understanding the concept.Ã‚Â Thanks.

By coefficient I mean absolute known value such as 2, as opposed to an unknown variable such as x.Ã‚Â

We are trying to find half life here, not an age.Ã‚Â Now if I'm trying to find x and then I assign x an estimation (please bear with me I understand it is a curve and therefore you say you know the whole curve) --by so doing I am changing the unknown variable I am trying to solve into a known coefficient.Ã‚Â I then use that estimated coefficient to solve the unknown variable.Ã‚Â

It is like taking a middle of the river measurement in calculating the HL. By that I mean you measure the amount of waterÃ‚Â that passes throughÃ‚Â a 4 ft long by 400 foot width of river and estimate how much water has passed through the river--when you don't know how long the river has been flowing in the first place.Ã‚Â

As well you mustÃ‚Â assume that the rate today is the rate in the distant past.Ã‚Â SO you in effect assign absolute value on the rate of 100 years ago when you really don't know it, and you assign absolute value on the water flowing through the 4Ã‚Â x 400 section of the river by estimating how long the water has been flowing through the river.Ã‚Â Otherwise you can't solve the equation.

By a 4 x 400 ft section of river--this an analogy of the present day measurements (coordinates) and trigonometrical curves of U238 decay counts.Ã‚Â U238Ã‚Â has been allegedly decaying for 4.6 years, when the "statistically probable" half life is 4. 46 BA.Ã‚Â Therefore you have to have an absolute value on the HL to calculate the HL.

This is circular and therefore invalid.

Yes, I was wrong to say 50/50 or 1:1 because U238 is in magnitude of 10^9 and U234 is 10^5.Ã‚Â But still they are half lives so 99% plus purity in the natural uranium is very high.Ã‚Â It seems there should also be lead 206 mixed in and there should be more than under a percentage point.Ã‚Â And the lead should be mixed homogeneouslyÃ‚Â within the confines of the natural uranium--IF the the HL is correct and the age of the earth.Ã‚Â

It's just not lining up.

Thank you skeptic.Ã‚Â I hope we can continue this discussion.

I wanted to correct one thing I said

...and you assign absolute value on the water flowing through the 4Ã‚Â x 400 section of the river by estimating how long the water has been flowing through the river.

What I should have said is you assign an absolute value on the water THAT HAS PASSED through...the section of river by estimating [a time passage factor].

I understand you have a graphed curve as a result of the coordinates through the passage of time which it took to find the coordinates. The problem comes in when you put the assumed HL of the isotope into the problem.

It is like saying we noticed the river slowed down over a few days mesurements in the cubic section we are measuring, but we don't know the length of the river, how long ago it rained, how long the river has been here, or what other factors have affected the flow of this river over the duration of it's existence.

You have a kg of U238 in front of you which has not yet decayed in 4.6 billion assumed years of earth's existence--so you must assume this material here is "the remainder which has not yet decayed." In other words U238 has been decaying for 4.6 billion years and it's HL is about the same--so you assume that half the original supply of u238 has decayed by now. There is no way to calculate that this supply you have now is going to give you the same reading it would have then, or if the counts are attributable to some other unknown factor.

### #5

Posted 18 October 2009 - 08:42 PM

Thanks skeptic, it's what I thought--an exponent for the lack of ability to type it like shown in text books. I'm not specialized in math or physics so I wanted to make sure it wasn't some other notation I wasn't familiar with.

I understand what you are saying about proportionality. I was talking within the measured quantity he gave me (i.e. kg of U238) but I failed to mention it. But the point of my question still is valid--the count slows down in a few minutes.

I understand the concept and I'm not confusing the two. If it seemed that I did it was my fault I did not express it. I was putting it in question form to see if I was indeed understanding the concept. Thanks.

By coefficient I mean absolute known value such as 2, as opposed to an unknown variable such as x.

We are trying to find half life here, not an age. Now if I'm trying to find x and then I assign x an estimation (please bear with me I understand it is a curve and therefore you say you know the whole curve) --by so doing I am changing the unknown variable I am trying to solve into a known coefficient. I then use that estimated coefficient to solve the unknown variable.

The equation for exponential decay is C=C0 * (1/2) ^ (t/T)

t = time since the start of the experiment

C0 = the number of counts at time t=0

C= the number of counts after time t

T = the half life

In the experiment that you described in your opening post we have t, C, and C0 so we can solve for T. We know that the equation applies because we can chart C and show that an exponential curve fits (

*Then you can do a curve fit and show (1) that the decay curve is exponential*)). Since we have 1 equation and 1 unknown it's possible to solve for the half life (

*(2) what the time constant (half-life) is."*)

Rearanging the equation so we can solve for T

C=C0 * (1/2) ^ (t/T)

C/C0 = (1/2) ^ (t/T)

ln(C/C0) = (t/T) ln((1/2))

ln(C/C0) = (t/T) ln((1/2))

T =t*ln((1/2)/ ln(C/C0)

Just measure the number of decays at the start of the experiment, start your stopwatch, then measure the decays after a certain amount of time t, plug in those values, and get T. There's no estimated coefficient.

Yes, I was wrong to say 50/50 or 1:1 because U238 is in magnitude of 10^9 and U234 is 10^5. But still they are half lives so 99% plus purity in the natural uranium is very high. It seems there should also be lead 206 mixed in and there should be more than under a percentage point. And the lead should be mixed homogeneously within the confines of the natural uranium--IF the the HL is correct and the age of the earth.

It's just not lining up.

Thank you skeptic. I hope we can continue this discussion.

You are misunderstanding what wikipedia is saying about uranium. The 99.27 percent U-238, 0.72 percent U-235, and 0.0055 percent U-234 figures are not talking about purity of deposits, they are talking about ratios of abundance. In other words, if you were to grab a single atom of uranium, there's a 99.27 chance that atom would be 238, .72 percent chance it would be 235, and .0055 chance it would be 234. It doesn't matter how many non-uranium atoms you grabbed in addition to that single uranium atom, they don't affect the chances of that single uranium atom being a 238,235 or 234 atom.

Or for an example, 1 kg of pure uranium is going to be made up of 99.27% U-238, .72% U-235, and .0055% U-234, now mix that 1 kg of uranium with 10 kg of lead, the 1kg of uranium is still going to made up of 99.27% U-238, .72% U-235, and .0055% U-234.

### #6

Posted 19 October 2009 - 02:15 AM

I also hope we could continue this discussion.

We are trying to find half life here, not an age. Now if I'm trying to find x and then I assign x an estimation (please bear with me I understand it is a curve and therefore you say you know the whole curve) --by so doing I am changing the unknown variable I am trying to solve into a known coefficient. I then use that estimated coefficient to solve the unknown variable.

Well, numbers said it right, you donÃ‚Â´t have to assume any coefficient. At first the half life is unknown, then you measure it by counting the decaying atoms.

By doing this again and again, you get the half life with a little bit of statistical error.

IÃ‚Â´m sorry for not getting your point, but we estimate nothing in this process, we just measure the half life.

One point you could make is: is the measured half life always constant?

Is that what you are getting at?

A river isnÃ‚Â´t a good analogy but letÃ‚Â´s stick with it. Ok, you have a river, you measure how many water flows through a certain part. You could do this at different days at different seasons and get different values.It is like taking a middle of the river measurement in calculating the HL. By that I mean you measure the amount of water that passes through a 4 ft long by 400 foot width of river and estimate how much water has passed through the river--when you don't know how long the river has been flowing in the first place.

Maybe its constant within an error limit, maybe the amount of water gets less and less. In the latter case you could plot all the measurements into a graph and maybe you get a certain type of curve. If its an exponential decay curve you could determine the half life. But thats were the analogy gets bad, because there are a lot of factors which could affect the flow rate of a river. There is no known factor which could affect the decay of an atom (under normal conditions).

As well you must assume that the rate today is the rate in the distant past. SO you in effect assign absolute value on the rate of 100 years ago when you really don't know it, and you assign absolute value on the water flowing through the 4 x 400 section of the river by estimating how long the water has been flowing through the river. Otherwise you can't solve the equation. By a 4 x 400 ft section of river--this an analogy of the present day measurements (coordinates) and trigonometrical curves of U238 decay counts. U238 has been allegedly decaying for 4.6 years, when the "statistically probable" half life is 4. 46 BA. Therefore you have to have an absolute value on the HL to calculate the HL. This is circular and therefore invalid.

I donÃ‚Â´t get it. Why canÃ‚Â´t I solve the equation? And why is it invalid? I have an amount of a radioactive isotope. I measure the decay. The decay gets less an less over time. I plot the decay rate vs. the time of measurement, see its getting down exponentially and could calculate the half life (which is nothing more than a factor of "steepness" of the curve). Where is this circular?

Wether or not the half life is really a constant is a different question. By fitting the curve to the measured values I see, that its following an exponential slope, so I can say in the distance of the measurement it is a constant.

Now we have to solve the problem if it was always a constant, ok?

Yes, I was wrong to say 50/50 or 1:1 because U238 is in magnitude of 10^9 and U234 is 10^5. But still they are half lives so 99% plus purity in the natural uranium is very high. It seems there should also be lead 206 mixed in and there should be more than under a percentage point. And the lead should be mixed homogeneously within the confines of the natural uranium--IF the the HL is correct and the age of the earth.

No Problem, I get a lot of things wrong, too. ItÃ‚Â´s human

As I tried to say with the radium example. You get all the other isotopes in the decay chain down to lead together with uranium. The percent values in the wiki are only the distributions between the uranium isotopes. In nature uranium exists mostly as pitchblende which is an oxide of uranium. The crystal structure of pitchblende is completely destroyed because of the radioactivity and the high fraction of foreign ions like for example lead. Within pitchblende the distribution of the different uranium isotopes is 99.27 percent U-238, 0.72 percent U-235, and 0.0055 percent U-234 but there are a lot of other metal ions in it.

I donÃ‚Â´t know which explanations you like better, numbers or mine but maybe it gets clearer by explaining it from different angles.

Hope that helped.

greetings

### #7 Guest_tharock220_*

Posted 19 October 2009 - 04:38 PM

1. So if it the count is slowing down by 100 every few minutes, has it always slowed down at the same rate? How would we know?

2.How would we know what we stated with for a coefficient in this calculation--unless you had an already assumed starting point in time. He states "(2) what the time constant (half-life) is..." as one of the coefficients for a calculation. SO he is in effect figuring the assumed half life to figure the half life. Am I right or wrong?

2. "Natural uranium is about 99.27 percent U-238, 0.72 percent U-235, and 0.0055 percent U-234." Wikipedia Why do we find natural uranium almost 100% U238--there should be detectable amounts of some isotopes that would show the decay process all the way to lead206 if it is truly been decaying.

3. If the half life of U238 is truly 4.46 BA then shouldn't we see natural uranium 50/50 with a U234 mixture?

Radioactive ecay is a differntial. You can use your measurements to solve for the proportionality constant. Then you solve the equation for .5 on the left side.

### #8

Posted 19 October 2009 - 05:50 PM

C=C0 * (1/2) ^ (t/T)

First of all I don't know the notations in these problems. * and ^

I understand the concept of exponential increase and a trigonometrical curve--the curve is trigonometrical isn't it?

But I see the the T (half life) in the equation. So are they using the half life IN the equation to figure the half life? That would be circular. (t/T) would be:

passed time since the beginning of the experiment / the half life. Or is this the answer/ product/quotient/sum?

### #9

Posted 19 October 2009 - 07:47 PM

Thanks guys.

C=C0 * (1/2) ^ (t/T)

First of all I don't know the notations in these problems. * and ^

* = multiplication. It's used instead of x because many equations use x as a variable

/ = division. 2/3 is 2 divided by 3.

^ = exponent. 3^2 is 3 squared. (1/2) ^ (t/T) is .5 raised to the power of t/T

ln = natural log, it's how you get T on one side of the equation. Basically if

*(base)^(exponent)=result*then

*exponent * ln(base) = ln(result)*

Trigonometry usually refers to cosine, sine, tangent, secant etc. I'd think it'd be more accurate to say the curve is exponential but that seems like a very minor issue.I understand the concept of exponential increase and a trigonometrical curve--the curve is trigonometrical isn't it?

I showed you how do do it in my post, you rearrange the equation to solve for T. That was the whole point of your question, how to solve for T, wasn't it?But I see the the T (half life) in the equation. So are they using the half life IN the equation to figure the half life? That would be circular. (t/T) would be:

passed time since the beginning of the experiment / the half life. Or is this the answer/ product/quotient/sum?

Given an equation such as volume = length * width * height, if we know the volume, the length, and the width we can solve for the height=volume/(length*width), no circularity involved. Same thing with the half life equation. We know C, C0, and t so we can solve for T.

if C=C0 * (1/2) ^ (t/T) is confusing just use T =t*ln(1/2)/ ln(C/C0). They are exactly the same equations.

### #10

Posted 20 October 2009 - 04:15 AM

It is like saying we noticed the river slowed down over a few days mesurements in the cubic section we are measuring, but we don't know the length of the river, how long ago it rained, how long the river has been here, or what other factors have affected the flow of this river over the duration of it's existence.

You have a kg of U238 in front of you which has not yet decayed in 4.6 billion assumed years of earth's existence--so you must assume this material here is "the remainder which has not yet decayed." In other words U238 has been decaying for 4.6 billion years and it's HL is about the same--so you assume that half the original supply of u238 has decayed by now. There is no way to calculate that this supply you have now is going to give you the same reading it would have then, or if the counts are attributable to some other unknown factor.

Hi AFJ

Radioactive decay rates for given isotopes have never been observed to vary beyond the precision of the instruments, in spite of countless measurements for over 100 years.

Not only that, we have some measurements of radioactive decay that span very long time periods:

- The Oklo reactor was a natural fission reactor that went critical about 2 billion years ago. Detailed analysis of its fission products show that the equations of nuclear physics worked the same then as they do today.

- Decay rates have been measured with great precision in supernovae 1987a and 1991t, at 169,000 and 60 million light years distance - and the rates are the same in these distant stars as they are on earth today.

So, the long-term unvarying rates of radioactive decay are not just assumed from extrapolation, they have been directly observed and experimentally confirmed.

Cheers

SeeJay

### #11

Posted 21 October 2009 - 04:20 PM

Hi AFJ

Radioactive decay rates for given isotopes have never been observed to vary beyond the precision of the instruments, in spite of countless measurements for over 100 years.

Not only that, we have some measurements of radioactive decay that span very long time periods:

- The Oklo reactor was a natural fission reactor that went critical about 2 billion years ago. Detailed analysis of its fission products show that the equations of nuclear physics worked the same then as they do today.

- Decay rates have been measured with great precision in supernovae 1987a and 1991t, at 169,000 and 60 million light years distance - and the rates are the same in these distant stars as they are on earth today.

So, the long-term unvarying rates of radioactive decay are not just assumed from extrapolation, they have been directly observed and experimentally confirmed.

Cheers

SeeJay

Yes, thank you SeeJay,

This is not my strong subject area--really kind of new for me --but a challenge.

In regards to your comments--I realize that everything looks great on paper, as it does in business also. But on the ground (just as in business) it is usually different.

Take the example of this article by Andrew Snelling. It shows the real nitty gritty of dealing with ore that has been open but dated anyway.

Even within the crystal lattice there are substitutions. Snelling writes, "...but even at the microscopic level within pitchblende grains themselves. Fig. 8 illustrates how Pb and Ca are both substituting for U in the UO2 cubic lattice."

From grain to grain in the pitchblende there is notable change. So the picture that everything corresponds well is not reality. "In addition, the pitchblende grains donÃ¢â‚¬â„¢t have uniform compositions so that Ã¢â‚¬ËœdatingÃ¢â‚¬â„¢ of sub-sections of any grain would tend to yield widely divergent U/Pb ratios and therefore varying Ã¢â‚¬ËœagesÃ¢â‚¬â„¢ within that single grain."

Snelling writes in conclusion, "The above evidence conclusively demonstrates that the U/Pb system, including its intermediate daughter products, especially Ra and Rn, has been so open with repeated large scale migrations of the elements that it is impossible to be sure of the precise status/history of any piece of pitchblende selected for dating...."

I do not "suspect" the scientists for dating this even though Snelling lays a pretty good case for the ore being a very open system. But it does make me wonder how many cases are really like this in the field. Even the ages they came up with within the several samples were greatly divergent--from 450 to 1200 mya.

Regards,

AFJ

### #12 Guest_tharock220_*

Posted 21 October 2009 - 06:03 PM

Let's say you observe a change in how much of a particular isotope is present at two separate times.

Then you have an equation that looks like this

Y=Xe^-kt where X is the initial amount, Y is the final amount, k is the decay constant. You divide Y by X so your equation looks like this now.

Y/X=e^-kt. Take the natural log of both sides and your left with

ln(Y/X)=-kt

Since Y/X is going to be a decimal the natural log will be negative. You're trying to solve for k. So k=ln(Y/X)/t.

After that you can set up your original equation as

.5=e^-kt and solve for time. It should be

.693/k = t.

Any time minerals in an open system are dated it's noted. That said, systems such as buried rocks that are not subject to geological reworking are closed.

### #13

Posted 22 October 2009 - 01:27 AM

Yes, thank you SeeJay,

This is not my strong subject area--really kind of new for me --but a challenge.

In regards to your comments--I realize that everything looks great on paper, as it does in business also. But on the ground (just as in business) it is usually different.

Take the example of this article by Andrew Snelling. It shows the real nitty gritty of dealing with ore that has been open but dated anyway.

Even within the crystal lattice there are substitutions. Snelling writes, "...but even at the microscopic level within pitchblende grains themselves. Fig. 8 illustrates how Pb and Ca are both substituting for U in the UO2 cubic lattice."

From grain to grain in the pitchblende there is notable change. So the picture that everything corresponds well is not reality. "In addition, the pitchblende grains donÃ¢â‚¬â„¢t have uniform compositions so that Ã¢â‚¬ËœdatingÃ¢â‚¬â„¢ of sub-sections of any grain would tend to yield widely divergent U/Pb ratios and therefore varying Ã¢â‚¬ËœagesÃ¢â‚¬â„¢ within that single grain."

Snelling writes in conclusion, "The above evidence conclusively demonstrates that the U/Pb system, including its intermediate daughter products, especially Ra and Rn, has been so open with repeated large scale migrations of the elements that it is impossible to be sure of the precise status/history of any piece of pitchblende selected for dating...."

I do not "suspect" the scientists for dating this even though Snelling lays a pretty good case for the ore being a very open system. But it does make me wonder how many cases are really like this in the field. Even the ages they came up with within the several samples were greatly divergent--from 450 to 1200 mya.

Regards,

AFJ

Hi AFJ

You have moved on to the subject of whether samples used for radiometric dating can be "contaminated" so as to give false readings.

(Your previous questions, and my response, were about the constancy of radioactive decay rates, and how amply they are justified by direct evidence.)

And the partial answer to your question is: Sure, samples can give false readings if they are not selected and prepared with sufficient care. In the article you linked, none of the references cited by Snelling (that I could find) said "Here is an exact date for this pitchblende"; they all said "These samples are a bit too messy to get reliable dates from", thus Snelling and the scientists agree on those particular samples.

The full answer to your question is, the samples that give bad dates (and go undetected) are a tiny minority. We know this because dates determined by radiometric methods are cross-checked against numerous other methods that use different physical principles (such as counting annual ice-layers, using different radioactive isotopes, or thermoluminescence). The methods all agree with each other - all these methods yield the same ages. So if one of them is wrong then all the others are wrong too, in completely different ways (because they use different methods), yet they just happen to yield the same ages by chance. As you can see, this is absurdly unlikely.

Cheers

SeeJay

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