Radiometric Dating: Can anyone make sense out of this?

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• The following was posted by an anonymous poster to the Forbes article on the Jo Hovind case. It s the only message on Forbes that the poster is credited with.
Message 1 of 96 , Oct 5, 2012
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The following was posted by an anonymous poster to the Forbes article on the Jo Hovind case. It's the only message on Forbes that the poster is credited with.

I got the impression that it is some kind of attempt to impeach certain dating methods.

If you can figure out what it is and have some kind of concise rebuttal, you may want to post it to the Forbes article or I can do it for you.

Here it is (I've added some spacing to make it easier to read):

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http://www.forbes.com/sites/peterjreilly/2012/10/05/young-earth-creationists-whipsawed-by-irs/

From: Zuma Musa
Date: Friday, October 5, 2012

The reliability of percentage remaining (50% of the remaining rule) that has been used by scientists for the relative half-lives elapsed in responding to radiometric dating method is in question.

Refer to the right hand side of the table in the website address,

http://en.wikipedia.org/wiki/Half-life.

A list of percentage remaining that corresponds to the number of the relative half-lives elapsed are presented as follows:

No. of half lives;
Fraction remaining;
Percentage remaining

01/1-100%
11/2-50% (=100% above x 50%)
21/4-25% (=50% as above x 50%)
31/8-12.5% (=25% as above x 50%)
so on and so forth
n1/2^n- (100%)/(2^n)

Using the above principle, we could arrive with weird and illogical conclusion below that would place the reliability of radiometric dating method into question:

If anyone of atoms, let's say, atom A, has been selected from a parent isotope, let's say, lutetium, to test the radioactive decay, this atom would surely have 50% of its atomic nucleus to be activated in radioactive decay in accordance to the 50% remaining rule as mentioned above.

The rule has turned up to find favour in selecting an atom if one would examine the possible decay from parent isotope since it might not be possible if there would be more than one atom is selected as mentioned below:

If any two atoms, let's say, atoms A and B, would be selected to test the decay, atom A might not respond to radioactive decay due to the existence of atom B in accordance to the 50% remaining rule. Or in other words, there would only be one atom responds to decay if there are two.

If any three atoms, let's say, atoms A, B and C, would be selected to test the decay, atom A might not respond to radioactive decay due to the existence of atoms B and C in accordance to 50% remaining rule.

If any four atoms, let say, atoms A, B, C and D, would be selected to test the decay, atom A would have much lesser chance to respond to decay due to the existence of atoms B, C and D.

There would turn up to have 2 atoms to respond if there are four as a result of 50% remaining rule is applied.

If there is a piece of 10,000 kg big rock[, let's say, 10^(a billion) atoms], 50% of the big rock (turns up to be 0.5×10^(a billion) atoms would not activate in radioactive decay and these would cause the above four selected atoms, i.e. atoms A, B, C and D, to have even much lesser chance to respond to decay due to the possible present of many half lives in the future as a result of the existence of numerous atoms.

As a result of the wide spread of the 50% inactive atoms within this piece of big rock, it is easily to destroy a piece a rock so as to locate a small portion that does not respond to decay due to it might need to wait for many half lives later in order to respond to decay as a result of the present of numerous atoms in accordance to 50% remaining rule.

This is not true since scientists could anyhow pick up any rock, let's say, lutetium, and yet still could locate decay emitted from it and this has placed 50% remaining rule into query.

If there is a gigantic mountain with 5,000 km height, 50% of this mountain would not respond to radioactive decay.

This mountain certainly consists of a huge sum of atoms when huge volume is covered. As 50% of inactive atoms would have spread throughout the whole mountain as a result of 50% remaining rule applied, it would turn up that it would be easily to locate a small portion of rock from the mountain that would not respond to radioactive decay.

However, that is not true when scientists would pick up any substance, let's say, Carbon-14, from environments for examination since they could easily locate a small portion that would respond to decay.

This has placed the reliability of 50% remaining rule into question as a result of the ease in locating a small portion of substance that would respond to decay despite its immense size.

The main problem here lies on scientists have placed 50% remaining rule on each half life and that half life is meant to be a very long years.

For example, for Carbon-14, it would take 5730 years for the 50% of the initial remaining to turn up to lose its capability in radioactive decay in order to have 50% of what has remained after the initial remaining to activate radioactive dating.

What if actual result of decay would not follow the sequence of 50% remaining rule in which it would take a shorter period to become inactive in decay instead of that 5730 years, using 5730 years as a base to presume that the decay would last in every half year would simply falsify the age that would be computed through radioactive dating method.

What if the so-called, radioactive decay, would not cause any decay but it would restart its initial operation after numerous years later, the reliability of radiometric dating method is in question.

The following is the extract from the last paragraph that is located in the website address,

[But in general, the heavier something is, the shorter its half-life (it's easier for stuff to tunnel out).]

The percentage remaining (50% of the remaining) to the responding to the number of half-lives elapsed contradicts the phrase, the heavier something is the shorter its half-life, as stated above.

This is by virtue of the biggest the rock the heaviest it is and the biggest the rock the wide spread will be the 50% of the non-activation of nucleus to be in decay and it would lead to the longer the half-life due to the application of 50% remaining rule as spelt out above and this leads to the contradiction of the statement as stated in this website in which the heavier would lead to shorter half-life.

What if this 50% remaining rule as mentioned above would have applied to Carbon-14 (the Parent Isotope), the following condition would appear:

Years Half livesPercentage Remaining

0-0-100%
57301-50% (100%*50%)
11460 (=5730*2)2-25% [50% (the above)*50%]
17190 (=5730*3)3-12.5% [25% (the above)*50%]
22920 (=5730*4)4-6.25% [25% (the above)*50%]
and so on and so forth
4,500,000,0008379888^(-1)x10^(-251397)

Note that the above years have been computed up to 4.5 billion years due to the scientists suggest the age of the earth to be that.

From the 50% remaining rule that has been computed for Carbon-14 above, it could come to the conclusion that 50 atoms out of 100 would remain active in radioactive decay in 5730 years and the rest would turn up to have lost their value in radioactive decay.

25 atoms out of 100 would remain active in decay by 11460 years and the rest would turn up to have lost their decay.

12.5 atoms out of 100 would remain active in decay and the balance would turn up to have lost their decay by 17190 years.

6.25 atoms out of 100 would remain active in decay and leaving the balance to have lost their decay by 22920 years.

1 atom out of 8×10^(251397) would remain active in decay and the balance would have lost their capability in radioactive decay by 4.5 billion years.

As 1 atom for Carbon-14 out of 8×10^(251397) would remain in active by 4.5 billion years in accordance to 50% remaining rule, it implies that it would need to get large amount of atoms from Carbon-14 so as to detect the existence of radioactive decay.

This is not true in science since it is easily to locate Carbon-14 that would emit radioactive decay and this has put the reliability of 50% remaining rule into query.

Some might support that the 50% remaining rule is subjected to exponential progress.

Let's assume that what they say is correct and presume that the half lives for Carbon-14 in 4.5 billion years would be shortened by 80% as the result of exponential progress.

The percentage remaining would turn up to be (100-80)%x8x10^(251397) and that is equal to 16×10^(251396).

Or in other words, only 1 atom would respond to decay out of 16×10^(251396) and the rest of them should have turned up to have lost their value in decay.

The ease to locate Carbon-14 that would respond to decay currently has put the reliability of radiometric dating method into question.

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• Amazing! Thanks Todd ________________________________ From: Todd Greene To: Maury_and_Baty@yahoogroups.com Sent: Wednesday, December 5,
Message 96 of 96 , Dec 5, 2012
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Amazing! Thanks Todd

From: Todd Greene <greeneto@...>
To: Maury_and_Baty@yahoogroups.com
Sent: Wednesday, December 5, 2012 7:33 AM
Subject: [M & B] Re: Real-world example of time dilation in a gravitational well
Biggest Black Hole Blast Discovered: Most Powerful Quasar Outflow Ever Found
(Science Daily, 11/26/2012)
http://www.sciencedaily.com/releases/2012/11/121128093712.htm

Excerpt:

Astronomers using ESO's Very Large Telescope (VLT) have discovered a quasar with the most energetic outflow ever seen, at least five times more powerful than any that have been observed to date. Quasars are extremely bright galactic centres powered by supermassive black holes. Many blast huge amounts of material out into their host galaxies, and these outflows play a key role in the evolution of galaxies.

--- In mailto:Maury_and_Baty%40yahoogroups.com, Ray Ausban <rayausban@...> wrote:
>
> Thanks Pi!
>
>
>
> ________________________________
> From: "PIASAN@..." <PIASAN@...>
> To: mailto:Maury_and_Baty%40yahoogroups.com
> Sent: Tuesday, December 4, 2012 7:41 AM
> Subject: Re: [M & B] Re: Real-world example of time dilation in a gravitational well
>
>
> Â
>
> From: Ray Ausban
> And how far reaching is the event horizon?
> Â
> Â
> Pi:
> Depends on the mass of the object.Â  The equation for calculating gravitational time dilation can be found at:
> http://en.wikipedia.org/wiki/Gravitational_time_dilation
> Â
> In order to determine the distance to the event horizon, you would need to solve the equation for "r"
> Â
> Â
> Â
> Â
> Ray:
> Can it reach another galaxy?
> Â
> Pi:
> You would probably need most of the mass of the universe inside the event horizon to do that.Â  If you had the entire mass of the Milky Way, the event horizon would only be something like 0.5 light years distant. (IIRC)
> Â
> Â
> Â
> Ray:
> The gravity of such a black hole must be enormous. Do other galaxies orbit such black holes?
> Â
> Pi:
> It appears that most, if not all, large galaxies have supermassive black holes at their center.
> Â
> Â
> Â
>

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