Archive-Date: Tue, 01 May 2001 06:26:47 -0700 Sender: owner-mad-scientists@VJC.COM Message-ID: From: docfarmer@riyadbank.com.sa Reply-To: Mad-Scientists@Mad-Scientists.ORG To: Mad-Scientists@Mad-Scientists.ORG CC: adria@arizona.edu, joed@cws.org, Vanrijn25@aol.com, Vanrijn25@aol.com Subject: RE: A New Gold Standard...Size Info Date: Tue, 1 May 2001 15:55:24 +0300 MIME-Version: 1.0 Content-Type: multipart/alternative ; boundary="----_=_NextPart_001_01C0D23F.65938274" This message is in MIME format. Since your mail reader does not understand this format, some or all of this message may not be legible. ------_=_NextPart_001_01C0D23F.65938274 Content-Type: text/plain; charset="iso-8859-1" Everyone, I'm still holding out hope that one of you has a precise measurement for a single gold atom. Any clues on where to look would be helpful as well. By the way, does anyone out there know if there is an e-Mail address for John Cleese or Michael Palin of Monty Python fame? It is in relation to the new metric system. Thank you. Doc Farmer Policy Development Advisor Official Monster Raving Loony Party -----Original Message----- From: docfarmer@riyadbank.com.sa [mailto:docfarmer@riyadbank.com.sa] Sent: Saturday, 21 April 2001 16:13 To: Mad-Scientists@Mad-Scientists.ORG; adria@arizona.edu Subject: RE: A New Gold Standard...Echo, echo, echo... Adria, First some good news. I found the NIST calculated standard for Avogadro's number - 6.02214199 * 10 -^23. Also, I've found a more precise atomic weight for gold - 196.966543. As to the size of the atom, I've got two possibilities - 1.79 angstroms for the atomic radius, and 1.34 angstroms for the covalent radius. I've no idea what the difference is between the two (yes, I know, 0.45 angstroms, very funny) so if anyone out there can give me a clue on the difference between atomic and covalent radii, I'd be grateful. Thanks. Doc Farmer Policy Development Advisor Official Monster Raving Loony Party -----Original Message----- From: docfarmer@riyadbank.com.sa [mailto:docfarmer@riyadbank.com.sa] Sent: Friday, 20 April 2001 13:09 To: Mad-Scientists@Mad-Scientists.ORG; adria@arizona.edu Subject: RE: A New Gold Standard...Echo, echo, echo... Adria, First, let me explain the "Prof" prefix in my last message. Since your name doesn't give an indication of gender, I didn't want to call you Mr. or Ms. and possibly cause offence. I didn't want to just use your first name in the original message, either, because I felt it might be too familiar (hope you don't mind it in this one). Besides, your knowledge level suggests that if you ain't a Prof yet, you probably are destined to be one in the future. I accept your argument regarding limiting the significant digits up to a point. My problem here is that I've got to try and be as accurate as possible with this measurement, as it will be the basis for an entire new metric system. We all know what happened when the French screwed up the last one (and I am allowed to say that, because there is some French in my lineage) and I don't want that to happen here. Many thanks for the updated Avogadro's constant. My calculations agree with yours (by the way, I'm using MS Excel, which is why I'm getting a more defined number (I hesitate to say "accurate" - we're talking Microsoft here)) so I can't explain the difference in the original calculation. As to the dimensional measurement, I realise that fluctuations occur in the electron sphere, which may make this more difficult. Would it be more accurate to merely measure the centre of the atom (the proton/neutron cluster), or are there variances in that measurement as well. In either case, if you could look up the size of the atom I'd be most grateful. By the bye, I think atoms have been actually observed some years back. IBM Labs was I believe the first group to actually manipulate atoms (they arranged them into their logo), and if I'm not mistaken there have been studies since which have shown atoms not as snapshots but as "live" movies, showing the movement of the electron sphere. Ain't science amazing?!? My thanks again for your efforts. Doc Farmer Policy Development Advisor Official Monster Raving Loony Party p.s. I did read the entire message. p.p.s. It's 13:15 here in Riyadh. That means it must be around 04:15 your time. Hope this isn't insomnia you're suffering from. -----Original Message----- From: Adria Decker [ mailto:adria@arizona.edu ] Sent: Friday, 20 April 2001 12:38 To: Mad-Scientists@Mad-Scientists.ORG Subject: Re: A New Gold Standard...Echo, echo, echo... Well, my trusty TI-86 and the Windows calculator application beg to differ, but if that's what you get, that's what you get. The concept of significant figures is a scientific standard. You cannot claim to accurately know any more digits than what you start with, and if you do, you are being misleading. Almost every single constant in science is rounded to some degree. It would be ridiculous to claim otherwise. Sure, your remainders take up more significance when mulitplied by 10^23, but that just means they're more significantly WRONG. Sure, 10/3 is 3.33333333333 ad infinitum, but what if that 3 is really 3.443298473 and that 10 is really 9.67899987? The only digit you know with any accuracy is the first 3. You would only know for sure that those threes go on forever if you had 10.00000000(to infinity)/3.000000000(to infinty). This is a universally accepted standard in science. It's akin to trying to measure attometers with a meter stick or nanoseconds with a grandfather clock: you just can't do it and claim to be telling the truth. If I tried to measure 0.01 microliters with a graduated cylinder, I'd be laughed out of the lab. Here are the most accurate values I can find for Avogadro's number and the molecular weight of gold (they're a little different from what I said before, but they're from a different book and this one is more recent): Avogadro's number: 6.022136736x10^23 M.W. of Au: 196.96655 With these values, the mass of one atom of gold is calculated to be 3.2707087x10^-22 grams. We cannot go any further than that because our original tools of measurement have their limits of 8 digits and so we must admit our limits of calculation. Although I'm flattered by the title, I must admit I am still an undergraduate (6 units from my B.S.), lest my friends on this mailing list expose me for an imposter. :) I also must admit to being a molecular biologist, not a chemist, but I've taken more chemistry classes than any one person should ever have to. Jon D was right about atoms being fuzzy things. The current model of an atom itself is actually made up of a lot of empty space, consisting of a nucleus (protons and neutrons) and electrons orbiting around it. Electrons don't even orbit in a "sphere." However, we don't even really know that for sure, no one has ever really SEEN an atom to my knowledge, only experimentally assigned qualities. I would be reluctant to define dimensions at all, but if you want I can try to look up the size of its orbitals. Here I stop, because most of you have probably lost interest in my long-windedness and aren't reading anymore anyway. --Adria > docfarmer@riyadbank.com.sa wrote: > > Prof. Decker, > > Many thanks for your corrections. However, I get 3.27097*10^-22, not > 3.27047. Or, more accurately 0.000 000 000 000 000 000 000 032 709 > 706 985 555 555 etc. gram per atom. Also, I think I must disagree > with your premise that accuracy can only be claimed to 6 digits. If > the division creates continual remainders, then accuracy can continue > to the nth degree (see also pi). Granted, the last significant digits > continue ad infinitum as 5, so in scientific notation the number would > show as 3.2709706985556*10^-22. Considering that this number would be > multiplied by 10^23 to get the new mass constant, the remainders take > up more significance. > > I agree that if there are more accurate figures for the atomic weight > and Avoradro's number, I would be happy to use them. The more > accurate the data, the more useful the new constant will be in the > long run. I don't want to make the French mistake here (and not the > one from the end of Blazing Saddles, either (hee hee)). > > I'm still in need of assistance on the actual dimensions of the atom > as well, so if there are any chemists or physics gurus out there, I'd > be grateful for your help. > > Once again, my thanks. > > Doc Farmer > Policy Development Advisor > Official Monster Raving Loony Party > > -----Original Message----- > From: Adria Decker [ mailto:adria@arizona.edu ] > Sent: Wednesday, 18 April 2001 19:37 > To: Mad-Scientists@Mad-Scientists.ORG > Subject: Re: A New Gold Standard...Echo, echo, echo... > > Well, first of all, you need to switch around your division. > You want > to find out grams/atom, not atoms/gram. Second of all, and since this > > is the only way I can get close to the number you got I'll assume it's > > what you did, it's 10^23, not 10^-23. Third of all, you're ignoring > significant figures, which is a bad thing to do in science, especially > > if you're coming up with some sort of measurement standard. What you > would actually get is 3.27x10^-22, because you can't wind up with more > > significant digits than you started with or else you're claiming more > accuracy than you actually have. If you want more accuracy, use > 6.02257 > X 10^23 for Avogadro's number and 196.9665 for the molecular weight of > > gold. Now you get 3.27047x10^-22 grams per atom, because you can only > > claim accuracy to 6 digits as we only have Avogadro's number to 6 > digits, even though we have gold to 7. I got these values out of my > organic chemistry textbook, so if they're wrong, blame them. I'm sure > there are more accurate values for both the molecular weight of gold > and > Avogadro's number, but since I'm late for my physics class, you're on > your own. :) > Woo! I'm actually putting my college education to use! > > --Adria Decker > > > docfarmer@riyadbank.com.sa wrote: > > > > Joe, > > > > Thanks for that. So, if I've done my maths correctly, a single atom > > > of gold should weigh 0.000 000 000 000 000 000 000 000 305 538 756 > 345 > > 178 gram. Could someone please confirm this? > > > > As to getting the actual dimensions of a single atom, I don't think > > your calculation will work. Since gold is not arranged in a linear > > format, but more of a > > > > x x x x x x x x x x > > x x x x x x x x x > > x x x x x x x x x x > > > > format, the offsets would skew the results. Does anyone have a > > measurement calculation on a single atom? I await your responses > with > > interest. Thanks again. > > > > Doc Farmer > > Policy Development Advisor > > Official Monster Raving Loony Party > > > > -----Original Message----- > > From: Joe D [ mailto:joed@cws.org ] > > Sent: Tuesday, 17 April 2001 21:28 > > To: Mad-Scientists@Mad-Scientists.ORG > > Subject: Re: A New Gold Standard...Echo, echo, echo... > > > > On Tue, Apr 17, 2001 at 08:04:37AM +0300, docfarmer@riyadbank.com.sa > > > wrote: > > > Dr. X, > > > > > > Your perspicacity is astounding (not to mention your flair for the > > > > pedantesque). I wish to use Gold 197 (the "stable" isotope) for > the > > > > > baseline measurements on my new metric system. As for my > > inappropriate use > > > of "rest", let me clarify. I meant a single atom in a stable > > structure (no > > > horse jokes here, please) without large fluctuations in the > electron > > sphere. > > > The dimensions and mass of such an atom would not change > > significantly over > > > the course of measurement (Holy Heisenburg!). > > > > Back of the envelope stuff here... > > > > 6.02 * 10^23 atoms of gold 197 weigh 197 grams. There may be a more > > > accurate representation of Avogadro's Number out there - it's been a > > > long > > time since high school chemistry. > > > > You could then use the density of gold to find out how many cubic > > centimeters 197 grams of gold occupies, and from that get the volume > > > of one > > atom. > > > > More or less... > > > > Joe D > > > > *************************************************************************** * The information transmitted in this e-Mail, and any files transmitted * * with it, is confidential and intended solely for the use of the * * individual(s) to whom it is addressed. Any review, retransmission, * * dissemination or other use of or taking action in reliance upon this * * information by persons or entities other than the intended recipient(s) * * is prohibited. Any views or opinions expressed are solely those of * * the author, and do not necessarily represent those of Riyad Bank. If * * you have received this message in error, please notify the sender and * * the system manager at postmaster@RiyadBank.com.sa and delete the material * * from your computer. * * * * This footnote confirms that this message and any associated attachments * * have been scanned by MIMESweeper for content security and the presence * * of computer viruses. * *************************************************************************** ------_=_NextPart_001_01C0D23F.65938274 Content-Type: text/html; charset="iso-8859-1" RE: A New Gold Standard...Echo, echo, echo...
Everyone,
 
I'm still holding out hope that one of you has a precise measurement for a single gold atom.  Any clues on where to look would be helpful as well.
 
By the way, does anyone out there know if there is an e-Mail address for John Cleese or Michael Palin of Monty Python fame?  It is in relation to the new metric system.  Thank you.
 
Doc Farmer
Policy Development Advisor
Official Monster Raving Loony Party
 
-----Original Message-----
From: docfarmer@riyadbank.com.sa [mailto:docfarmer@riyadbank.com.sa]
Sent: Saturday, 21 April 2001 16:13
To: Mad-Scientists@Mad-Scientists.ORG; adria@arizona.edu
Subject: RE: A New Gold Standard...Echo, echo, echo...

Adria,
 
First some good news.  I found the NIST calculated standard for Avogadro's number - 6.02214199 * 10 -^23.  Also, I've found a more precise atomic weight for gold - 196.966543.  As to the size of the atom, I've got two possibilities - 1.79 angstroms for the atomic radius, and 1.34 angstroms for the covalent radius.  I've no idea what the difference is between the two (yes, I know, 0.45 angstroms, very funny) so if anyone out there can give me a clue on the difference between atomic and covalent radii, I'd be grateful.  Thanks.
 
Doc Farmer
Policy Development Advisor
Official Monster Raving Loony Party
-----Original Message-----
From: docfarmer@riyadbank.com.sa [mailto:docfarmer@riyadbank.com.sa]
Sent: Friday, 20 April 2001 13:09
To: Mad-Scientists@Mad-Scientists.ORG; adria@arizona.edu
Subject: RE: A New Gold Standard...Echo, echo, echo...

Adria,

First, let me explain the "Prof" prefix in my last message.  Since your name doesn't give an indication of gender, I didn't want to call you Mr. or Ms. and possibly cause offence.  I didn't want to just use your first name in the original message, either, because I felt it might be too familiar (hope you don't mind it in this one).  Besides, your knowledge level suggests that if you ain't a Prof yet, you probably are destined to be one in the future.

I accept your argument regarding limiting the significant digits up to a point.  My problem here is that I've got to try and be as accurate as possible with this measurement, as it will be the basis for an entire new metric system.  We all know what happened when the French screwed up the last one (and I am allowed to say that, because there is some French in my lineage) and I don't want that to happen here.

Many thanks for the updated Avogadro's constant.  My calculations agree with yours (by the way, I'm using MS Excel, which is why I'm getting a more defined number (I hesitate to say "accurate" - we're talking Microsoft here)) so I can't explain the difference in the original calculation.

As to the dimensional measurement, I realise that fluctuations occur in the electron sphere, which may make this more difficult.  Would it be more accurate to merely measure the centre of the atom (the proton/neutron cluster), or are there variances in that measurement as well.  In either case, if you could look up the size of the atom I'd be most grateful.

By the bye, I think atoms have been actually observed some years back.  IBM Labs was I believe the first group to actually manipulate atoms (they arranged them into their logo), and if I'm not mistaken there have been studies since which have shown atoms not as snapshots but as "live" movies, showing the movement of the electron sphere.  Ain't science amazing?!?

My thanks again for your efforts.

Doc Farmer
Policy Development Advisor
Official Monster Raving Loony Party

p.s.            I did read the entire message.

p.p.s.  It's 13:15 here in Riyadh.  That means it must be around 04:15 your time.  Hope this isn't insomnia you're suffering from.

-----Original Message-----
From: Adria Decker [mailto:adria@arizona.edu]
Sent: Friday, 20 April 2001 12:38
To: Mad-Scientists@Mad-Scientists.ORG
Subject: Re: A New Gold Standard...Echo, echo, echo...


Well, my trusty TI-86 and the Windows calculator application beg to
differ, but if that's what you get, that's what you get.
The concept of significant figures is a scientific standard.  You cannot
claim to accurately know any more digits than what you start with, and
if you do, you are being misleading.  Almost every single constant in
science is rounded to some degree.  It would be ridiculous to claim
otherwise.  Sure, your remainders take up more significance when
mulitplied by 10^23, but that just means they're more significantly
WRONG.  Sure, 10/3 is 3.33333333333 ad infinitum, but what if that 3 is
really 3.443298473 and that 10 is really 9.67899987?  The only digit you
know with any accuracy is the first 3.  You would only know for sure
that those threes go on forever if you had 10.00000000(to
infinity)/3.000000000(to infinty).  This is a universally accepted
standard in science.  It's akin to trying to measure attometers with a
meter stick or nanoseconds with a grandfather clock:  you just can't do
it and claim to be telling the truth.  If I tried to measure 0.01
microliters with a graduated cylinder, I'd be laughed out of the lab.
Here are the most accurate values I can find for Avogadro's number and
the molecular weight of gold (they're a little different from what I
said before, but they're from a different book and this one is more
recent):
Avogadro's number: 6.022136736x10^23
M.W. of Au:  196.96655
With these values, the mass of one atom of gold is calculated to be
3.2707087x10^-22 grams.  We cannot go any further than that because our
original tools of measurement have their limits of 8 digits and so we
must admit our limits of calculation.

Although I'm flattered by the title, I must admit I am still an
undergraduate (6 units from my B.S.), lest my friends on this mailing
list expose me for an imposter.  :) I also must admit to being a
molecular biologist, not a chemist, but I've taken more chemistry
classes than any one person should ever have to.

Jon D was right about atoms being fuzzy things.  The current model of an
atom itself is actually made up of a lot of empty space, consisting of a
nucleus (protons and neutrons) and electrons orbiting around it.
Electrons don't even orbit in a "sphere."   However, we don't even
really know that for sure, no one has ever really SEEN an atom to my
knowledge, only experimentally assigned qualities.  I would be reluctant
to define dimensions at all, but if you want I can try to look up the
size of its orbitals.

Here I stop, because most of you have probably lost interest in my
long-windedness and aren't reading anymore anyway.

--Adria

> docfarmer@riyadbank.com.sa wrote:
>
> Prof. Decker,
>
> Many thanks for your corrections.  However, I get 3.27097*10^-22, not
> 3.27047.  Or, more accurately 0.000 000 000 000 000 000 000 032 709
> 706 985 555 555 etc. gram per atom.  Also, I think I must disagree
> with your premise that accuracy can only be claimed to 6 digits.  If
> the division creates continual remainders, then accuracy can continue
> to the nth degree (see also pi).  Granted, the last significant digits
> continue ad infinitum as 5, so in scientific notation the number would
> show as 3.2709706985556*10^-22.  Considering that this number would be
> multiplied by 10^23 to get the new mass constant, the remainders take
> up more significance.
>
> I agree that if there are more accurate figures for the atomic weight
> and Avoradro's number, I would be happy to use them.  The more
> accurate the data, the more useful the new constant will be in the
> long run.  I don't want to make the French mistake here (and not the
> one from the end of Blazing Saddles, either (hee hee)).
>
> I'm still in need of assistance on the actual dimensions of the atom
> as well, so if there are any chemists or physics gurus out there, I'd
> be grateful for your help.
>
> Once again, my thanks.
>
> Doc Farmer
> Policy Development Advisor
> Official Monster Raving Loony Party
>
> -----Original Message-----
> From: Adria Decker [mailto:adria@arizona.edu]
> Sent: Wednesday, 18 April 2001 19:37
> To: Mad-Scientists@Mad-Scientists.ORG
> Subject: Re: A New Gold Standard...Echo, echo, echo...
>
>         Well, first of all, you need to switch around your division.
> You want
> to find out grams/atom, not atoms/gram.  Second of all, and since this
>
> is the only way I can get close to the number you got I'll assume it's
>
> what you did, it's 10^23, not 10^-23.  Third of all, you're ignoring
> significant figures, which is a bad thing to do in science, especially
>
> if you're coming up with some sort of measurement standard.  What you
> would actually get is 3.27x10^-22, because you can't wind up with more
>
> significant digits than you started with or else you're claiming more
> accuracy than you actually have.  If you want more accuracy, use
> 6.02257
> X 10^23 for Avogadro's number and 196.9665 for the molecular weight of
>
> gold.  Now you get 3.27047x10^-22 grams per atom, because you can only
>
> claim accuracy to 6 digits as we only have Avogadro's number to 6
> digits, even though we have gold to 7.  I got these values out of my
> organic chemistry textbook, so if they're wrong, blame them. I'm sure
> there are more accurate values for both the molecular weight of gold
> and
> Avogadro's number, but since I'm late for my physics class, you're on
> your own.  :)
> Woo! I'm actually putting my college education to use!
>
> --Adria Decker
>
> > docfarmer@riyadbank.com.sa wrote:
> >
> > Joe,
> >
> > Thanks for that.  So, if I've done my maths correctly, a single atom
>
> > of gold should weigh 0.000 000 000 000 000 000 000 000 305 538 756
> 345
> > 178 gram.  Could someone please confirm this?
> >
> > As to getting the actual dimensions of a single atom, I don't think
> > your calculation will work.  Since gold is not arranged in a linear
> > format, but more of a
> >
> > x x x x x x x x x x
> >  x x x x x x x x x
> > x x x x x x x x x x
> >
> > format, the offsets would skew the results.  Does anyone have a
> > measurement calculation on a single atom?  I await your responses
> with
> > interest.  Thanks again.
> >
> > Doc Farmer
> > Policy Development Advisor
> > Official Monster Raving Loony Party
> >
> > -----Original Message-----
> > From: Joe D [mailto:joed@cws.org]
> > Sent: Tuesday, 17 April 2001 21:28
> > To: Mad-Scientists@Mad-Scientists.ORG
> > Subject: Re: A New Gold Standard...Echo, echo, echo...
> >
> > On Tue, Apr 17, 2001 at 08:04:37AM +0300, docfarmer@riyadbank.com.sa
>
> > wrote:
> > > Dr. X,
> > >
> > > Your perspicacity is astounding (not to mention your flair for the
>
> > > pedantesque).  I wish to use Gold 197 (the "stable" isotope) for
> the
> >
> > > baseline measurements on my new metric system.  As for my
> > inappropriate use
> > > of "rest", let me clarify.  I meant a single atom in a stable
> > structure (no
> > > horse jokes here, please) without large fluctuations in the
> electron
> > sphere.
> > > The dimensions and mass of such an atom would not change
> > significantly over
> > > the course of measurement (Holy Heisenburg!).
> >
> > Back of the envelope stuff here...
> >
> > 6.02 * 10^23 atoms of gold 197 weigh 197 grams.  There may be a more
>
> > accurate representation of Avogadro's Number out there - it's been a
>
> > long
> > time since high school chemistry.
> >
> > You could then use the density of gold to find out how many cubic
> > centimeters 197 grams of gold occupies, and from that get the volume
>
> > of one
> > atom.
> >
> > More or less...
> >
> > Joe D
> >
> >



***************************************************************************
* The information transmitted in this e-Mail, and any files transmitted *
* with it, is confidential and intended solely for the use of the *
* individual(s) to whom it is addressed. Any review, retransmission, *
* dissemination or other use of or taking action in reliance upon this *
* information by persons or entities other than the intended recipient(s) *
* is prohibited. Any views or opinions expressed are solely those of *
* the author, and do not necessarily represent those of Riyad Bank. If *
* you have received this message in error, please notify the sender and *
* the system manager at postmaster@RiyadBank.com.sa and delete the material *
* from your computer. *
* *
* This footnote confirms that this message and any associated attachments *
* have been scanned by MIMESweeper for content security and the presence *
* of computer viruses. *
***************************************************************************
 ------_=_NextPart_001_01C0D23F.65938274-- ================================================================================ Archive-Date: Tue, 01 May 2001 06:46:26 -0700 Sender: owner-mad-scientists@VJC.COM Message-ID: From: docfarmer@riyadbank.com.sa Reply-To: Mad-Scientists@Mad-Scientists.ORG To: Mad-Scientists@Mad-Scientists.ORG CC: adria@arizona.edu, joed@cws.org, Vanrijn25@aol.com, Vanrijn25@aol.com Subject: RE: A New Gold Standard...Size Info Date: Tue, 1 May 2001 15:55:24 +0300 MIME-Version: 1.0 Content-Type: multipart/alternative ; boundary="----_=_NextPart_001_01C0D23F.65938274" This message is in MIME format. Since your mail reader does not understand this format, some or all of this message may not be legible. ------_=_NextPart_001_01C0D23F.65938274 Content-Type: text/plain; charset="iso-8859-1" Everyone, I'm still holding out hope that one of you has a precise measurement for a single gold atom. Any clues on where to look would be helpful as well. By the way, does anyone out there know if there is an e-Mail address for John Cleese or Michael Palin of Monty Python fame? It is in relation to the new metric system. Thank you. Doc Farmer Policy Development Advisor Official Monster Raving Loony Party -----Original Message----- From: docfarmer@riyadbank.com.sa [mailto:docfarmer@riyadbank.com.sa] Sent: Saturday, 21 April 2001 16:13 To: Mad-Scientists@Mad-Scientists.ORG; adria@arizona.edu Subject: RE: A New Gold Standard...Echo, echo, echo... Adria, First some good news. I found the NIST calculated standard for Avogadro's number - 6.02214199 * 10 -^23. Also, I've found a more precise atomic weight for gold - 196.966543. As to the size of the atom, I've got two possibilities - 1.79 angstroms for the atomic radius, and 1.34 angstroms for the covalent radius. I've no idea what the difference is between the two (yes, I know, 0.45 angstroms, very funny) so if anyone out there can give me a clue on the difference between atomic and covalent radii, I'd be grateful. Thanks. Doc Farmer Policy Development Advisor Official Monster Raving Loony Party -----Original Message----- From: docfarmer@riyadbank.com.sa [mailto:docfarmer@riyadbank.com.sa] Sent: Friday, 20 April 2001 13:09 To: Mad-Scientists@Mad-Scientists.ORG; adria@arizona.edu Subject: RE: A New Gold Standard...Echo, echo, echo... Adria, First, let me explain the "Prof" prefix in my last message. Since your name doesn't give an indication of gender, I didn't want to call you Mr. or Ms. and possibly cause offence. I didn't want to just use your first name in the original message, either, because I felt it might be too familiar (hope you don't mind it in this one). Besides, your knowledge level suggests that if you ain't a Prof yet, you probably are destined to be one in the future. I accept your argument regarding limiting the significant digits up to a point. My problem here is that I've got to try and be as accurate as possible with this measurement, as it will be the basis for an entire new metric system. We all know what happened when the French screwed up the last one (and I am allowed to say that, because there is some French in my lineage) and I don't want that to happen here. Many thanks for the updated Avogadro's constant. My calculations agree with yours (by the way, I'm using MS Excel, which is why I'm getting a more defined number (I hesitate to say "accurate" - we're talking Microsoft here)) so I can't explain the difference in the original calculation. As to the dimensional measurement, I realise that fluctuations occur in the electron sphere, which may make this more difficult. Would it be more accurate to merely measure the centre of the atom (the proton/neutron cluster), or are there variances in that measurement as well. In either case, if you could look up the size of the atom I'd be most grateful. By the bye, I think atoms have been actually observed some years back. IBM Labs was I believe the first group to actually manipulate atoms (they arranged them into their logo), and if I'm not mistaken there have been studies since which have shown atoms not as snapshots but as "live" movies, showing the movement of the electron sphere. Ain't science amazing?!? My thanks again for your efforts. Doc Farmer Policy Development Advisor Official Monster Raving Loony Party p.s. I did read the entire message. p.p.s. It's 13:15 here in Riyadh. That means it must be around 04:15 your time. Hope this isn't insomnia you're suffering from. -----Original Message----- From: Adria Decker [ mailto:adria@arizona.edu ] Sent: Friday, 20 April 2001 12:38 To: Mad-Scientists@Mad-Scientists.ORG Subject: Re: A New Gold Standard...Echo, echo, echo... Well, my trusty TI-86 and the Windows calculator application beg to differ, but if that's what you get, that's what you get. The concept of significant figures is a scientific standard. You cannot claim to accurately know any more digits than what you start with, and if you do, you are being misleading. Almost every single constant in science is rounded to some degree. It would be ridiculous to claim otherwise. Sure, your remainders take up more significance when mulitplied by 10^23, but that just means they're more significantly WRONG. Sure, 10/3 is 3.33333333333 ad infinitum, but what if that 3 is really 3.443298473 and that 10 is really 9.67899987? The only digit you know with any accuracy is the first 3. You would only know for sure that those threes go on forever if you had 10.00000000(to infinity)/3.000000000(to infinty). This is a universally accepted standard in science. It's akin to trying to measure attometers with a meter stick or nanoseconds with a grandfather clock: you just can't do it and claim to be telling the truth. If I tried to measure 0.01 microliters with a graduated cylinder, I'd be laughed out of the lab. Here are the most accurate values I can find for Avogadro's number and the molecular weight of gold (they're a little different from what I said before, but they're from a different book and this one is more recent): Avogadro's number: 6.022136736x10^23 M.W. of Au: 196.96655 With these values, the mass of one atom of gold is calculated to be 3.2707087x10^-22 grams. We cannot go any further than that because our original tools of measurement have their limits of 8 digits and so we must admit our limits of calculation. Although I'm flattered by the title, I must admit I am still an undergraduate (6 units from my B.S.), lest my friends on this mailing list expose me for an imposter. :) I also must admit to being a molecular biologist, not a chemist, but I've taken more chemistry classes than any one person should ever have to. Jon D was right about atoms being fuzzy things. The current model of an atom itself is actually made up of a lot of empty space, consisting of a nucleus (protons and neutrons) and electrons orbiting around it. Electrons don't even orbit in a "sphere." However, we don't even really know that for sure, no one has ever really SEEN an atom to my knowledge, only experimentally assigned qualities. I would be reluctant to define dimensions at all, but if you want I can try to look up the size of its orbitals. Here I stop, because most of you have probably lost interest in my long-windedness and aren't reading anymore anyway. --Adria > docfarmer@riyadbank.com.sa wrote: > > Prof. Decker, > > Many thanks for your corrections. However, I get 3.27097*10^-22, not > 3.27047. Or, more accurately 0.000 000 000 000 000 000 000 032 709 > 706 985 555 555 etc. gram per atom. Also, I think I must disagree > with your premise that accuracy can only be claimed to 6 digits. If > the division creates continual remainders, then accuracy can continue > to the nth degree (see also pi). Granted, the last significant digits > continue ad infinitum as 5, so in scientific notation the number would > show as 3.2709706985556*10^-22. Considering that this number would be > multiplied by 10^23 to get the new mass constant, the remainders take > up more significance. > > I agree that if there are more accurate figures for the atomic weight > and Avoradro's number, I would be happy to use them. The more > accurate the data, the more useful the new constant will be in the > long run. I don't want to make the French mistake here (and not the > one from the end of Blazing Saddles, either (hee hee)). > > I'm still in need of assistance on the actual dimensions of the atom > as well, so if there are any chemists or physics gurus out there, I'd > be grateful for your help. > > Once again, my thanks. > > Doc Farmer > Policy Development Advisor > Official Monster Raving Loony Party > > -----Original Message----- > From: Adria Decker [ mailto:adria@arizona.edu ] > Sent: Wednesday, 18 April 2001 19:37 > To: Mad-Scientists@Mad-Scientists.ORG > Subject: Re: A New Gold Standard...Echo, echo, echo... > > Well, first of all, you need to switch around your division. > You want > to find out grams/atom, not atoms/gram. Second of all, and since this > > is the only way I can get close to the number you got I'll assume it's > > what you did, it's 10^23, not 10^-23. Third of all, you're ignoring > significant figures, which is a bad thing to do in science, especially > > if you're coming up with some sort of measurement standard. What you > would actually get is 3.27x10^-22, because you can't wind up with more > > significant digits than you started with or else you're claiming more > accuracy than you actually have. If you want more accuracy, use > 6.02257 > X 10^23 for Avogadro's number and 196.9665 for the molecular weight of > > gold. Now you get 3.27047x10^-22 grams per atom, because you can only > > claim accuracy to 6 digits as we only have Avogadro's number to 6 > digits, even though we have gold to 7. I got these values out of my > organic chemistry textbook, so if they're wrong, blame them. I'm sure > there are more accurate values for both the molecular weight of gold > and > Avogadro's number, but since I'm late for my physics class, you're on > your own. :) > Woo! I'm actually putting my college education to use! > > --Adria Decker > > > docfarmer@riyadbank.com.sa wrote: > > > > Joe, > > > > Thanks for that. So, if I've done my maths correctly, a single atom > > > of gold should weigh 0.000 000 000 000 000 000 000 000 305 538 756 > 345 > > 178 gram. Could someone please confirm this? > > > > As to getting the actual dimensions of a single atom, I don't think > > your calculation will work. Since gold is not arranged in a linear > > format, but more of a > > > > x x x x x x x x x x > > x x x x x x x x x > > x x x x x x x x x x > > > > format, the offsets would skew the results. Does anyone have a > > measurement calculation on a single atom? I await your responses > with > > interest. Thanks again. > > > > Doc Farmer > > Policy Development Advisor > > Official Monster Raving Loony Party > > > > -----Original Message----- > > From: Joe D [ mailto:joed@cws.org ] > > Sent: Tuesday, 17 April 2001 21:28 > > To: Mad-Scientists@Mad-Scientists.ORG > > Subject: Re: A New Gold Standard...Echo, echo, echo... > > > > On Tue, Apr 17, 2001 at 08:04:37AM +0300, docfarmer@riyadbank.com.sa > > > wrote: > > > Dr. X, > > > > > > Your perspicacity is astounding (not to mention your flair for the > > > > pedantesque). I wish to use Gold 197 (the "stable" isotope) for > the > > > > > baseline measurements on my new metric system. As for my > > inappropriate use > > > of "rest", let me clarify. I meant a single atom in a stable > > structure (no > > > horse jokes here, please) without large fluctuations in the > electron > > sphere. > > > The dimensions and mass of such an atom would not change > > significantly over > > > the course of measurement (Holy Heisenburg!). > > > > Back of the envelope stuff here... > > > > 6.02 * 10^23 atoms of gold 197 weigh 197 grams. There may be a more > > > accurate representation of Avogadro's Number out there - it's been a > > > long > > time since high school chemistry. > > > > You could then use the density of gold to find out how many cubic > > centimeters 197 grams of gold occupies, and from that get the volume > > > of one > > atom. > > > > More or less... > > > > Joe D > > > > *************************************************************************** * The information transmitted in this e-Mail, and any files transmitted * * with it, is confidential and intended solely for the use of the * * individual(s) to whom it is addressed. Any review, retransmission, * * dissemination or other use of or taking action in reliance upon this * * information by persons or entities other than the intended recipient(s) * * is prohibited. Any views or opinions expressed are solely those of * * the author, and do not necessarily represent those of Riyad Bank. If * * you have received this message in error, please notify the sender and * * the system manager at postmaster@RiyadBank.com.sa and delete the material * * from your computer. * * * * This footnote confirms that this message and any associated attachments * * have been scanned by MIMESweeper for content security and the presence * * of computer viruses. * *************************************************************************** ------_=_NextPart_001_01C0D23F.65938274 Content-Type: text/html; charset="iso-8859-1" RE: A New Gold Standard...Echo, echo, echo...
Everyone,
 
I'm still holding out hope that one of you has a precise measurement for a single gold atom.  Any clues on where to look would be helpful as well.
 
By the way, does anyone out there know if there is an e-Mail address for John Cleese or Michael Palin of Monty Python fame?  It is in relation to the new metric system.  Thank you.
 
Doc Farmer
Policy Development Advisor
Official Monster Raving Loony Party
 
-----Original Message-----
From: docfarmer@riyadbank.com.sa [mailto:docfarmer@riyadbank.com.sa]
Sent: Saturday, 21 April 2001 16:13
To: Mad-Scientists@Mad-Scientists.ORG; adria@arizona.edu
Subject: RE: A New Gold Standard...Echo, echo, echo...

Adria,
 
First some good news.  I found the NIST calculated standard for Avogadro's number - 6.02214199 * 10 -^23.  Also, I've found a more precise atomic weight for gold - 196.966543.  As to the size of the atom, I've got two possibilities - 1.79 angstroms for the atomic radius, and 1.34 angstroms for the covalent radius.  I've no idea what the difference is between the two (yes, I know, 0.45 angstroms, very funny) so if anyone out there can give me a clue on the difference between atomic and covalent radii, I'd be grateful.  Thanks.
 
Doc Farmer
Policy Development Advisor
Official Monster Raving Loony Party
-----Original Message-----
From: docfarmer@riyadbank.com.sa [mailto:docfarmer@riyadbank.com.sa]
Sent: Friday, 20 April 2001 13:09
To: Mad-Scientists@Mad-Scientists.ORG; adria@arizona.edu
Subject: RE: A New Gold Standard...Echo, echo, echo...

Adria,

First, let me explain the "Prof" prefix in my last message.  Since your name doesn't give an indication of gender, I didn't want to call you Mr. or Ms. and possibly cause offence.  I didn't want to just use your first name in the original message, either, because I felt it might be too familiar (hope you don't mind it in this one).  Besides, your knowledge level suggests that if you ain't a Prof yet, you probably are destined to be one in the future.

I accept your argument regarding limiting the significant digits up to a point.  My problem here is that I've got to try and be as accurate as possible with this measurement, as it will be the basis for an entire new metric system.  We all know what happened when the French screwed up the last one (and I am allowed to say that, because there is some French in my lineage) and I don't want that to happen here.

Many thanks for the updated Avogadro's constant.  My calculations agree with yours (by the way, I'm using MS Excel, which is why I'm getting a more defined number (I hesitate to say "accurate" - we're talking Microsoft here)) so I can't explain the difference in the original calculation.

As to the dimensional measurement, I realise that fluctuations occur in the electron sphere, which may make this more difficult.  Would it be more accurate to merely measure the centre of the atom (the proton/neutron cluster), or are there variances in that measurement as well.  In either case, if you could look up the size of the atom I'd be most grateful.

By the bye, I think atoms have been actually observed some years back.  IBM Labs was I believe the first group to actually manipulate atoms (they arranged them into their logo), and if I'm not mistaken there have been studies since which have shown atoms not as snapshots but as "live" movies, showing the movement of the electron sphere.  Ain't science amazing?!?

My thanks again for your efforts.

Doc Farmer
Policy Development Advisor
Official Monster Raving Loony Party

p.s.            I did read the entire message.

p.p.s.  It's 13:15 here in Riyadh.  That means it must be around 04:15 your time.  Hope this isn't insomnia you're suffering from.

-----Original Message-----
From: Adria Decker [mailto:adria@arizona.edu]
Sent: Friday, 20 April 2001 12:38
To: Mad-Scientists@Mad-Scientists.ORG
Subject: Re: A New Gold Standard...Echo, echo, echo...


Well, my trusty TI-86 and the Windows calculator application beg to
differ, but if that's what you get, that's what you get.
The concept of significant figures is a scientific standard.  You cannot
claim to accurately know any more digits than what you start with, and
if you do, you are being misleading.  Almost every single constant in
science is rounded to some degree.  It would be ridiculous to claim
otherwise.  Sure, your remainders take up more significance when
mulitplied by 10^23, but that just means they're more significantly
WRONG.  Sure, 10/3 is 3.33333333333 ad infinitum, but what if that 3 is
really 3.443298473 and that 10 is really 9.67899987?  The only digit you
know with any accuracy is the first 3.  You would only know for sure
that those threes go on forever if you had 10.00000000(to
infinity)/3.000000000(to infinty).  This is a universally accepted
standard in science.  It's akin to trying to measure attometers with a
meter stick or nanoseconds with a grandfather clock:  you just can't do
it and claim to be telling the truth.  If I tried to measure 0.01
microliters with a graduated cylinder, I'd be laughed out of the lab.
Here are the most accurate values I can find for Avogadro's number and
the molecular weight of gold (they're a little different from what I
said before, but they're from a different book and this one is more
recent):
Avogadro's number: 6.022136736x10^23
M.W. of Au:  196.96655
With these values, the mass of one atom of gold is calculated to be
3.2707087x10^-22 grams.  We cannot go any further than that because our
original tools of measurement have their limits of 8 digits and so we
must admit our limits of calculation.

Although I'm flattered by the title, I must admit I am still an
undergraduate (6 units from my B.S.), lest my friends on this mailing
list expose me for an imposter.  :) I also must admit to being a
molecular biologist, not a chemist, but I've taken more chemistry
classes than any one person should ever have to.

Jon D was right about atoms being fuzzy things.  The current model of an
atom itself is actually made up of a lot of empty space, consisting of a
nucleus (protons and neutrons) and electrons orbiting around it.
Electrons don't even orbit in a "sphere."   However, we don't even
really know that for sure, no one has ever really SEEN an atom to my
knowledge, only experimentally assigned qualities.  I would be reluctant
to define dimensions at all, but if you want I can try to look up the
size of its orbitals.

Here I stop, because most of you have probably lost interest in my
long-windedness and aren't reading anymore anyway.

--Adria

> docfarmer@riyadbank.com.sa wrote:
>
> Prof. Decker,
>
> Many thanks for your corrections.  However, I get 3.27097*10^-22, not
> 3.27047.  Or, more accurately 0.000 000 000 000 000 000 000 032 709
> 706 985 555 555 etc. gram per atom.  Also, I think I must disagree
> with your premise that accuracy can only be claimed to 6 digits.  If
> the division creates continual remainders, then accuracy can continue
> to the nth degree (see also pi).  Granted, the last significant digits
> continue ad infinitum as 5, so in scientific notation the number would
> show as 3.2709706985556*10^-22.  Considering that this number would be
> multiplied by 10^23 to get the new mass constant, the remainders take
> up more significance.
>
> I agree that if there are more accurate figures for the atomic weight
> and Avoradro's number, I would be happy to use them.  The more
> accurate the data, the more useful the new constant will be in the
> long run.  I don't want to make the French mistake here (and not the
> one from the end of Blazing Saddles, either (hee hee)).
>
> I'm still in need of assistance on the actual dimensions of the atom
> as well, so if there are any chemists or physics gurus out there, I'd
> be grateful for your help.
>
> Once again, my thanks.
>
> Doc Farmer
> Policy Development Advisor
> Official Monster Raving Loony Party
>
> -----Original Message-----
> From: Adria Decker [mailto:adria@arizona.edu]
> Sent: Wednesday, 18 April 2001 19:37
> To: Mad-Scientists@Mad-Scientists.ORG
> Subject: Re: A New Gold Standard...Echo, echo, echo...
>
>         Well, first of all, you need to switch around your division.
> You want
> to find out grams/atom, not atoms/gram.  Second of all, and since this
>
> is the only way I can get close to the number you got I'll assume it's
>
> what you did, it's 10^23, not 10^-23.  Third of all, you're ignoring
> significant figures, which is a bad thing to do in science, especially
>
> if you're coming up with some sort of measurement standard.  What you
> would actually get is 3.27x10^-22, because you can't wind up with more
>
> significant digits than you started with or else you're claiming more
> accuracy than you actually have.  If you want more accuracy, use
> 6.02257
> X 10^23 for Avogadro's number and 196.9665 for the molecular weight of
>
> gold.  Now you get 3.27047x10^-22 grams per atom, because you can only
>
> claim accuracy to 6 digits as we only have Avogadro's number to 6
> digits, even though we have gold to 7.  I got these values out of my
> organic chemistry textbook, so if they're wrong, blame them. I'm sure
> there are more accurate values for both the molecular weight of gold
> and
> Avogadro's number, but since I'm late for my physics class, you're on
> your own.  :)
> Woo! I'm actually putting my college education to use!
>
> --Adria Decker
>
> > docfarmer@riyadbank.com.sa wrote:
> >
> > Joe,
> >
> > Thanks for that.  So, if I've done my maths correctly, a single atom
>
> > of gold should weigh 0.000 000 000 000 000 000 000 000 305 538 756
> 345
> > 178 gram.  Could someone please confirm this?
> >
> > As to getting the actual dimensions of a single atom, I don't think
> > your calculation will work.  Since gold is not arranged in a linear
> > format, but more of a
> >
> > x x x x x x x x x x
> >  x x x x x x x x x
> > x x x x x x x x x x
> >
> > format, the offsets would skew the results.  Does anyone have a
> > measurement calculation on a single atom?  I await your responses
> with
> > interest.  Thanks again.
> >
> > Doc Farmer
> > Policy Development Advisor
> > Official Monster Raving Loony Party
> >
> > -----Original Message-----
> > From: Joe D [mailto:joed@cws.org]
> > Sent: Tuesday, 17 April 2001 21:28
> > To: Mad-Scientists@Mad-Scientists.ORG
> > Subject: Re: A New Gold Standard...Echo, echo, echo...
> >
> > On Tue, Apr 17, 2001 at 08:04:37AM +0300, docfarmer@riyadbank.com.sa
>
> > wrote:
> > > Dr. X,
> > >
> > > Your perspicacity is astounding (not to mention your flair for the
>
> > > pedantesque).  I wish to use Gold 197 (the "stable" isotope) for
> the
> >
> > > baseline measurements on my new metric system.  As for my
> > inappropriate use
> > > of "rest", let me clarify.  I meant a single atom in a stable
> > structure (no
> > > horse jokes here, please) without large fluctuations in the
> electron
> > sphere.
> > > The dimensions and mass of such an atom would not change
> > significantly over
> > > the course of measurement (Holy Heisenburg!).
> >
> > Back of the envelope stuff here...
> >
> > 6.02 * 10^23 atoms of gold 197 weigh 197 grams.  There may be a more
>
> > accurate representation of Avogadro's Number out there - it's been a
>
> > long
> > time since high school chemistry.
> >
> > You could then use the density of gold to find out how many cubic
> > centimeters 197 grams of gold occupies, and from that get the volume
>
> > of one
> > atom.
> >
> > More or less...
> >
> > Joe D
> >
> >



***************************************************************************
* The information transmitted in this e-Mail, and any files transmitted *
* with it, is confidential and intended solely for the use of the *
* individual(s) to whom it is addressed. Any review, retransmission, *
* dissemination or other use of or taking action in reliance upon this *
* information by persons or entities other than the intended recipient(s) *
* is prohibited. Any views or opinions expressed are solely those of *
* the author, and do not necessarily represent those of Riyad Bank. If *
* you have received this message in error, please notify the sender and *
* the system manager at postmaster@RiyadBank.com.sa and delete the material *
* from your computer. *
* *
* This footnote confirms that this message and any associated attachments *
* have been scanned by MIMESweeper for content security and the presence *
* of computer viruses. *
***************************************************************************
 ------_=_NextPart_001_01C0D23F.65938274--