Archive-Date: Sat, 14 Apr 2001 01:25:30 -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 Subject: RE: A New Gold Standard...Echo, echo, echo... Date: Sat, 14 Apr 2001 11:07:36 +0300 MIME-Version: 1.0 Content-Type: multipart/alternative; boundary="----_=_NextPart_001_01C0C4BC.42D4240E" 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_01C0C4BC.42D4240E Content-Type: text/plain; charset="ISO-8859-1" Hey Anybody! Is anybody out there that can answer the questions below, or point me in the right direction? Thanks. Doc -----Original Message----- From: docfarmer@riyadbank.com.sa [mailto:docfarmer@riyadbank.com.sa] Sent: Wednesday, 21 March 2001 10:01 To: Mad-Scientists@Mad-Scientists.ORG Subject: A New Gold Standard... Greetings! I'd like some information pertaining to gold (AU - Atomic Number 79). * What is the specific size of a single atom of gold (in nanometers or picometers)? * What is the exact physical weight of a single atom of gold (in milligrams or nanograms)? * How about one billion (1,000,000,000) atoms of gold as well? I'm finalising the new Loony Metric system, which will be based on the size and weight of the gold atom. If possible, please provide documentary references to the figures above, so that I won't run into any conflicts with the ISO. Many thanks in advance for your help. Doc Farmer Policy Development Advisor Official Monster Raving Loony Party *************************************************************************** * 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_01C0C4BC.42D4240E Content-Type: text/html; charset="ISO-8859-1"
Hey Anybody!
 
Is anybody out there that can answer the questions below, or point me in the right direction?  Thanks.
 
Doc
-----Original Message-----
From: docfarmer@riyadbank.com.sa [mailto:docfarmer@riyadbank.com.sa]
Sent: Wednesday, 21 March 2001 10:01
To: Mad-Scientists@Mad-Scientists.ORG
Subject: A New Gold Standard...

Greetings!
 
I'd like some information pertaining to gold (AU - Atomic Number 79). 
I'm finalising the new Loony Metric system, which will be based on the size and weight of the gold atom.  If possible, please provide documentary references to the figures above, so that I won't run into any conflicts with the ISO.  Many thanks in advance for your help.
 
Doc Farmer
Policy Development Advisor
Official Monster Raving Loony Party
 
 
 


***************************************************************************
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* 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) *
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* you have received this message in error, please notify the sender and *
* the system manager at postmaster@RiyadBank.com.sa and delete the material *
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* *
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------_=_NextPart_001_01C0C4BC.42D4240E-- ================================================================================ Archive-Date: Sat, 14 Apr 2001 20:50:46 -0700 Sender: owner-mad-scientists@VJC.COM Message-ID: From: docfarmer@riyadbank.com.sa Reply-To: Mad-Scientists@Mad-Scientists.ORG To: x@euthanasia.fsnet.co.uk CC: Mad-Scientists@Mad-Scientists.ORG Subject: RE: A New Gold Standard...Echo, echo, echo... Date: Sun, 15 Apr 2001 06:41:34 +0300 MIME-Version: 1.0 Content-Type: multipart/alternative; boundary="----_=_NextPart_001_01C0C55F.233FA534" 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_01C0C55F.233FA534 Content-Type: text/plain; charset="ISO-8859-1" Dr. X, Thank you. You've shown me that my request should have been more specific. Allow me to clarify. The size should be calculated on a single atom at rest, in a 1g environment at sea level, at a precise temperature of 0 degrees Celsius. Weight (or more accurately, Mass) should be under the same environment. I was only aware of one isotope of gold (the stable one you wear in jewellery), but if there are others please let me know. I know that the atomic radius is 1.79 Angstroms and the covalent radius is 1.34 Angstroms, but I'm unsure what that means (I've got a Periodic Table on my Visor, but I'm not familiar with all the terminology). Again, my thanks. Doc Farmer Policy Development Advisor Official Monster Raving Loony Party -----Original Message----- From: x [mailto:x@euthanasia.fsnet.co.uk] Sent: Saturday, 14 April 2001 18:00 To: docfarmer@riyadbank.com.sa Subject: Re: A New Gold Standard...Echo, echo, echo... size depends on how excited the atom is as the electron orbit will change. the weight is dependent on which isotope you are weighing and of course the gravity enviroment in which you are doing the weighing. anyweigh enough with the atom weighing get your man over to china and wei up some chuk kam before america spoils everything. yours greedily dr.x. ----- Original Message ----- From: docfarmer@riyadbank.com.sa To: Mad-Scientists@Mad-Scientists.ORG Sent: Saturday, April 14, 2001 9:07 AM Subject: RE: A New Gold Standard...Echo, echo, echo... Hey Anybody! Is anybody out there that can answer the questions below, or point me in the right direction? Thanks. Doc -----Original Message----- From: docfarmer@riyadbank.com.sa [mailto:docfarmer@riyadbank.com.sa] Sent: Wednesday, 21 March 2001 10:01 To: Mad-Scientists@Mad-Scientists.ORG Subject: A New Gold Standard... Greetings! I'd like some information pertaining to gold (AU - Atomic Number 79). * What is the specific size of a single atom of gold (in nanometers or picometers)? * What is the exact physical weight of a single atom of gold (in milligrams or nanograms)? * How about one billion (1,000,000,000) atoms of gold as well? I'm finalising the new Loony Metric system, which will be based on the size and weight of the gold atom. If possible, please provide documentary references to the figures above, so that I won't run into any conflicts with the ISO. Many thanks in advance for your help. Doc Farmer Policy Development Advisor Official Monster Raving Loony Party *************************************************************************** * 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_01C0C55F.233FA534 Content-Type: text/html; charset="ISO-8859-1" Content-Transfer-Encoding: quoted-printable
Dr.=20 X,
 
Thank=20 you.  You've shown me that my request should have been more = specific. =20 Allow me to clarify.
 
The=20 size should be calculated on a single atom at rest, in a 1g environment = at=20 sea level, at a precise temperature of 0 degrees Celsius.  = Weight (or=20 more accurately, Mass) should be under the same environment.  = I was=20 only aware of one isotope of gold (the stable one you wear in = jewellery), but if=20 there are others please let me know.  I know that the atomic = radius is 1.79=20 Angstroms and the covalent radius is 1.34 Angstroms, but I'm unsure = what that=20 means (I've got a Periodic Table on my Visor, but I'm not familiar with = all the=20 terminology). 
 
Again,=20 my thanks.
 
Doc=20 Farmer
Policy=20 Development Advisor
Official Monster Raving Loony Party
 
 
-----Original Message-----
From: x=20 [mailto:x@euthanasia.fsnet.co.uk]
Sent: Saturday, 14 April = 2001=20 18:00
To: docfarmer@riyadbank.com.sa
Subject: Re: = A New=20 Gold Standard...Echo, echo, echo...

size depends on how excited = the atom is=20 as the electron orbit will change.
 
the weight is dependent on = which isotope=20 you are weighing and of course the gravity enviroment in which you = are doing=20 the weighing.
 
anyweigh enough with = the atom=20 weighing get your man over to china and wei up = some chuk kam =20 before america spoils everything.
 
          &nb= sp;           &nb= sp;  =20 yours greedily   dr.x.
----- Original Message -----
From:=20 docfarmer@riyadbank.com.sa=20
Sent: Saturday, April 14, = 2001 9:07=20 AM
Subject: RE: A New Gold=20 Standard...Echo, echo, echo...

Hey Anybody!
 
Is=20 anybody out there that can answer the questions below, or point me = in the=20 right direction?  Thanks.
 
Doc
-----Original Message-----
From: docfarmer@riyadbank.com.sa=20 [mailto:docfarmer@riyadbank.com.sa]
Sent: Wednesday, 21 = March=20 2001 10:01
To: Mad-Scientists@Mad-Sci= entists.ORG
Subject:=20 A New Gold Standard...

Greetings!
 
I'd like some information pertaining to gold (AU - = Atomic Number=20 79). 
  • What is the specific size of a single atom of gold (in = nanometers=20 or picometers)?=20
  • What is=20 the exact physical weight of a single atom of gold (in = milligrams or=20 nanograms)? 
  • How about one billion (1,000,000,000) atoms of gold as = well?
I'm finalising the new Loony Metric system, which will = be based on=20 the size and weight of the gold atom.  If possible, please = provide=20 documentary references to the figures above, so that I won't run = into any=20 conflicts with the ISO.  Many thanks in advance for your=20 help.
 
Doc Farmer
Policy Development Advisor
Official Monster Raving Loony Party
 
 
 


*******************************************************= ********************
*=20 The information transmitted in this e-Mail, and any files = transmitted=20 *
* with it, is confidential and intended solely for the use = of the=20 *
* individual(s) to whom it is addressed. Any review, = retransmission,=20 *
* dissemination or other use of or taking action in reliance = upon=20 this *
* information by persons or entities other than the = intended=20 recipient(s) *
* is prohibited. Any views or opinions = expressed are=20 solely those of *
* the author, and do not necessarily = represent those=20 of Riyad Bank. If *
* you have received this message in error, = please=20 notify the sender and *
* the system manager at=20 postmaster@RiyadBank.com.sa and delete the material *
* from = your=20 computer. *
* *
* This footnote confirms that this message = and any=20 associated attachments *
* have been scanned by MIMESweeper = for content=20 security and the presence *
* of computer viruses.=20 = *
*******************************************************************= ********
= ------_=_NextPart_001_01C0C55F.233FA534-- ================================================================================ Archive-Date: Mon, 16 Apr 2001 22:13:47 -0700 Sender: owner-mad-scientists@VJC.COM Message-ID: From: docfarmer@riyadbank.com.sa Reply-To: Mad-Scientists@Mad-Scientists.ORG To: x@euthanasia.fsnet.co.uk, Mad-Scientists@Mad-Scientists.ORG Subject: RE: A New Gold Standard...Echo, echo, echo... Date: Tue, 17 Apr 2001 08:04:37 +0300 MIME-Version: 1.0 Content-Type: multipart/alternative; boundary="----_=_NextPart_001_01C0C6FD.3D67FEA8" 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_01C0C6FD.3D67FEA8 Content-Type: text/plain; charset="iso-8859-1" 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!). As to your time machine reference, I'm a bit confused (so what else is new?). Last I checked, the price of gold had dropped significantly already. My thanks for your help and your humour. Doc Farmer Policy Development Advisor Official Monster Raving Loony Party -----Original Message----- From: x [mailto:x@euthanasia.fsnet.co.uk] Sent: Tuesday, 17 April 2001 03:39 To: docfarmer@riyadbank.com.sa Subject: Re: A New Gold Standard...Echo, echo, echo... gold 198 is used in various medical scenarios, and must be considered being 'heavy' and out here. how will you get the atom to 'rest'? are you keeping the time machine quiet until the bottom drops out of the gold market? yours heavily, dr.x. ----- Original Message ----- From: docfarmer@riyadbank.com.sa To: x@euthanasia.fsnet.co.uk Cc: Mad-Scientists@Mad-Scientists.ORG Sent: Sunday, April 15, 2001 4:41 AM Subject: RE: A New Gold Standard...Echo, echo, echo... Dr. X, Thank you. You've shown me that my request should have been more specific. Allow me to clarify. The size should be calculated on a single atom at rest, in a 1g environment at sea level, at a precise temperature of 0 degrees Celsius. Weight (or more accurately, Mass) should be under the same environment. I was only aware of one isotope of gold (the stable one you wear in jewellery), but if there are others please let me know. I know that the atomic radius is 1.79 Angstroms and the covalent radius is 1.34 Angstroms, but I'm unsure what that means (I've got a Periodic Table on my Visor, but I'm not familiar with all the terminology). Again, my thanks. Doc Farmer Policy Development Advisor Official Monster Raving Loony Party -----Original Message----- From: x [mailto:x@euthanasia.fsnet.co.uk] Sent: Saturday, 14 April 2001 18:00 To: docfarmer@riyadbank.com.sa Subject: Re: A New Gold Standard...Echo, echo, echo... size depends on how excited the atom is as the electron orbit will change. the weight is dependent on which isotope you are weighing and of course the gravity enviroment in which you are doing the weighing. anyweigh enough with the atom weighing get your man over to china and wei up some chuk kam before america spoils everything. yours greedily dr.x. ----- Original Message ----- From: docfarmer@riyadbank.com.sa To: Mad-Scientists@Mad-Scientists.ORG Sent: Saturday, April 14, 2001 9:07 AM Subject: RE: A New Gold Standard...Echo, echo, echo... Hey Anybody! Is anybody out there that can answer the questions below, or point me in the right direction? Thanks. Doc -----Original Message----- From: docfarmer@riyadbank.com.sa [mailto:docfarmer@riyadbank.com.sa] Sent: Wednesday, 21 March 2001 10:01 To: Mad-Scientists@Mad-Scientists.ORG Subject: A New Gold Standard... Greetings! I'd like some information pertaining to gold (AU - Atomic Number 79). * What is the specific size of a single atom of gold (in nanometers or picometers)? * What is the exact physical weight of a single atom of gold (in milligrams or nanograms)? * How about one billion (1,000,000,000) atoms of gold as well? I'm finalising the new Loony Metric system, which will be based on the size and weight of the gold atom. If possible, please provide documentary references to the figures above, so that I won't run into any conflicts with the ISO. Many thanks in advance for your help. Doc Farmer Policy Development Advisor Official Monster Raving Loony Party *************************************************************************** * 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_01C0C6FD.3D67FEA8 Content-Type: text/html; charset="iso-8859-1"
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!). 
 
As to your time machine reference, I'm a bit confused (so what else is new?).  Last I checked, the price of gold had dropped significantly already. 
 
My thanks for your help and your humour. 
 
Doc Farmer
Policy Development Advisor
Official Monster Raving Loony Party
-----Original Message-----
From: x [mailto:x@euthanasia.fsnet.co.uk]
Sent: Tuesday, 17 April 2001 03:39
To: docfarmer@riyadbank.com.sa
Subject: Re: A New Gold Standard...Echo, echo, echo...

gold 198 is used in various medical scenarios, and must be considered  being 'heavy' and out here.  how will you get the atom to 'rest'?  are you keeping the time machine quiet until the bottom drops out of the gold market?  
              yours heavily, dr.x.
 
 
----- Original Message -----
Sent: Sunday, April 15, 2001 4:41 AM
Subject: RE: A New Gold Standard...Echo, echo, echo...

Dr. X,
 
Thank you.  You've shown me that my request should have been more specific.  Allow me to clarify.
 
The size should be calculated on a single atom at rest, in a 1g environment at sea level, at a precise temperature of 0 degrees Celsius.  Weight (or more accurately, Mass) should be under the same environment.  I was only aware of one isotope of gold (the stable one you wear in jewellery), but if there are others please let me know.  I know that the atomic radius is 1.79 Angstroms and the covalent radius is 1.34 Angstroms, but I'm unsure what that means (I've got a Periodic Table on my Visor, but I'm not familiar with all the terminology). 
 
Again, my thanks.
 
Doc Farmer
Policy Development Advisor
Official Monster Raving Loony Party
 
 
-----Original Message-----
From: x [mailto:x@euthanasia.fsnet.co.uk]
Sent: Saturday, 14 April 2001 18:00
To: docfarmer@riyadbank.com.sa
Subject: Re: A New Gold Standard...Echo, echo, echo...

size depends on how excited the atom is as the electron orbit will change.
 
the weight is dependent on which isotope you are weighing and of course the gravity enviroment in which you are doing the weighing.
 
anyweigh enough with the atom weighing get your man over to china and wei up some chuk kam  before america spoils everything.
 
                          yours greedily   dr.x.
----- Original Message -----
Sent: Saturday, April 14, 2001 9:07 AM
Subject: RE: A New Gold Standard...Echo, echo, echo...

Hey Anybody!
 
Is anybody out there that can answer the questions below, or point me in the right direction?  Thanks.
 
Doc
-----Original Message-----
From: docfarmer@riyadbank.com.sa [mailto:docfarmer@riyadbank.com.sa]
Sent: Wednesday, 21 March 2001 10:01
To: Mad-Scientists@Mad-Scientists.ORG
Subject: A New Gold Standard...

Greetings!
 
I'd like some information pertaining to gold (AU - Atomic Number 79). 
  • What is the specific size of a single atom of gold (in nanometers or picometers)?
  • What is the exact physical weight of a single atom of gold (in milligrams or nanograms)? 
  • How about one billion (1,000,000,000) atoms of gold as well?
I'm finalising the new Loony Metric system, which will be based on the size and weight of the gold atom.  If possible, please provide documentary references to the figures above, so that I won't run into any conflicts with the ISO.  Many thanks in advance for your help.
 
Doc Farmer
Policy Development Advisor
Official Monster Raving Loony Party
 
 
 


***************************************************************************
* 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_01C0C6FD.3D67FEA8-- ================================================================================ Archive-Date: Tue, 17 Apr 2001 11:29:14 -0700 Sender: owner-mad-scientists@VJC.COM Date: Tue, 17 Apr 2001 13:27:42 -0500 From: Joe D Reply-To: Mad-Scientists@Mad-Scientists.ORG To: Mad-Scientists@Mad-Scientists.ORG Subject: Re: A New Gold Standard...Echo, echo, echo... Message-ID: <20010417132742.B7422@splut.cws.org> References: MIME-Version: 1.0 Content-Type: text/plain; charset=us-ascii 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 ================================================================================ Archive-Date: Wed, 18 Apr 2001 01:04:54 -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, joed@cws.org Subject: RE: A New Gold Standard...Echo, echo, echo... Date: Wed, 18 Apr 2001 10:56:08 +0300 MIME-Version: 1.0 Content-Type: multipart/alternative ; boundary="----_=_NextPart_001_01C0C7DE.5EBF9A7A" 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_01C0C7DE.5EBF9A7A Content-Type: text/plain; charset="iso-8859-1" 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_01C0C7DE.5EBF9A7A Content-Type: text/html; charset="iso-8859-1" Content-Transfer-Encoding: quoted-printable RE: A New Gold Standard...Echo, echo, echo...

Joe,

Thanks for that.  So, if I've done my maths correctl= y, 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 d= on'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 an= yone have a measurement calculation on a single atom?  I await your re= sponses 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@riyad= bank.com.sa wrote:
> Dr. X,
>
> Your perspicacity is astounding (not to mention you= r flair for the
> pedantesque).  I wish to use Gold 197 (the &qu= ot;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 fluctuation= s in the electron sphere.
> The dimensions and mass of such an atom would not c= hange 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.  The= re 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 ma= ny cubic
centimeters 197 grams of gold occupies, and from that ge= t 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_01C0C7DE.5EBF9A7A-- ================================================================================ Archive-Date: Wed, 18 Apr 2001 09:39:58 -0700 Sender: owner-mad-scientists@VJC.COM Message-ID: <3ADDC2AE.7EC7005B@arizona.edu> Date: Wed, 18 Apr 2001 09:37:02 -0700 From: Adria Decker Reply-To: Mad-Scientists@Mad-Scientists.ORG MIME-Version: 1.0 To: Mad-Scientists@Mad-Scientists.ORG Subject: Re: A New Gold Standard...Echo, echo, echo... References: Content-Type: text/plain; charset=us-ascii Content-Transfer-Encoding: 7bit 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. * > *************************************************************************** ================================================================================ Archive-Date: Wed, 18 Apr 2001 10:27:14 -0700 Sender: owner-mad-scientists@VJC.COM Date: Wed, 18 Apr 2001 12:25:39 -0500 From: Joe D Reply-To: Mad-Scientists@Mad-Scientists.ORG To: Mad-Scientists@Mad-Scientists.ORG Subject: Re: A New Gold Standard...Echo, echo, echo... Message-ID: <20010418122539.A11287@splut.cws.org> References: MIME-Version: 1.0 Content-Type: text/plain; charset=us-ascii On Wed, Apr 18, 2001 at 10:56:08AM +0300, docfarmer@riyadbank.com.sa wrote: > 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. An atom is a fuzzy thing. I don't know if anyone's tried to measure it. The approach I was taking is that X number of atoms occupy Y volume of space, so the average volume per atom is Y/X. There's a lot of empty space between atoms, so it may not be a strictly accurate approach, but it has an appealing simplicity. Joe D ================================================================================ Archive-Date: Wed, 18 Apr 2001 16:51:35 -0700 Sender: owner-mad-scientists@VJC.COM From: Vanrijn25@aol.com Reply-To: Mad-Scientists@Mad-Scientists.ORG Message-ID: <6e.9b8bd0c.280f8220@aol.com> Date: Wed, 18 Apr 2001 19:49:52 EDT Subject: Re: A New Gold Standard...Echo, echo, echo... To: Mad-Scientists@mad-scientists.org MIME-Version: 1.0 Content-Type: text/plain; charset="US-ASCII" Content-Transfer-Encoding: 7bit hello, well done adria! i dont receive alot of the mad scientist discussions anymore but this was refreshing. attention to significant digits is vital in making sound decisions based on numeric data. also, i've found that too much accuracy tends to leave things open to severe scrutiny. FYI..to find the size of an atom of Gold, look up what valence layer it occupies and the same book should tell you how big that is. cause your right: you can't extrapolate the size of an atom from the size of say..a Kilo of the same stuff. be good now, bob ================================================================================ Archive-Date: Thu, 19 Apr 2001 22:33:27 -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, adria@arizona.edu Subject: RE: A New Gold Standard...Echo, echo, echo... Date: Fri, 20 Apr 2001 08:24:24 +0300 MIME-Version: 1.0 Content-Type: multipart/alternative; boundary="----_=_NextPart_001_01C0C95B.7F639154" 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_01C0C95B.7F639154 Content-Type: text/plain; charset="iso-8859-1" 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_01C0C95B.7F639154 Content-Type: text/html; charset="iso-8859-1" Content-Transfer-Encoding: quoted-printable RE: A New Gold Standard...Echo, echo, echo...

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_01C0C95B.7F639154-- ================================================================================ Archive-Date: Fri, 20 Apr 2001 02:40:36 -0700 Sender: owner-mad-scientists@VJC.COM Message-ID: <3AE00374.DB8A6180@arizona.edu> Date: Fri, 20 Apr 2001 02:37:56 -0700 From: Adria Decker Reply-To: Mad-Scientists@Mad-Scientists.ORG MIME-Version: 1.0 To: Mad-Scientists@Mad-Scientists.ORG Subject: Re: A New Gold Standard...Echo, echo, echo... References: Content-Type: text/plain; charset=us-ascii Content-Transfer-Encoding: 7bit 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. * > > > *************************************************************************** ================================================================================ Archive-Date: Fri, 20 Apr 2001 03:17:57 -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, adria@arizona.edu Subject: RE: A New Gold Standard...Echo, echo, echo... Date: Fri, 20 Apr 2001 13:09:02 +0300 MIME-Version: 1.0 Content-Type: multipart/alternative ; boundary="----_=_NextPart_001_01C0C983.4341C47A" 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_01C0C983.4341C47A Content-Type: text/plain; charset="iso-8859-1" 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_01C0C983.4341C47A Content-Type: text/html; charset="iso-8859-1" Content-Transfer-Encoding: quoted-printable RE: A New Gold Standard...Echo, echo, echo...

Adria,

First, let me explain the "Prof" prefix in my l= ast 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, becaus= e I felt it might be too familiar (hope you don't mind it in this one).&nbs= p; 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 Fren= ch 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 wh= y 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 fluctua= tions occur in the electron sphere, which may make this more difficult.&nbs= p; 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 mo= st grateful.

By the bye, I think atoms have been actually observed som= e years back.  IBM Labs was I believe the first group to actually mani= pulate 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 a= s "live" movies, showing the movement of the electron sphere.&nbs= p; Ain't science amazing?!?

My thanks again for your efforts.

Doc Farmer
Policy Development Advisor
Official Monster Raving Loony Party

p.s.         &nbs= p;  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 su= ffering 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 applicat= ion beg to
differ, but if that's what you get, that's what you get.=
The concept of significant figures is a scientific stand= ard.  You cannot
claim to accurately know any more digits than what you s= tart with, and
if you do, you are being misleading.  Almost every = single constant in
science is rounded to some degree.  It would be rid= iculous to claim
otherwise.  Sure, your remainders take up more sign= ificance when
mulitplied by 10^23, but that just means they're more si= gnificantly
WRONG.  Sure, 10/3 is 3.33333333333 ad infinitum, b= ut what if that 3 is
really 3.443298473 and that 10 is really 9.67899987?&nbs= p; The only digit you
know with any accuracy is the first 3.  You would o= nly know for sure
that those threes go on forever if you had 10.00000000(t= o
infinity)/3.000000000(to infinty).  This is a unive= rsally accepted
standard in science.  It's akin to trying to measur= e attometers with a
meter stick or nanoseconds with a grandfather clock:&nbs= p; you just can't do
it and claim to be telling the truth.  If I tried t= o measure 0.01
microliters with a graduated cylinder, I'd be laughed ou= t of the lab.
Here are the most accurate values I can find for Avogadr= o'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 calcu= lated to be
3.2707087x10^-22 grams.  We cannot go any further t= han that because our
original tools of measurement have their limits of 8 dig= its and so we
must admit our limits of calculation.

Although I'm flattered by the title, I must admit I am st= ill an
undergraduate (6 units from my B.S.), lest my friends on= this mailing
list expose me for an imposter.  :) I also must adm= it 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 ar= ound 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 a= n atom to my
knowledge, only experimentally assigned qualities. = I would be reluctant
to define dimensions at all, but if you want I can try t= o look up the
size of its orbitals.

Here I stop, because most of you have probably lost inter= est 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 00= 0 000 000 000 032 709
> 706 985 555 555 etc. gram per atom.  Also, I t= hink I must disagree
> with your premise that accuracy can only be claimed= to 6 digits.  If
> the division creates continual remainders, then acc= uracy can continue
> to the nth degree (see also pi).  Granted, the= last significant digits
> continue ad infinitum as 5, so in scientific notati= on the number would
> show as 3.2709706985556*10^-22.  Considering t= hat this number would be
> multiplied by 10^23 to get the new mass constant, t= he 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 mis= take here (and not the
> one from the end of Blazing Saddles, either (hee he= e)).
>
> I'm still in need of assistance on the actual dimen= sions of the atom
> as well, so if there are any chemists or physics gu= rus 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= ...
>
>         Wel= l, first of all, you need to switch around your division.
> You want
> to find out grams/atom, not atoms/gram.  Secon= d of all, and since this
>
> is the only way I can get close to the number you g= ot I'll assume it's
>
> what you did, it's 10^23, not 10^-23.  Third o= f 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 s= tandard.  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 yo= u'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 at= om, 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, bl= ame them. I'm sure
> there are more accurate values for both the molecul= ar weight of gold
> and
> Avogadro's number, but since I'm late for my physic= s class, you're on
> your own.  :)
> Woo! I'm actually putting my college education to u= se!
>
> --Adria Decker
>
> > docfarmer@riyadbank.com.sa wrote:
> >
> > Joe,
> >
> > Thanks for that.  So, if I've done my mat= hs 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 t= his?
> >
> > As to getting the actual dimensions of a singl= e atom, I don't think
> > your calculation will work.  Since gold i= s 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.&nb= sp; 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 [m= ailto: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, docf= armer@riyadbank.com.sa
>
> > wrote:
> > > Dr. X,
> > >
> > > Your perspicacity is astounding (not to m= ention your flair for the
>
> > > pedantesque).  I wish to use Gold 19= 7 (the "stable" isotope) for
> the
> >
> > > baseline measurements on my new metric sy= stem.  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 f= luctuations in the
> electron
> > sphere.
> > > The dimensions and mass of such an atom w= ould not change
> > significantly over
> > > the course of measurement (Holy Heisenbur= g!).
> >
> > 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 o= ut 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 fr= om 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_01C0C983.4341C47A-- ================================================================================ Archive-Date: Sat, 21 Apr 2001 11:30:01 -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, adria@arizona.edu Subject: RE: A New Gold Standard...Echo, echo, echo... Date: Sat, 21 Apr 2001 16:12:52 +0300 MIME-Version: 1.0 Content-Type: multipart/alternative; boundary="----_=_NextPart_001_01C0CA66.1D1AFADA" 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_01C0CA66.1D1AFADA Content-Type: text/plain; charset="iso-8859-1" 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_01C0CA66.1D1AFADA Content-Type: text/html; charset="iso-8859-1" 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
> >
> >



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* you have received this message in error, please notify the sender and *
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* *
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