Author Topic: Calculating number of atoms inside of a nanoparticle and on the surface  (Read 6482 times)

Offline PeterXXL

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In order to calculate the amount of stabilizing / capping agents that is needed, it should depend on the size of the nanoparticles, as the molecules of the stabilizers attach to the outside of the nanoparticxle.

If we assume that the nanoparticles are spheres, then the volume (V) and area (A) are...

A = 4 x Pi x r2
V = ( 4 x Pi x r3 ) / 3

So for nanoparticles that are 10 nm in diameter we have...

A = 314 nm2
V = 524 nm3

And the covalent radius (the radius for atoms in molecules) for Silver is 145 pm +/- 5 pm (= 140 to 150 pm) , so this means that we also can calculate the number of the nanoparticles as...

Number of atoms in the nanoparticle = V / covalent radius
Number of atoms on the surface of the nanoparticle = A / covalent radius

So for a 10 nm nanoparticle it means...

Atoms in the nanoparticle = 524 / 0.145 (or 0.140 to 150) = 3614 (3493 to 3743)
Atoms on the surface of the nanoparticle = 314 /  0.145 (or 0.140 to 0.150) = 2166 (2093 to 2243)

And the ratio between the surface area and volume for the 10 nm particle is 2166 / 3614 = 60%. Which means, that 60% of the silver atoms are on the surface of a spherical nanoparticle of the size of 10 nm in diameter.

Based on this, we should be able to calculate how much stabilizers we need to add to, as they are attached to the surface atoms of the nanoparticles only.

The larger the nanoparticles are, the less percentage of them are on the surface.

Offline wgpeters

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I dont think so.  Your formulas are incorrect.

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Offline PeterXXL

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I dont think so.  Your formulas are incorrect.

In what way are the formulas wrong?

I did a check with the online calculator that I just found here...

http://www.cleavebooks.co.uk/scol/calsph.htm

Entering the value 5 as radius confirms the surface area and volume.



Offline wgpeters

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To find out how many volumes fit in a larger volume, you obviously have to divide volume by volume, not radius.  Likewise with areas. 
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Offline PeterXXL

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To find out how many volumes fit in a larger volume, you obviously have to divide volume by volume, not radius.  Likewise with areas.

Ok, so for a 10 nm diameter nanoparticle we have a radius of 5 nm, and the covalent atom radius is 0.145, and the volume is 524 nm3 then, the percentage of atoms on the surface must have the same ratio as the volume of thickness of atoms on the surface to entire volume of the nanoparticle, like…

O = Outer radius of the nanoparticle (outside of the atoms on the surface) = 5 nm
I = Inner radius of the nanoparticle (inside of the atoms on the surface) = 5 – 0.145 = 4.855 nm, so…

Volume of the entire nanoparticle = 524 nm3
Volume of the “inner” part of the nanoparticle (excluding 1 layer of atoms) =

= ( 4 x Pi x I3 ) / 3 = 479 nm3 

So the outer layer of atoms on the surface have a volume of 45 nm3  (524 – 479) 

Surface ratio =  45 / 524 = 8.6%

So less than 9% of the atoms are in fact on the surface of a 10 nm diameter spherical silver nanoparticle.

Offline kephra

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If I take a bucket and fill it with golf balls, is the bucket really full? 
Colloidal Silver is only a bargain if you make it yourself.

Offline PeterXXL

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If I take a bucket and fill it with golf balls, is the bucket really full?


It's for sure full of golf balls, but there's space in-between the golf balls there ;)


On a molecular level, if the bucket is the nanoparticle, then there's no room for other atoms in that nanoparticle, assuming that that the nanoparticle is spherical.

Offline kephra

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On a molecular level, if the bucket is the nanoparticle, then there's no room for other atoms in that nanoparticle, assuming that that the nanoparticle is spherical.
Nevertheless, all the volume of the nanoparticle is not used up. 
Colloidal Silver is only a bargain if you make it yourself.

Offline kephra

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Take a look at this: https://en.wikipedia.org/wiki/Atomic_packing_factor
BTW, covalent bonding radius is not the correct term, its metallic bonding radius.  The values are the same though in this case.
Covalent bonding radius refers to the bond length between unlike atoms, as in sodium chloride for example.
Colloidal Silver is only a bargain if you make it yourself.

Offline PeterXXL

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Take a look at this: https://en.wikipedia.org/wiki/Atomic_packing_factor
BTW, covalent bonding radius is not the correct term, its metallic bonding radius.  The values are the same though in this case.
Covalent bonding radius refers to the bond length between unlike atoms, as in sodium chloride for example.


Thanks for the link. As they refer to crystallography, I wonder if this also is the case for nanoparticles as colloids or not; are atoms packed closer as crystals than in soluble form?

Offline kephra

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The internal makeup of a particle does not depend on whether it is suspended in water or not.
Crystalline simply means the atoms are arranged in an orderly manner, instead of randomly placed.  A dissolved substance cannot be crystalline. 

Since silver nanoparticles are not soluble, your last question does not apply to them.

Applying what you learned so far, what is your new estimate for the percentage of surface atoms on a 10nm particle?
Colloidal Silver is only a bargain if you make it yourself.

Offline PeterXXL

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The internal makeup of a particle does not depend on whether it is suspended in water or not.
Crystalline simply means the atoms are arranged in an orderly manner, instead of randomly placed.  A dissolved substance cannot be crystalline. 

Since silver nanoparticles are not soluble, your last question does not apply to them.

Applying what you learned so far, what is your new estimate for the percentage of surface atoms on a 10nm particle?


So ref. to https://en.wikipedia.org/wiki/Atomic_packing_factor

…atoms are packed, and the majority of metals have an “atomic packing factor” of 0.68 to 0.74 so this mean that we can assume that the actual size between each atom in the nanoparticle is 0.68
to 0.74 as small. So a silver atom which has an atomic radius of 0.145 we get 0.145 x 0.68 = 0.0986 to 0.145 x 0.74 = 0.1073. And based on that…

For a nanoparticle of 10 nm in diameter we have...

Surface Area = 314 nm2
Volume = 524 nm3
Atomic radius = 0.145 nm

Atomic radius corrected for “atomic packing factor” = 0.0986 or 0.1073 nm
O = Outer radius of the nanoparticle (outside of the atoms on the surface) = 5 nm
I = Inner radius of the nanoparticle (inside of the atoms on the surface) = 5 – 0.0986 = 4.9014 to 5 – 0.1073 = 4.8927 nm, so…

Volume of the entire nanoparticle = 524 nm3
Volume of the “inner” part of the nanoparticle (excluding 1 layer of atoms) = ( 4 x Pi x I3 ) / 3 = 490 to 493 nm3

So the outer layer of atoms on the surface have a volume of 31 to 34 nm3  (524 – 490 to 493) .

Compared to the total volume of the nanoparticle, this is  31 / 524 to 34 / 524 = 5.9% to 6.5% of the total volume.
« Last Edit: July 21, 2015, 12:35:15 AM by PeterXXL »

Offline kephra

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Um no.  The packing factor says that only 68 to 74% of the volume of the nanoparticle would actually be silver atoms, the rest is empty space.  So the nanoparticle will have fewer atoms, not more.
Go back to the bucket of golf balls, and see all the empty space that cannot hold another golf ball.
Colloidal Silver is only a bargain if you make it yourself.

Offline PeterXXL

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Um no.  The packing factor says that only 68 to 74% of the volume of the nanoparticle would actually be silver atoms, the rest is empty space.  So the nanoparticle will have fewer atoms, not more.
Go back to the bucket of golf balls, and see all the empty space that cannot hold another golf ball.

Ok, so for a 10 nm diameter nanoparticle we have a volume of 524 nm3 but only 68 - 74% of that consists of atoms, making the volume of atoms to 524 x 0.68 or 524 x 0.74 = 356 to 524 nm3.

And if we divide that with the volume of a silver atom, which have a radius of 0.145 nm, equal to an atomic volume of 1.28 nm3, we get 356 to 524 / 1.28 = 278 to 409 atoms in the nanoparticle.

And on the surface we have a layer of atoms with the volume of 490 to 493 nm3 (according to my previous post), but adjusted for the atomic packing factor, multiplied with 0.68 to 0.74, equal to 333 to 365 nm3

So the surface to total volume ratio (corrected for the atomic packing factor) is 333 to 365 / 356 to 524 = 64 to 94%, which confirms that most of the silver atoms atoms on a 10 nm diameter spherical nanoparticle are actually on the surface.

Offline kephra

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Quote
And if we divide that with the volume of a silver atom, which have a radius of 0.145 nm, equal to an atomic volume of 1.28 nm3, we get 356 to 524 / 1.28 = 278 to 409 atoms in the nanoparticle.
Not even close. 
Colloidal Silver is only a bargain if you make it yourself.