Calculating number of atoms inside of a nanoparticle and on the surface

[PeterXXL]

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.

[wgpeters]

I dont think so.  Your formulas are incorrect.

[PeterXXL]

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.

[wgpeters]

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

[PeterXXL]

[kephra]

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

[PeterXXL]

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.

[kephra]

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. 

[kephra]

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.

[PeterXXL]

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?

[kephra]

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?

[PeterXXL]

[kephra]

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.

[PeterXXL]

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.

[kephra]

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.