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 nm

^{3} 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 nm

^{3}.

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 nm

^{3}, 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 nm

^{3} (according to my previous post), but adjusted for the atomic packing factor, multiplied with 0.68 to 0.74, equal to 333 to 365 nm

^{3}.

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.