Question

Number of atoms in a nanoparticle?

Silver nanoparticles are included in the material sports socks are made from. Silver nanoparticles can kill bacteria. A silver atom has a diameter of 2.8 * 10 ^ -10 m. Give the maximum and minimum number of silver atoms in the thickness of a nanoparticle.

Answer 1

Odd question ... the term nanoparticle doesn't have a specific definition, only a typical range so the specific diameter quoted doesn't help us much.

Explanation:

Nanoparticle would normally describe something between 10 - 100 manometers wide, so we could estimate there would be between 40 and 400 atoms, but that may not be very helpful to you.

Maybe someone else can improve on this so I'll request a double-check.

Answer 2

The number of silver atoms in a nanoparticle can range from
`"30 atoms"` to `3.0 × 10^7 color(white)(l)"atoms"`.

Explanation:

Assume that the atoms are hard spheres.

They fill the nanoparticles, which have diameters ranging from 1 nm to 100 nm.

However, the atomic spheres do not fill all the available space, because there are holes in the spaces between the spheres.

Even when the atoms are packed most efficiently, the atoms will occupy
only 74 % of the available volume.

Minimum number of silver atoms

(a) Calculate the volume of the smallest nanoparticle

`V_text(nano) = 4/3πr^3 = 1/6πd^3`

`V_text(nano) = 1/6π(1 × 10^"-9"color(white)(l) "m")^3 = 5.2 ×10^"-28" color(white)(l)"m"^3`

(b) Calculate the occupied volume

`V_text(occupied) = 0.74 × 5.2 ×10^"-28" color(white)(l)"m"^3 = 3.9 ×10^"-28" color(white)(l)"m"^3"`

(c) Calculate the volume of an atom

`V_text(atom) = 1/6πd^3 = 1/6π(2.8 × 10^"-10"color(white)(l)"m"^3) = 1.15 ×10^"-29" color(white)(l)"m"^3`

(d) Calculate the number of atoms

`"No. of atoms" = V_text(occupied)/V_text(atom) =3.9 ×10^"-28" color(red)(cancel(color(black)("m"^3))) × "1 atom"/(1.15 ×10^"-29" color(red)(cancel(color(black)("m"^3)))) = "30 atoms"`

Maximum number of silver atoms

(a) Calculate the volume of the largest nanoparticle

`V_text(max)/V_text(min) = (color(red)(cancel(color(black)(1/6π)))d_text(max)^3)/(color(red)(cancel(color(black)(1/6π)))d_text(min)^3) = ((100 color(red)(cancel(color(black)("nm"))))/(1 color(red)(cancel(color(black)("nm")))))^3 = 1.0 × 10^6`

The volume of the largest nanoparticle is `1.0 × 10^6` times the volume of the smallest nanoparticle.

(b) Calculate the number of atoms

`"No. of atoms" = "30 atoms" ×1.0 × 10^6 = 3.0 × 10^7 color(white)(l)"atoms"`