Specialty Chemistry Forums > Materials and Nanochemistry forum

Why are there more atoms on surface when radius decreases

(1/2) > >>

ethyl ethanoate:
Hello,

If the surface to volume ratio of a sphere with radius r is (4πr2) / (4/3 * πr3) = 3/r,

why is it that there are more atoms on the surface when the radius decreases? And why are there more atoms on surface when the number of atoms increase?

Borek:
So basically you are asking why 3/r grows when r gets smaller?

What is larger: 1/2 or 1/3? 3/2 or 3/3?

Corribus:
Also, there aren't more atoms on the surface when the particle gets smaller. Same as the fact that the surface area gets smaller as the radius gets smaller. Rather, the fraction of total atoms in the particle that are on the surface increases as the radius decreases. This is related to the concept called the specific surface area, the surface area of a solid divided by its mass. Since the surface area of a sphere scales as r2 and the mass scales as r3 (mass is proportional to volume), then specific surface area scales as 1/r, has (SI) units of m-1, and gets larger as r gets smaller. Since the nanoparticle mass can be thought of as in terms of "number of atoms", then the number of atoms on the surface divided by the total number of atoms in the particle also gets larger as r gets smaller. You can see it is the case by imagining a "particle" of exactly 1 atom. In this case, 100% of the atoms are on the surface. And in the case of a huge particle a mile wide, the fraction of particles on the surface is effectively zero.

jeffmoonchop:
Check the graph. The Y axis is not number of atoms on surface. Its the percentage of atoms on the surface.

ethyl ethanoate:
Sorry, I was just so confused because in the script it was basically talking about the ratio of surface area to volume. It showed that it was 3/r and then concluded from the equation that "with decreasing radius the ratio of surface area to volume increases and more and more atoms are on the surface".


--- Quote from: Corribus on October 20, 2021, 06:15:05 PM ---Also, there aren't more atoms on the surface when the particle gets smaller. Same as the fact that the surface area gets smaller as the radius gets smaller. Rather, the fraction of total atoms in the particle that are on the surface increases as the radius decreases. This is related to the concept called the specific surface area, the surface area of a solid divided by its mass. Since the surface area of a sphere scales as r2 and the mass scales as r3 (mass is proportional to volume), then specific surface area scales as 1/r, has (SI) units of m-1, and gets larger as r gets smaller. Since the nanoparticle mass can be thought of as in terms of "number of atoms", then the number of atoms on the surface divided by the total number of atoms in the particle also gets larger as r gets smaller. You can see it is the case by imagining a "particle" of exactly 1 atom. In this case, 100% of the atoms are on the surface. And in the case of a huge particle a mile wide, the fraction of particles on the surface is effectively zero.

--- End quote ---

Ah okay I get it now, so basically if we decrease r that basically means increasing number of atoms?
For example if we imagine one particle as exactly one atom, then if we decrease its radius, it splits up into many tiny atoms which still occupies the same volume (like the picture above) which leads to the fact that the ratio of the atoms on the surface to the total number of atoms is very very little?

Is my understanding correct?

Navigation

[0] Message Index

[#] Next page