Finding number of atoms and volume

  • Thread starter Thread starter fernanhen
  • Start date Start date
  • Tags Tags
    Atoms Volume
Click For Summary
To determine the number of atoms in 1 cm³ of copper, the density and mass of a copper atom are used in calculations. The density of copper is 8,920 kg/m³, and the mass of a copper atom is 5.30 x 10^-25 kg. The initial approach involves using the formula for density (mass/volume) and converting units appropriately to find the number of atoms per cm³. The discussion emphasizes the need for correct unit conversions and suggests using unitary conversion factors to simplify the calculations. The focus remains on solving part (a) first, as it is essential for completing parts (b) and (c).
fernanhen
Messages
9
Reaction score
0

Homework Statement



The mass of a copper atom is 5.30 10-25 kg, and the density of copper is 8 920 kg/m3 .
(a) Determine the number of atoms in 1 cm3 of copper.

_______Cu—atom/cm3



(b) Visualize the one cubic centimeter as formed by stacking up identical cubes, with one copper atom at the center of each. Determine the volume of each cube.


_______cm3/Cu—atom




(c) Find the edge dimension of each cube, which represents an estimate for the spacing between atoms.

_________cm



Homework Equations



Density=mass/Volume



The Attempt at a Solution




Well, the book the book we are using for this class "Physics for Scientists and Engineers" by Serway has no right being used in an Introductory Physics class.

I re-read the relevant chapter THREE times and still haven't figured out how to solve this.

Therefore, I used my intuition alone.

Using the info I have:

3.06*10^(-25)kg= (8,920kg/m^3)*1cm^3 = M=D*V

I canceled the volumes but that won't left over any missing variables.

I googled for alternatives and my textbook absolutely does not point us in any direction regarding this question. Please help.

Thanks in advance.
 
Physics news on Phys.org
You need the answer from a) to do b) and c). So focus on that first. It is essentially a unit conversion problem.
 
lewando said:
You need the answer from a) to do b) and c). So focus on that first. It is essentially a unit conversion problem.

Yes, I just figured that this morning. Thanks.

But now I am having trouble with that.

I equated:

(8920kg/m^3)*(m/cm)*(cm-atom)/kg=Cu-atom/cm^3

Then 8920/m^3*1m/100*5.24*10^(-27)=Cu-atom/cm^3

But now what? Does my setup even look right?
 
fernanhen said:
I equated:

(8920kg/m^3)*(m/cm)*(cm-atom)/kg=Cu-atom/cm^3

Then 8920/m^3*1m/100*5.24*10^(-27)=Cu-atom/cm^3

But now what? Does my setup even look right?

Not really. You need to inventory all of your unitary conversion factors and apply them correctly.
1 = 1 Cu atom / 5.30 10-25 kg
1 = 1m3 / 1x106 cm3
1 = 8920 kg / 1m3

You are trying to convert 1 cm3 [of Cu] to Cu atoms. If you had a unitary conversion factor
in the form of:
1 = X Cu atoms / Y cm3 you would be close to being finished with part a). Can you try this using the above unitary factors?
 
Last edited:
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

Similar threads

  • · Replies 16 ·
Replies
16
Views
1K
  • · Replies 16 ·
Replies
16
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
15
Views
2K
  • · Replies 9 ·
Replies
9
Views
9K
  • · Replies 5 ·
Replies
5
Views
6K
  • · Replies 1 ·
Replies
1
Views
2K