Finding number of atoms and volume

[fernanhen]

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.

 

[lewando]

You need the answer from a) to do b) and c). So focus on that first. It is essentially a unit conversion problem.

 

[fernanhen]

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?

 

[lewando]

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?

 

[Nickie]

Dear student,

I understand your frustration and confusion with this problem. It can be challenging to apply concepts from a textbook to a real-world scenario. However, as a scientist, it is important to use logical reasoning and problem-solving skills to find a solution.

First, let's start by identifying the given information and what we are trying to find. We are given the mass of a copper atom (5.30*10^-25 kg) and the density of copper (8,920 kg/m^3). We are trying to find the number of atoms in 1 cm^3 of copper, the volume of each cube, and the edge dimension of each cube.

To solve for the number of atoms, we can use the formula for density: density = mass/volume. We know the density and the volume (1 cm^3), so we can rearrange the equation to solve for mass: mass = density * volume. Plugging in the values, we get mass = (8,920 kg/m^3) * (1 cm^3) = 8,920*10^-6 kg. Now, to find the number of atoms, we can divide the mass by the mass of one atom (5.30*10^-25 kg). This gives us (8,920*10^-6 kg)/(5.30*10^-25 kg) = 1.68*10^19 atoms.

To find the volume of each cube, we can use the given information that there is one atom at the center of each cube. This means that the volume of the cube is equal to the volume of the atom. The volume of an atom can be approximated as a sphere, with the formula V = (4/3)*π*r^3, where r is the radius of the atom. We can use the mass and density of the copper atom to calculate the radius using the formula for density: density = mass/volume. Rearranging for volume, we get volume = mass/density. Plugging in the values, we get volume = (5.30*10^-25 kg)/(8,920 kg/m^3) = 5.94*10^-30 m^3. Now, we can solve for the radius using the volume formula: r = (3*volume)/(4*π)^(1/3). Plugging in the values, we get r = (3*5.94*10^-30 m