### Re: Answer keys w/ explanation?

Posted:

**Sat Jul 07, 2018 3:30 am**Hello,

In order to find the atomic number density I used the equation you have linked:

N=ρNa/M

The values I used where ρ = 16g/cm^3; Na = 6.022 x 10^23 nuclei/mole; and M = 181

The value I received for N = 5.32 x 10^22 atoms of tantalum/cm^3

The cubed root of this value is approximately 3.76 x 10^7 atoms/cm which I would imagine is just how many diameters of the atom (including the electrons) you need to line next to one another to make 1 cm worth of atoms.

In order to find the distance between each center, you simply divide the cm value by the atomic value, or re-arrange by flipping:

1 cm/3.76 x 10^7 atoms

= 2.66x 10^-8 cm

Now because this is the distance between each nuclei, then we can assume it is also the diameter of the atom, including the electrons at rest state (no excitation). As well, we must also say that the diameter of the nucleus is 1.70 x 10^-12

In the question, it says that the nuclei need to take up about 1.0% of the area. This can be modeled by the equation

T = Nn (1.70 x 10^-12)/ Na (2.66 x 10^-8)

Where Nn is the number of nuclei; Na is the number of atoms; and T is the thickness of the foil

Because we want 1.0% of the area to be blocked by nuclei, the we sub the value Nn = 1 and Na =100

The end result is T = 6.39 x 10^-7 cm

This is clearly incorrect since we cannot have different amounts of nuclei and atoms. So what I did was subtract the diameter of the nuclei from the diameter from the atom so we can assume that Na is the free space between the two nuclei. The problem is there is to much of a difference in units between the two values, so when I subtract and round, it remains the same value.

The question on hand now is… is it right?

Ameen Aydan

In order to find the atomic number density I used the equation you have linked:

N=ρNa/M

The values I used where ρ = 16g/cm^3; Na = 6.022 x 10^23 nuclei/mole; and M = 181

The value I received for N = 5.32 x 10^22 atoms of tantalum/cm^3

The cubed root of this value is approximately 3.76 x 10^7 atoms/cm which I would imagine is just how many diameters of the atom (including the electrons) you need to line next to one another to make 1 cm worth of atoms.

In order to find the distance between each center, you simply divide the cm value by the atomic value, or re-arrange by flipping:

1 cm/3.76 x 10^7 atoms

= 2.66x 10^-8 cm

Now because this is the distance between each nuclei, then we can assume it is also the diameter of the atom, including the electrons at rest state (no excitation). As well, we must also say that the diameter of the nucleus is 1.70 x 10^-12

In the question, it says that the nuclei need to take up about 1.0% of the area. This can be modeled by the equation

T = Nn (1.70 x 10^-12)/ Na (2.66 x 10^-8)

Where Nn is the number of nuclei; Na is the number of atoms; and T is the thickness of the foil

Because we want 1.0% of the area to be blocked by nuclei, the we sub the value Nn = 1 and Na =100

The end result is T = 6.39 x 10^-7 cm

This is clearly incorrect since we cannot have different amounts of nuclei and atoms. So what I did was subtract the diameter of the nuclei from the diameter from the atom so we can assume that Na is the free space between the two nuclei. The problem is there is to much of a difference in units between the two values, so when I subtract and round, it remains the same value.

The question on hand now is… is it right?

Ameen Aydan