Answer keys w/ explanation?

 Posts: 93
 Joined: Sun Apr 15, 2018 7:33 pm
 Real name: Ameen Aydan
Re: Answer keys w/ explanation?
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 rearrange 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 rearrange 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
 Rich Feldman
 Posts: 1308
 Joined: Mon Dec 21, 2009 11:59 pm
 Real name: Rich Feldman
 Location: Santa Clara County, CA, USA
Re: Answer keys w/ explanation?
First half looks correct, up to and including your value for nuclear diameter: 1.70E12 cm (or 1.36E12 if we use the smaller value for r0). Applause for your typography with rhos and square root symbols. The only ones I remember how to get without menus are Ω and °.
Let's see if someone else steps up to review the next step Ameen presented for review:
probability of hitting some nucleus when shooting at metal foil.
How many nuclei in 1 cm^2 of foil, if it were 0.1 cm thick? What's the total of their target areas? The book says to ignore overlap, so all nuclei are fully visible to the shooter.
Let's see if someone else steps up to review the next step Ameen presented for review:
probability of hitting some nucleus when shooting at metal foil.
How many nuclei in 1 cm^2 of foil, if it were 0.1 cm thick? What's the total of their target areas? The book says to ignore overlap, so all nuclei are fully visible to the shooter.
All models are wrong; some models are useful.  George Box
 Dennis P Brown
 Posts: 1854
 Joined: Sun May 20, 2012 2:46 pm
 Real name: Dennis P Brown
 Location: Glen Arm, MD
Re: Answer keys w/ explanation?
A hint: consider a single layer of atoms (each atom is about 2.7 angstrom in diameter as you calculated) and view these as circular disks. You know that a nucleus is 104 the size of this atom so the nuclear disk size is what? To get 1% coverage, one needs 1% of the total area desired (assume 1 cm2) being nuclei so simply calculate the number of these tiny disks to get 1% of 1 cm2 (hint again: it is a very big number.) Since one can only use more layers to achieve this then how many layers provide that number of nuclei ? The answer is obvious here (I am essentially giving it.)

 Posts: 93
 Joined: Sun Apr 15, 2018 7:33 pm
 Real name: Ameen Aydan
Re: Answer keys w/ explanation?
The method i'm using now is off of a website that talks about cross section and scattering. On there, there is an equation that represents this. What I thought of doing was rearranging it so i]I can solve for the thickness. Here is the website:
The original equation is:
Rs/Ri = Na L ρ σ/A 10^3 Kg
When rearranged, I obtain the equation:
L = Rs A 10^3 Kg/ Ri Na ρ σ
I'm not completely sure this is correct if anyone would like to double check. The value I used are as follow:
Rs (scattered particles) = 1
Ri (particle current) = 100
Na (Avogadro's number) = 6.022 x 10^23
A (atomic mass) = 181
ρ (density of the material) = 16g/cm^3
σ (cross section) = 2.26 x 10^24 cm^2
For the sigma value, I wasn't sure weather to use barns or the area of one atom, so I used the area since the first equation on the website was σ = πr^2
at the end my value was 8.31 x 10^5 = L or T. I can't tell weather it is metres or centimetres because on the website they use the MKS system and no the CGS system.
As soon as this is confirmed, I will attempt your method Denis.
Ameen Aydan
The original equation is:
Rs/Ri = Na L ρ σ/A 10^3 Kg
When rearranged, I obtain the equation:
L = Rs A 10^3 Kg/ Ri Na ρ σ
I'm not completely sure this is correct if anyone would like to double check. The value I used are as follow:
Rs (scattered particles) = 1
Ri (particle current) = 100
Na (Avogadro's number) = 6.022 x 10^23
A (atomic mass) = 181
ρ (density of the material) = 16g/cm^3
σ (cross section) = 2.26 x 10^24 cm^2
For the sigma value, I wasn't sure weather to use barns or the area of one atom, so I used the area since the first equation on the website was σ = πr^2
at the end my value was 8.31 x 10^5 = L or T. I can't tell weather it is metres or centimetres because on the website they use the MKS system and no the CGS system.
As soon as this is confirmed, I will attempt your method Denis.
Ameen Aydan

 Posts: 93
 Joined: Sun Apr 15, 2018 7:33 pm
 Real name: Ameen Aydan
Re: Answer keys w/ explanation?
the website is
http://hyperphysics.phyastr.gsu.edu/hb ... ec.html#c1
sorry for some reason it wasn't loading. The equation is:
http://hyperphysics.phyastr.gsu.edu/hb ... ec.html#c1
sorry for some reason it wasn't loading. The equation is:
 Rich Feldman
 Posts: 1308
 Joined: Mon Dec 21, 2009 11:59 pm
 Real name: Rich Feldman
 Location: Santa Clara County, CA, USA
Re: Answer keys w/ explanation?
Ameen, you're inviting confusion by trying to apply an allstepsinone formula that you don't understand.
Andrew and Dennis and I are walking you through a more stepbystep approach, based on fundamentals.
When you get the answer that way, then you can confidently take apart the Rs/Ri formula from hyperphysics.
For example, you just learned how to get atoms/cm^3 from grams/cm^3, grams/mole, and atoms/mole.
You know the size of a Ta nucleus, so you can get its side view (disk) area in cm^2. And in m^2 and barns, for exercise.
You know how many nuclei are in a 1 cm cube of metal.
If the nonnuclear material were invisible, the cube would cast a squareshaped partial shadow, because some of the light (figuratively) is blocked by nuclei. How faint is the shadow?
Halliday's reasonable directive to "ignore overlap" means the square partial shadow is made up of N ittybitty nucleus shadows. If they collectively occupy more than 1% of the square, then going through a 1 cm thickness of material was too much.
Andrew and Dennis and I are walking you through a more stepbystep approach, based on fundamentals.
When you get the answer that way, then you can confidently take apart the Rs/Ri formula from hyperphysics.
For example, you just learned how to get atoms/cm^3 from grams/cm^3, grams/mole, and atoms/mole.
You know the size of a Ta nucleus, so you can get its side view (disk) area in cm^2. And in m^2 and barns, for exercise.
You know how many nuclei are in a 1 cm cube of metal.
If the nonnuclear material were invisible, the cube would cast a squareshaped partial shadow, because some of the light (figuratively) is blocked by nuclei. How faint is the shadow?
Halliday's reasonable directive to "ignore overlap" means the square partial shadow is made up of N ittybitty nucleus shadows. If they collectively occupy more than 1% of the square, then going through a 1 cm thickness of material was too much.
All models are wrong; some models are useful.  George Box

 Posts: 93
 Joined: Sun Apr 15, 2018 7:33 pm
 Real name: Ameen Aydan
Re: Answer keys w/ explanation?
Hello,
Sorry about the late reply I had family things to attend to, it is the summer after all.
After considering what Rich and Dennis said I crunched some numbers.
1.0% of 1 cm^2 is 0.01 cm^2. From what I understand you want me to find how many nuclei fit in the 0.01 cm^2 which can also be written as 10^2 cm^2
After calculating the area of one nuclei with a radius of 8.48 x 10^13 cm, I got 2.26 x 10^24 cm^2
Now I divide the 1.0% area by the area of the nuclei:
10^2 cm^2 / 2.26 x 10^24 cm^2 = 4.42 x 10^21 nuclei per 10^2 cm^2 or 1.0% of 1 cm^2
Now I divide the value attained by how many tantalum atoms there are in on cm^2 (which is basically how many atoms in one cm squared):
4.42 x 10^21 / 1.41 x 10^15 = 3.13 x 10^6
This number as I would imagine represents the number of layers of atoms needed to make the foil, so now I multiply the diameter of this number by the diameter of the atom (including electrons).
(3.31 x 10^6) x (2.66 x 10^8 cm) = 8.33 x 10^2 cm or 0.833 mm thickness
I believe I have finished the question and am confident that this is the right answer!
Ameen Aydan
Sorry about the late reply I had family things to attend to, it is the summer after all.
After considering what Rich and Dennis said I crunched some numbers.
1.0% of 1 cm^2 is 0.01 cm^2. From what I understand you want me to find how many nuclei fit in the 0.01 cm^2 which can also be written as 10^2 cm^2
After calculating the area of one nuclei with a radius of 8.48 x 10^13 cm, I got 2.26 x 10^24 cm^2
Now I divide the 1.0% area by the area of the nuclei:
10^2 cm^2 / 2.26 x 10^24 cm^2 = 4.42 x 10^21 nuclei per 10^2 cm^2 or 1.0% of 1 cm^2
Now I divide the value attained by how many tantalum atoms there are in on cm^2 (which is basically how many atoms in one cm squared):
4.42 x 10^21 / 1.41 x 10^15 = 3.13 x 10^6
This number as I would imagine represents the number of layers of atoms needed to make the foil, so now I multiply the diameter of this number by the diameter of the atom (including electrons).
(3.31 x 10^6) x (2.66 x 10^8 cm) = 8.33 x 10^2 cm or 0.833 mm thickness
I believe I have finished the question and am confident that this is the right answer!
Ameen Aydan
 Rich Feldman
 Posts: 1308
 Joined: Mon Dec 21, 2009 11:59 pm
 Real name: Rich Feldman
 Location: Santa Clara County, CA, USA
Re: Answer keys w/ explanation?
Full credit for that one. At least, your bottom line matches mine from last week.
[edit] The step about layers is handy, and leads to the right answer. But it's not necessary. It might reinforce the usually wrong idea that atoms really do crystallize in neat xyz cubic lattices. My suggestion to look at cube root of atoms/cm^3 number, was to give a rough idea of atom spacing and size, as a sanity check. More about that another day.
Back to blocking. You correctly figured how many nuclei you need to block 0.01 cm^2  let''s call the number X. We can also go back to your correct computation of N, the number of nuclei per cm^3 of solid metal, and not bother taking its cube root. Suppose the thickness of your tantalum sheet is t, from front to back. Do you see that each square cm of the sheet has a metal volume of t cm^3, and contains t * N nuclei, no matter how the atoms are microscopically arranged? Then we want t * N = X; solve for t. [\edit]
On your more general topic, explanations, here is a suggestion for working physics problems.
Spreadsheet calculator programs are very handy for performing & documenting the calculations.
Especially if you ever want to repeat them with different inputs.
As an example, here's the worksheet I made for this exercise.
The inputs are entered in blueshaded cells.
All other numbers are computed by simple onestep formulas in their cells. Unfortunately, to see a formula you need to open the spreadsheet program and select the cell.
I manually changed some text to green, for some of the intermediate answers that match yours. Spreadsheet or not, I think all unit conversions (such as between grams and kg, or Hz and MHz) should be kept outside of the formulas that handle the main work. One formula you quoted from hyperphysics was unacceptably sloppy in that regard, inviting confusion. When I'm working with practical units like nH or picoseconds, and need to apply familiar electronic formulas, I make separate cells for the values in SI units. Cells are cheap.
More columns were added later, because of an interest in nucleus sizes for the first time in decades. First, to do tantalum with a different r0 value. Then to do some other elements. Whole columns of inputs and formulas were copied and pasted from the first column.
[edit] The step about layers is handy, and leads to the right answer. But it's not necessary. It might reinforce the usually wrong idea that atoms really do crystallize in neat xyz cubic lattices. My suggestion to look at cube root of atoms/cm^3 number, was to give a rough idea of atom spacing and size, as a sanity check. More about that another day.
Back to blocking. You correctly figured how many nuclei you need to block 0.01 cm^2  let''s call the number X. We can also go back to your correct computation of N, the number of nuclei per cm^3 of solid metal, and not bother taking its cube root. Suppose the thickness of your tantalum sheet is t, from front to back. Do you see that each square cm of the sheet has a metal volume of t cm^3, and contains t * N nuclei, no matter how the atoms are microscopically arranged? Then we want t * N = X; solve for t. [\edit]
On your more general topic, explanations, here is a suggestion for working physics problems.
Spreadsheet calculator programs are very handy for performing & documenting the calculations.
Especially if you ever want to repeat them with different inputs.
As an example, here's the worksheet I made for this exercise.
The inputs are entered in blueshaded cells.
All other numbers are computed by simple onestep formulas in their cells. Unfortunately, to see a formula you need to open the spreadsheet program and select the cell.
I manually changed some text to green, for some of the intermediate answers that match yours. Spreadsheet or not, I think all unit conversions (such as between grams and kg, or Hz and MHz) should be kept outside of the formulas that handle the main work. One formula you quoted from hyperphysics was unacceptably sloppy in that regard, inviting confusion. When I'm working with practical units like nH or picoseconds, and need to apply familiar electronic formulas, I make separate cells for the values in SI units. Cells are cheap.
More columns were added later, because of an interest in nucleus sizes for the first time in decades. First, to do tantalum with a different r0 value. Then to do some other elements. Whole columns of inputs and formulas were copied and pasted from the first column.
All models are wrong; some models are useful.  George Box

 Posts: 93
 Joined: Sun Apr 15, 2018 7:33 pm
 Real name: Ameen Aydan
Re: Answer keys w/ explanation?
Hello again,
While reading the two books, I always came across these very weird values that were basically inverse. Some examples are sec^1 or cm^3 or even K^1. My closest guess is that it corresponds to a smaller value such as cm^3 is really just mm^3 or that sec^1 is just millisecond. I'm not sure if this is really right or not so can anybody tell me? I tried looking for it on the internet and came across definition that said it represented wave function such as hertz or something that wasn't very helpful. Are there any links or resources you can provide that may explain this to me?
Thank You,
Ameen Aydan
While reading the two books, I always came across these very weird values that were basically inverse. Some examples are sec^1 or cm^3 or even K^1. My closest guess is that it corresponds to a smaller value such as cm^3 is really just mm^3 or that sec^1 is just millisecond. I'm not sure if this is really right or not so can anybody tell me? I tried looking for it on the internet and came across definition that said it represented wave function such as hertz or something that wasn't very helpful. Are there any links or resources you can provide that may explain this to me?
Thank You,
Ameen Aydan
 Richard Hull
 Moderator
 Posts: 12204
 Joined: Fri Jun 15, 2001 1:44 pm
 Real name: Richard Hull
Re: Answer keys w/ explanation?
I always despised this notation, but it creeps into a lot of scientific math.... sec^1 means " per sec" and cm^3 means "per cubic centimeter"....Sadly, after a while you get used to it.
Richard Hull
Richard Hull
Progress may have been a good thing once, but it just went on too long.  Yogi Berra
Fusion is the energy of the future....and it always will be
Retired now...Doing only what I want and not what I should...every day is a saturday.
Fusion is the energy of the future....and it always will be
Retired now...Doing only what I want and not what I should...every day is a saturday.