Answer keys w/ explanation?

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Ameen Aydan
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Answer keys w/ explanation?

Post by Ameen Aydan »

hello,

I just received my books Nuclear Radiation Physics by Ralph E. Lapp and introduction to Nuclear Physics by David Halliday as well as some inherited books like Principles of Electronic Circuits or Radiologic Science for Technologists.

looking through the books, it is very hard to grasp for a 14 year old to say the least and i'm having some trouble understanding even basic concepts. But at the end of they day, I manage. Now the place i'm having most difficulty is the problems at the end of the chapters, namely the nuclear physics ones. The problem is that the first book (Lapp) has the answers, but no explanation as to how I can get there (the answer). With the second book (Halliday), it has questions, but no answers? It's very hard to find the answers on the internet if not impossible most often.

So I thought, we can collectively answer them together, helping one another. Basically you post your question and someone strolling by will answer! Fun!

All I ask is that you put down the book its from, what edition it is, the page number, and of course, the question.
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Re: Answer keys w/ explanation?

Post by Ameen Aydan »

Just to start this rolling, I will ask the first question from the book Introductory Nuclear Physics by David Halliday 2nd edition.

The question is from page 22 of the first chapter Basic Nuclear Concepts

A Tantalum foil (A=181) has 1.0% of its projected area blocked out by nuclei. How thick is the foil? Assume no over overlapping. The density of tantalum is 16 g/cm3 (cubed).

The part i'm stuck on is when it says 1.0% of its area is blocked out by nuclei. What does that mean?
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Re: Answer keys w/ explanation?

Post by Andrew Seltzman »

This is the rutherford scattering problem:
http://hyperphysics.phy-astr.gsu.edu/hb ... escat.html

It means assume the atom is free space except for the nucleus which is solid sphere. Calculate the radius of the nucleii and thus the area each nucleii is blocking. Find how thick a foil is required to have nucleii block 1% of the area of the foil.

Watch this video:
https://ocw.mit.edu/courses/chemistry/5 ... f-nucleus/
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Rich Feldman
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Re: Answer keys w/ explanation?

Post by Rich Feldman »

Great idea for using the forum.
Let's just beware of lazy students asking other people to solve their homework problems. :-)
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Re: Answer keys w/ explanation?

Post by John Futter »

Rich
if all is right who would ask this question of a 14 year old
ammen is learning well done and help will come

Ameen do not take us for a ride
the gods might become very angry
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Richard Hull
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Re: Answer keys w/ explanation?

Post by Richard Hull »

At 14 it is tough to think like a college grad trained in such things extant within nuclear physics. Learning never ends, but a firm foundation helps a lot and at 14, it typically just isn't there yet unless incredibly gifted. Many of the incredibly gifted can be somewhat self taught by 14.

Word problems in physics, as presented here, need a foundation to interpret them or a class using the book in question with a good teacher to help build that foundation.

For those who would listen to those wise men who came before us, I offer two essays on studies. the famous Francis Bacon essay, followed by one written by Samuel Johnson. I have always cherished them. They are both are found at the following URL

http://www.psy.gla.ac.uk/~steve/best/BaconJohnson.pdf

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Rich Feldman
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Re: Answer keys w/ explanation?

Post by Rich Feldman »

Oops! I sincerely apologize for a sentence that's been misinterpreted, Ameen.
That homework comment wasn't aimed at _your_ question!

Your excellent cover letter, and proposal, clearly justified presenting a problem out of a textbook.
On other forums, not so much here, we sometimes get people seeking homework solutions without saying so.

The tantalum foil question taught me something technical.
We know that from A=181 and the bulk density, and a familiar physics constant, we can figure the numerical density of atoms. On the order of 10^22 per cm^3 for most solid elements, IIRC.
Then Internet searching, instead of a forum, explained that we can figure the nucleus size from A and another physics constant: the typical density of atomic nuclei. Does Halliday's book mention "liquid drop model" ? When you get the area that's geometrically blocked by one tantalum nucleus, in cm^2, you can also tell us what the value is in barns. :-)

p.s. Can someone in the trade tell us if SI basic units (aka MKS) are displacing centimeters and grams, in nuclear science and engineering work? The SI value for density of tantalum is 16,000 kg per m^3. The editors of hyperphysics site (e.g. link given above by Andrew) apparently have chosen to use meters and kilograms.
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Re: Answer keys w/ explanation?

Post by Ameen Aydan »

Hello

It is true that I am not very gifted, the most I've ever done was a fast track through my grade 8 math course straight to grade 9 math. I would imagine that it is not that big of a deal in the fusor community. This summer I am devoting my time to studying the knowledge needed for a fusion reactor. I am taking online courses in hopes of learning university grade mathematics such as complex number and integrals since from my understanding, you need this foundation to make it here. Lot's of the children here that I have seen do not have a good background or any at all really. It is just that they have strong funding which paves the way for an easy build. But this is only my assumption so don't even consider what I said. And at the end of the day, though I may not be smart, I am a dreamer and a hard worker, which makes one hell of a good person for this type of project

As for the answer? Well, I tried my best:

The equation from the book r = r0 A^1/3 gives the radius of the nucleus where r0 is a constant that is equal to about 1.5 x 10^-13 cm

Solving for the radius: Solving for the volume gives us 2.55 x 10^-36 cm^3
r = (1.5 x 10^-13 cm)(3√181)
r = (1.5 x 10^-13 cm)(5.66)
Therefore r = 8.48 x10^-13 cm

Since the volume of the nucleus makes up only 1.0% of the foil than 99.0% is empty space.
(2.55 x 10^-36 cm^3)(1 x 10^2) solving for the radius of a nucleus with given radius
= 2.55 x 10^-34 → on the other side gives us r = 3.94 x 10^-12 cm

(3.94 x 10^-12)(2)
= 7.88 x 10^-12 cm

This means that the thickness is 7.88 x 10^-12 cm.

This, without a doubt, is completely incorrect. seeing that the question mentioned the density of tantalum, it clearly indicates that it must be used. The only is I have no idea where. Can anybody indicate where I went wrong in my calculations? This just goes to show how ungifted I am.

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Rich Feldman
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Re: Answer keys w/ explanation?

Post by Rich Feldman »

Don't be so modest, Ameen. You already know more about nuclear science than 99.9% of the world's 14yo's.

The equation from the book r = r0 A^1/3 gives the radius of the nucleus where r0 is a constant that is equal to about 1.5 x 10^-13 cm
Solving for the radius: Solving for the volume gives us 2.55 x 10^-36 cm^3
r = (1.5 x 10^-13 cm)(3√181)
r = (1.5 x 10^-13 cm)(5.66)
Therefore r = 8.48 x10^-13 cm


So far so good, including your answer for the volume of nucleus. Some other references, like the link from Andrew, give r0 as 1.2 x 10^-13 cm (a number often printed by computer programs as 1.2E-13). That would make the radius 6.79E-13, and volume 1.31E-36, for a very minor difference in the bottom line.

What is the area of a circular disk with the same radius as the tantalum nucleus?

Since the volume of the nucleus makes up only 1.0% of the foil than 99.0% is empty space.

That's where you went off track. You need a lot more zeros between the decimal point and the 1%. The meaning of the 1% in textbook problem will be clear after a couple more guided steps.

seeing that the question mentioned the density of tantalum, it clearly indicates that it must be used. The only is I have no idea where.

From that factor, and A, you can determine the number of tantalum atoms N per cubic centimeter of metal. Here's one reference:
https://www.nuclear-power.net/nuclear-p ... r-density/ The atomic mass, in grams per mole, is called M in their formula. Same number as A in this problem (proton + neutron count per atom).

What number do you get for N? Its cube root is the number of atoms per lineal cm, if they were in a regular cubic lattice. What's the corresponding distance between atoms, center to center?
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Re: Answer keys w/ explanation?

Post by Tom McCarthy »

It can be a lot of fun and rewarding (and tough!) to learn the theory behind the actual device we build. However, I wouldn't get too caught up in learning theory before actually making something. My Fusor isn't done yet, but one thing I'd do differently would be to get to work with my hands quicker, rather than waiting around to learn theorems and college info. That sort of knowledge is absolutely useful, but you'll pick it up and understand it better by building and making.

While the overriding sentiment today is that most technological advances are due to academia and building up from theory, an awful lot actually comes from tinkering and practical experimentation.

Anyways, you're awesome for taking this on.
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Re: Answer keys w/ explanation?

Post by Ameen Aydan »

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
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Rich Feldman
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Re: Answer keys w/ explanation?

Post by Rich Feldman »

First half looks correct, up to and including your value for nuclear diameter: 1.70E-12 cm (or 1.36E-12 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.
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Re: Answer keys w/ explanation?

Post by Dennis P Brown »

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 10-4 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.)
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Re: Answer keys w/ explanation?

Post by Ameen Aydan »

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:

Image

The original equation is:

Rs/Ri = Na L ρ σ/A 10^-3 Kg

When re-arranged, 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
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Re: Answer keys w/ explanation?

Post by Ameen Aydan »

the website is

http://hyperphysics.phy-astr.gsu.edu/hb ... ec.html#c1

sorry for some reason it wasn't loading. The equation is:

Image
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Re: Answer keys w/ explanation?

Post by Rich Feldman »

Ameen, you're inviting confusion by trying to apply an all-steps-in-one formula that you don't understand.

Andrew and Dennis and I are walking you through a more step-by-step 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 non-nuclear material were invisible, the cube would cast a square-shaped 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 itty-bitty nucleus shadows. If they collectively occupy more than 1% of the square, then going through a 1 cm thickness of material was too much.
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Re: Answer keys w/ explanation?

Post by Ameen Aydan »

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
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Re: Answer keys w/ explanation?

Post by Rich Feldman »

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 x-y-z 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 blue-shaded cells.
All other numbers are computed by simple one-step 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.
nuke2.PNG
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.
nukes.PNG
All models are wrong; some models are useful. -- George Box
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Re: Answer keys w/ explanation?

Post by Ameen Aydan »

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
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Re: Answer keys w/ explanation?

Post by Richard Hull »

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.

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