Answer keys w/ explanation?

 Posts: 37
 Joined: Sun Apr 15, 2018 7:33 pm
 Real name: Ameen Aydan
Answer keys w/ explanation?
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.
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.

 Posts: 37
 Joined: Sun Apr 15, 2018 7:33 pm
 Real name: Ameen Aydan
Re: Answer keys w/ explanation?
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?
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?
This is the rutherford scattering problem:
http://hyperphysics.phyastr.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 ... fnucleus/
http://hyperphysics.phyastr.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 ... fnucleus/
Andrew Seltzman
www.rtftechnologies.org
www.rtftechnologies.org
 Rich Feldman
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 Real name: Rich Feldman
 Location: Santa Clara County, CA, USA
Re: Answer keys w/ explanation?
Great idea for using the forum.
Let's just beware of lazy students asking other people to solve their homework problems.
Let's just beware of lazy students asking other people to solve their homework problems.
Richard Feldman

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Re: Answer keys w/ explanation?
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
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
 Richard Hull
 Site Admin
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 Real name: Richard Hull
Re: Answer keys w/ explanation?
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
Richard Hull
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
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.
 Rich Feldman
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 Joined: Mon Dec 21, 2009 11:59 pm
 Real name: Rich Feldman
 Location: Santa Clara County, CA, USA
Re: Answer keys w/ explanation?
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.
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.
Richard Feldman

 Posts: 37
 Joined: Sun Apr 15, 2018 7:33 pm
 Real name: Ameen Aydan
Re: Answer keys w/ explanation?
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.
Ameen Aydan
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.
Ameen Aydan
 Rich Feldman
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 Joined: Mon Dec 21, 2009 11:59 pm
 Real name: Rich Feldman
 Location: Santa Clara County, CA, USA
Re: Answer keys w/ explanation?
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 1.2E13). That would make the radius 6.79E13, and volume 1.31E36, 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.nuclearpower.net/nuclearp ... rdensity/ 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?
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 1.2E13). That would make the radius 6.79E13, and volume 1.31E36, 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.nuclearpower.net/nuclearp ... rdensity/ 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?
Richard Feldman

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Re: Answer keys w/ explanation?
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.
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.