GR9677 #19



Alternate Solutions 
fjornada 20091019 10:11:53  Yet another solution with energy conversion (using atomic systems). If you are used to working with atomic physics, it'll be quite quick, otherwise forget this post.
In atomic systems, energy is given by Hartree, distance by Bohr, and the best, .
One Hartree is twice the ground state of the H atom, so divide 5 Mev by 27 eV and you'll get the kinetic energy in a.u. Also, in a.u. .
Now make T=V and you'll find r ~ 5 E4 Bohrs. One Bohr is the classical electron orbit (~ 5E11 m). A rough guess around one Angstrom (1E10 m) works just fine.
  wittensdog 20091016 18:32:09  There is a way to do this without ever making explicit reference to the conversion between electronvolts and joules. Remember that an electronvolt is exactly what it sounds like, the electron charge multiplied by one volt. If you divide 1 eV by the electron charge, you have one volt, which is very nice because that is just 1 N * m / C , which is something that has a value of 1 in SI units.
Now, due to the col0umb potential, you're going to have a term of the form,
100 e^2 / ( 4 * pi * eps_0 * r )
on one side, and a term of the form,
E = 5 * 10 ^6 eV
on the other. So, what you can do is just divide by the electron charge, and get,
5 * 10^6 V = 100 e / (4 * pi * eps_0 * r)
The electron charge and eps_o are given on the front sheet, so now you just plug in numbers.
By dividing out the electron charge, you're basically working with electric potential, and not potential energy.   kcme 20070617 10:25:20  I would rather solve this with dimensional analysis. No calculations necessary
You know to be backscattered at such a large angle, the distance of closest approach is going to be very small, on the order of the size of the nucleus (). That eliminates choices C, D, and E.
So what about A? Well, is somewhere between 3 and 4, so at most choice A is something like 1.22*4 fm, or about 5 fm. That's smaller than a big nucleus like silver.
That leaves choice B, which at 29 fm strikes me as quite reasonable  

Comments 
arturodonjuan 20161017 16:50:10  There\'s a really nifty alternative to just \"plugging in the constants\" from the front page, which is all quite complicated. This requires remembering and , where is the finestructure constant.\r\n\r\nThe equation we\'re solving, which you should be able to write down in a millisecond, is (by conservation of energy),\r\n\r\n\r\n\r\nor, moving to the other side,\r\n\r\n\r\n\r\nas you can see, has popped out,\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\nThis is a hell of a lot easier to deal with.
arturodonjuan 20161017 16:51:15 
Oh my f***ing gosh this doesn\'t display at all like the preview did.

  pam d 20110911 19:21:18  Okay so the scattering angle of 180 degrees was just a bunch of smoke and mirrors hiding a simple conservation of energy freebie. FFFFFFFUUUUUUUUUUUU
justin_l 20131018 00:26:21 
well technically the 180 degrees was what told you you could directly apply conservation of energy. It means it was a headon collision.

  walczyk 20110403 02:22:02  You all should remember J/eV is which is equal to the charge of an electron q!! That should make this problem SO easy if you remember k is nearly   Phys4 20100803 16:50:03  If you set up the problem correctly like Yosun does, just do an order of magnitude calculation. You will get 14 and one can plainly see the other solutions have at least two orders of magnitude off. The small coefficients indicate that one does not have to worry about them influencing the order of magnitude.   fjornada 20091019 10:11:53  Yet another solution with energy conversion (using atomic systems). If you are used to working with atomic physics, it'll be quite quick, otherwise forget this post.
In atomic systems, energy is given by Hartree, distance by Bohr, and the best, .
One Hartree is twice the ground state of the H atom, so divide 5 Mev by 27 eV and you'll get the kinetic energy in a.u. Also, in a.u. .
Now make T=V and you'll find r ~ 5 E4 Bohrs. One Bohr is the classical electron orbit (~ 5E11 m). A rough guess around one Angstrom (1E10 m) works just fine.
  wittensdog 20091016 18:32:09  There is a way to do this without ever making explicit reference to the conversion between electronvolts and joules. Remember that an electronvolt is exactly what it sounds like, the electron charge multiplied by one volt. If you divide 1 eV by the electron charge, you have one volt, which is very nice because that is just 1 N * m / C , which is something that has a value of 1 in SI units.
Now, due to the col0umb potential, you're going to have a term of the form,
100 e^2 / ( 4 * pi * eps_0 * r )
on one side, and a term of the form,
E = 5 * 10 ^6 eV
on the other. So, what you can do is just divide by the electron charge, and get,
5 * 10^6 V = 100 e / (4 * pi * eps_0 * r)
The electron charge and eps_o are given on the front sheet, so now you just plug in numbers.
By dividing out the electron charge, you're basically working with electric potential, and not potential energy.
wittensdog 20091016 18:43:46 
So I guess the broader moral is that whenever you have an expression that has the electron charge, and also units of eV in it, you can pull the electron charge out from the term " eV " in order to make things simpler.

flyboy621 20101023 06:54:13 
This is the best solution.

dmerthe 20111109 22:08:03 
I like this solution. Simple and easy to remember.

  physicsisgod 20081028 15:21:50  Yosun's solution is right, but if you're like me and want to know the physics behind this, check this out:
http://hyperphysics.phyastr.gsu.edu/hbase/nuclear/rutsca2.html#c5
Maybe the other commenters already knew this, but this question is asking about Rutherford scattering, which is actually much more detailed and interesting than the solution suggests. ETS was just nice and gave us a 180 degree scattering angle, which simplifies things quite a bit. In fact, if you recognize this as Rutherford scattering at 180 degrees, you can just memorize the formula when the angle is 180 degrees,. and get the right answer exactly.   glkjap2 20080611 01:24:35  Quick way: The first thing we recall is that m_{\alpha}>>T so we can say . Alpha particle has four nucleons each with approximately 940 MeV mass so we can approximate that . So we find that approximately . Plugging 2p into the uncertainty relation we have . Using we find that which is closest to answer B.   kcme 20070617 10:25:20  I would rather solve this with dimensional analysis. No calculations necessary
You know to be backscattered at such a large angle, the distance of closest approach is going to be very small, on the order of the size of the nucleus (). That eliminates choices C, D, and E.
So what about A? Well, is somewhere between 3 and 4, so at most choice A is something like 1.22*4 fm, or about 5 fm. That's smaller than a big nucleus like silver.
That leaves choice B, which at 29 fm strikes me as quite reasonable
Moush 20100929 19:09:21 
I don't see how simply being reflected 180 degrees means that an extremely close nuclear approach is involved. Sure, if << then it's more likely that it gets deflected < 180 degrees. But if < or ~/>/>> then 180 deg reflections are likely. Proximity will be inversely proportional to its speed and probability will be directly proportional to its speed.
I guess getting to know ETS is part of the problem, knowing when to assume and how much one can without much worry...

  cyberdeathreaper 20070107 21:25:35  I second that comment by scottopoly  this seems like a very involved calculation to be doing by hand (translation = it seems timeconsuming). Are there any shortcuts in the calculation to consider? Or, perhaps another approach?
yosun 20070222 19:04:19 
Well, it's not very time consuming, since we are just solving for r... From baby's basic math, you should get . Note that in the approx, I did not square the q, since one of the q's cancel in the conversion ... The answer comes out to the right order, which is enough for this question.

  scottopoly 20061103 15:49:25  ah yes, the old "convert to SI".
Blake7 20070922 22:06:35 
eV to Joules and back is given no less than TWICE in the Table of Information at the beginning of the test.
You should easily see it in Planck's constant
h = 6.63E34 Joule*second = 4.14E15 eV* second
(6.63E34Joule = 4.14E15 eV)
and also again in the magnitude of the electron charge
e = 1.6o218 E19 Coulomb
(from the definition of "Electron Volt")
Drill those a couple of times and they'll be right there for you.

 




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