GR9277 #95



Alternate Solutions 
student2008 20081016 14:29:27  One can answer the question without solving it, recalling that the typical nuclear radius is femtometer, i.e. sm. Thus, the typical nuclear cross section . Actually, this is not a solution, but if you lose your way during exam, it might help:) And I found (C) exactly this way before using the Yosun's solution.  

Comments 
Fily 20110318 09:06:10  Of course answer above is correct,but one correction is required,(10E2/10E12)(1/10E20)(1/10E4) will give 10E26.   Imperate 20081017 06:42:18  It's obvious that the "useful area" (the area that will actually scatter if hit) is [latex] n \sigma A [/latex] were n is the number of target atoms per unit area, and A is the area (in some other GRE questions one has number density per volume and the thickness too but that's not the case here). The scattering probability is thus [latex] \frac{ n \sigma A}{A} =n \sigma[/latex]. Therefor if [latex] N_I [/latex] is the number of incident protons and [latex] N_S [/latex] is the number of scattered protons. One has [latex]N_S=n \sigma N_I [/latex] which is gives the total cross section as [latex] \sigma=\frac{N_S}{n N_I} [/latex]. Now to get differential cross section (per steradian) we just divide by the solid angle provided [latex] \sigma=\frac{N_S}{n N_I \Omega} [/latex]. Plug in the numbers.
Imperate 20081017 06:45:21 
oops....
It's obvious that the "useful area" (the area that will actually scatter if hit) is were n is the number of target atoms per unit area, and A is the area (in some other GRE questions one has number density per volume and the thickness too but that's not the case here). The scattering probability is thus Therefore if is the number of incident protons and is the number of scattered protons. One has which is gives the total cross section as .Now to get differential cross section (per steradian) we just divide by the solid angle provided Plug in the numbers.

  student2008 20081016 14:29:27  One can answer the question without solving it, recalling that the typical nuclear radius is femtometer, i.e. sm. Thus, the typical nuclear cross section . Actually, this is not a solution, but if you lose your way during exam, it might help:) And I found (C) exactly this way before using the Yosun's solution.
nyuko 20091030 22:58:05 
Sorry...I guess you have a typo, should be not ?
by the way, I feel a little uncomfortable with your method because just a factor of 3 or 4 in length, after taking square, you would get an extra in the area. And for this problem, an order difference would lead to another answer.
What's more, those are charged particles (well, and not only coulomb interactions can take part; there are many others...). It is possible to have a cross section much different from the particle size...

  ric 20061125 03:30:21  I think that the solution should sound like this.
Call the number of scattered events (protons in this case) per second. We are given the "superficial" density of the target protons/ and the number of protons incident per second . Calling the total cross section we have
T=N_B \,n_T\,\sigma.
We are then given the value protons per second for a solid angle sterad.
Thus the differential cross section is
cm/sterad
  ric 20061125 03:27:08  I think that the solution should sound like this.
Call the number of scattered events (protons in this case) per second. We are given the "superficial" density of the target protons/ and the number of protons incident per second . Calling the total cross section we have
.
We are then given the value protons per second for a solid angle sterad.
Thus the differential cross section is
cm/sterad
  ric 20061125 03:25:31  I think that the solution should sound like this.
Call the number of scattered events (protons in this case) per second. We are given the "superficial" density of the target protons/ and the number of protons incident per second . Calling the total cross section we have
T=N_B \,n_T\,\sigma.
We are then given the value protons per second for a solid angle sterad.
Thus the differential cross section is
cm/sterad
  ric 20061125 03:06:36    kevglynn 20061017 19:35:32  I hate to be a pain, but does anyone have a sound physics solution to this problem? I have found a couple of relations for the differential scattering cross section, but they haven't been very helpful.
kolahalb 20071111 07:53:06 
I do not intend to show ric's job is not good.But,this can be done by dimensional analysis and simple reasonings...
We must have cm^2/steradian.So,directly hit:
(1/(10^20))(1/(10^4))=(10^16) cm^2/steradian
Now,We must multiply the ratio between number of incident (I) and scattered (S) protons to (10^16) cm^2/steradian.Otherwise the units will not match.The fraction may be (I/S) or,(S/I).We do not know.
If it is (I/S),note that the diameter of the scattering nucleus becomes rougly 10^3 cm!!!
[As (pi*(r^2))~(10^6)]
The other option is to multiply (10^16) by (S/I) which preserves the units and gives a correct estimate of nulear dimension:
(10^13) cm=(10^15) m

flyboy621 20101023 05:09:00 
Looks like Ric has it right.

  Andresito 20060330 12:22:52  (1/10^20) is incorrect.
The correct ratio is (1/10^12)
grae313 20071102 21:35:56 
Yes, this typo should definitely be fixed! The answer should have a in the denominator, not two 's
Thank you, Yosun

 

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