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GR9277 #95
Problem
 GREPhysics.NET Official Solution Alternate Solutions
\prob{95}
A beam of $10^{12}$ protons per second is incident on a target containing $10^{20}$ nuclei per square cm. At an angle of 10 degrees, there are $10^2$ protons per second elastically scattered into a detector that subtends a solid angle of $10^{-4}$ steradians. What is the differential elastic scattering cross section, in units of sq cm per steradian?

1. $10^{-24}$
2. $10^{-25}$
3. $10^{-26}$
4. $10^{-27}$
5. $10^{-28}$

Advanced Topics$\Rightarrow$}Dimensional Analysis

The final units must be $cm^2/steradians$. One is given

$10^{12}protons/s$

$10^{20}nuclei/cm^2$

$10^2protons/s$

$10^{-4}steradians$

The combination $(10^2/10^{20})(1/10^{20})(1/10^{-4})$ gives the right units as well as answer choice (C).

Alternate Solutions
 student20082008-10-16 14:29:27 One can answer the question without solving it, recalling that the typical nuclear radius is femtometer, i.e. $10^{-13}$ sm. Thus, the typical nuclear cross section $\sigma\sim(1\,fm)^2\sim10^{-26} {sm}^2$. 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.Reply to this comment
Fily
2011-03-18 09:06:10
Of course answer above is correct,but one correction is required,(10E2/10E12)(1/10E20)(1/10E-4) will give 10E-26.
Imperate
2008-10-17 06:42:18
It's obvious that the "useful area" (the area that will actually scatter if hit) is $n \sigma A$ 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 $\frac{ n \sigma A}{A} =n \sigma$. Therefor if $N_I$ is the number of incident protons and $N_S$ is the number of scattered protons. One has $N_S=n \sigma N_I$ which is gives the total cross section as $\sigma=\frac{N_S}{n N_I}$. Now to get differential cross section (per steradian) we just divide by the solid angle provided $\sigma=\frac{N_S}{n N_I \Omega}$. Plug in the numbers.
 Imperate2008-10-17 06:45:21 oops.... It's obvious that the "useful area" (the area that will actually scatter if hit) is $n \sigma A$ 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 $\frac{ n \sigma A}{A} =n \sigma$ Therefore if $N_I$ is the number of incident protons and $N_S$is the number of scattered protons. One has $N_S=n \sigma N_I$ which is gives the total cross section as $\sigma=\frac{N_S}{n N_I}$.Now to get differential cross section (per steradian) we just divide by the solid angle provided $\sigma=\frac{N_S}{n N_I \Omega}$Plug in the numbers.
student2008
2008-10-16 14:29:27
One can answer the question without solving it, recalling that the typical nuclear radius is femtometer, i.e. $10^{-13}$ sm. Thus, the typical nuclear cross section $\sigma\sim(1\,fm)^2\sim10^{-26} {sm}^2$. 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.
 nyuko2009-10-30 22:58:05 Sorry...I guess you have a typo, should be $cm^2$ not $sm^2$ ? 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 $10^1$ 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
2006-11-25 03:30:21
I think that the solution should sound like this.

Call $T$ the number of scattered events (protons in this case) per second. We are given the "superficial" density of the target $n_T=10^{20}$ protons/$cm^2$ and the number of protons incident per second $N_B=10^{12}$. Calling $\sigma$ the total cross section we have

T=N_B \,n_T\,\sigma.

We are then given the value $T=10^2$ protons per second for a solid angle $\Delta\Omega=10^{-4}$ sterad.

Thus the differential cross section is

$\frac{\sigma}{\Delta\omega}=\frac{T}{N_B\,n_T\,Delta\Omega}=10^{-26}$ cm$^2$/sterad

ric
2006-11-25 03:27:08
I think that the solution should sound like this.

Call $T$ the number of scattered events (protons in this case) per second. We are given the "superficial" density of the target $n_T=10^{20}$ protons/$cm^2$ and the number of protons incident per second $N_B$. Calling $\sigma$ the total cross section we have

$T=N_B \,n_T\,\sigma$.

We are then given the value $T=10^2$ protons per second for a solid angle $\Delta\Omega=10^{-4}$ sterad.

Thus the differential cross section is

$\frac{\sigma}{\Delta\Omega}=\frac{T}{N_B\,n_T\,\Delta\Omega}=10^{26}$ cm$^2$/sterad

ric
2006-11-25 03:25:31
I think that the solution should sound like this.

Call $T$ the number of scattered events (protons in this case) per second. We are given the "superficial" density of the target $n_T=10^{20}$ protons/$cm^2$ and the number of protons incident per second $N_B$. Calling $\sigma$ the total cross section we have

T=N_B \,n_T\,\sigma.

We are then given the value $T=10^2$ protons per second for a solid angle $\Delta\Omega=10^{-4}$ sterad.

Thus the differential cross section is

$\frac{\sigma}{\Delta\omega}=\frac{T}{N_B\,n_T\,Delta\Omega}=10^{26}$ cm$^2$/sterad

ric
2006-11-25 03:06:36
kevglynn
2006-10-17 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.
 kolahalb2007-11-11 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
 flyboy6212010-10-23 05:09:00 Looks like Ric has it right.
Andresito
2006-03-30 12:22:52
(1/10^20) is incorrect.

The correct ratio is (1/10^12)
 grae3132007-11-02 21:35:56 Yes, this typo should definitely be fixed! The answer should have a $10^{12}$ in the denominator, not two $10^{20}$'s Thank you, Yosun

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