GREPhysics.NET: GR9277 #95

GR9277 #95

Problem

GREPhysics.NET Official Solution

\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?

$10^{-24}$
$10^{-25}$
$10^{-26}$
$10^{-27}$
$10^{-28}$

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

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.

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Comments

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.

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Imperate
2008-10-17 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
2008-10-17 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.

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student2008
2008-10-16 14:29:27

nyuko
2009-10-30 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...

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ric
2006-11-25 03:30:21

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

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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

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ric
2006-11-25 03:25:31

T=N_B \,n_T\,\sigma

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ric
2006-11-25 03:06:36

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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.

kolahalb
2007-11-11 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

flyboy621
2010-10-23 05:09:00

Looks like Ric has it right.

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Andresito
2006-03-30 12:22:52

(1/10^20) is incorrect.

The correct ratio is (1/10^12)

grae313
2007-11-02 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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Post A Comment!

You are replying to:

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

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