GREPhysics.NET: GR8677 #42

GR8677 #42

Problem

GREPhysics.NET Official Solution

Mechanics $\Rightarrow$ }Scattering Cross Section

The target nuclei per $cm^3$ is $\rho=N/V=1E20 nuclei/cm^3$ .

The scatterer thickness is $t=1E-1cm$ .

Passing through the scatterer, only $P=1E-6$ particles are scattered. (This is the probability of scattering.)

One can obtain to formula for the scattering cross section from ordinary dimensional analysis spiced up with some common sense (or recalling its definition back in Marion and Thornton's Dynamics book). It is:

$A = P\frac{V}{Nl}=\frac{P}{\rho l}=\frac{1E-6}{1E20 \times 1E-1}=10^{-6-20+1}=10^{-25}, <br />$

which is choice (C).

Alternate Solutions

eshaghoulian
2007-09-15 15:45:00

If you assume linearity in the variables and their inverses (and that there are no constants up front), then there is a unique solution. By looking at the answer choices, you can deduce that it must go as $1/\rho$ . Since the answer is in $cm^2$ , it must also go as $1/l$ . So we have $1/l*\rho=10^{-19}$ . It must, then, depend linearly on the final variable, P, to give you an answer choice in the list.

I guess this method is good if you remember seeing a formula that was linear in all its variables (or inverse variables), which would probably be applicable to a very small subset of people. However, on the GRE, I would suggest making this assumption if you have no other way of solving the problem. The answer choices are begging you to. Besides, you'd probably have to come up with a complicated formula (remember that you have to get the units right too, so squaring $0.1 cm$ gives you $0.01 cm^2$ , and you have to fix for this somewhere else in your formula) to land on one of the answer choices, especially because it says "the scattering cross section is" and not "the scattering cross section is approximately" (this rules out constants that aren't exactly $10^x$ for integer $x$ ).

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Comments

jw111
2008-09-01 14:42:52

Let's forget about Scattering Cross Section a moment, and image a fish who want to swim across the pool with half of it's cross section with a net.

If the fish is blind, the chance of the fish to reach the other end of the pool is clearly 50%. You get this answer by

the area of the net/total area(cross section) of pool = $\frac{\frac{1}{2}A}{A}=\frac{1}{2}$

Let's go back to Scattering Cross Section.

DEFINITION :
If a SINGLE nuclei can hold a net on it' hand (if it has) to catch the proton, the Scattering Cross Section is just the area of the net !

S0 for protons passing through this pool (the scatterer)

the chance $\frac{1}{10^6}$ = sum of total area of net/total area of pool =
volume*density*cross section/area =
$\frac{0.01*A*10^{20}*\alpha}{A}$

so the $\alpha$ is the answer.

jw111
2008-09-01 14:52:31

correction

0.1 (not 0.01)

phys2718
2008-10-16 12:03:24

This is the most incomprehensible explanation I have ever read. Yes, lets "image a fish who want to swim across the pool".

neon37
2008-11-02 21:45:09

I guess this analogy was good for ppl like me who mistook the scattering cross-section with the cross-section area.

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petr1243
2008-03-08 15:37:10

Just remember that probablility P= R/R0 is:

P=( $\sigma$ )(nAx)/A = ( $\sigma$ )(nx) giving us:

cross section $\sigma$ = P/nx

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eshaghoulian
2007-09-15 15:45:00

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Let's forget about Scattering Cross Section a moment, and image a fish who want to swim across the pool with half of it's cross section with a net. If the fish is blind, the chance of the fish to reach the other end of the pool is clearly 50%. You get this answer by the area of the net/total area(cross section) of pool = $\frac{\frac{1}{2}A}{A}=\frac{1}{2}$ Let's go back to Scattering Cross Section. DEFINITION : If a SINGLE nuclei can hold a net on it' hand (if it has) to catch the proton, the Scattering Cross Section is just the area of the net ! S0 for protons passing through this pool (the scatterer) the chance $\frac{1}{10^6}$ = sum of total area of net/total area of pool = volume*density*cross section/area = $\frac{0.01*A*10^{20}*\alpha}{A}$ so the $\alpha$ is the answer.

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