GREPhysics.NET: GR8677 #91

GR8677 #91

Problem

GREPhysics.NET Official Solution

Optics $\Rightarrow$ }Bragg Reflection

Bragg diffraction is basically the wavelength change of a wave impinging two adjacent layers of the crystal. More specifically, it relates the wavelength difference of the incident wave on the top layer to that of the reflected wave on the lower layer. The distance between layers (``lattice planes", in Solid-State-Speak) is $d$ , and the angle of incidence is $\theta$ . The change in wavelength is thus a simple geometry problem, the result being the celebrated Bragg Reflection Relation,

$2d \sin \theta = n \lambda, <br />$

where $n$ corresponds to the order of the reflection.

The problem supplies the following numbers

$\begin{eqnarray} n=1\\ <br /> d=3E-10\\ \theta = 30^\circ\\ m_e\approx 10E-31, <br /> \end{eqnarray}$

and if one recalls the de Broglie relation, $p = h/\lambda$ , one has, in general, $n\lambda = nh/{m v} = 2d \sin\theta$ . Plugging in some numbers, one finds that $\frac{6E-34}{10E-31v}=3E-10$ , since $\sin 30^\circ = 1/2$ , and where the approximation scheme to battle the no-calculators-allow requirement is shown. Thus, one has $v=0.25E7$ , which is very close to choice (D).

The problem .

Alternate Solutions

chrisfizzix
2008-10-03 14:32:17

I feel that the posted solution lacks a bit of transparency; this is the same solution but reformatted.

Bragg Scattering gives the result
$2 d \sin{\theta} = n \lambda$
and the DeBroglie relation gives
$p = \frac{h}{\lambda}$ .
First-order reflection lets us set $n = 1$ , also the relation usually fails if n > 1 anyway, so we can rearrange to write
$p = \frac{h}{2 d \sin{\theta} }$

The answer in E is greater than $c$ , so we can get rid of that. All of the other answers are non-relativistic (even for D, $\beta \approx 0.01$ ) so we can just use $p = mv$ .

Solving this, we get
$v = \frac{h}{2 d \sin{\theta} m_e}$
Plugging in values gives a result closest to D.

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Comments

student2008
2008-10-14 06:07:56

Since the orders of magnitude in answers differ a lot, we can even don't get use of $\theta$ & Bragg condition. All we use is $mv=p=\frac hd$ , since $d\sim \lambda$ (actually, in this problem $2\sin\theta=1$ , so we even obtain the exact result).

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chrisfizzix
2008-10-03 14:32:17

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madfish
2007-11-02 19:04:53

if one knows electrons travel very fast to give off x-rays, but not faster than the speed of light...one can elminate all but C & D

chrisfizzix
2008-10-03 14:13:43

The electron isn't radiating X-rays - the spacing between lattice planes in the crystal has been previously measured using X-rays. Thus, the threshold for X-ray emission by an electron has nothing to do with this problem.

Also, electrons need both high energy and some kind of accelerating force to emit X-rays, such as the magnetic field from a bending magnet in a synchrotron.

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You are replying to:

I feel that the posted solution lacks a bit of transparency; this is the same solution but reformatted. Bragg Scattering gives the result $2 d \sin{\theta} = n \lambda$ and the DeBroglie relation gives $p = \frac{h}{\lambda}$ . First-order reflection lets us set $n = 1$ , also the relation usually fails if n > 1 anyway, so we can rearrange to write $p = \frac{h}{2 d \sin{\theta} }$ The answer in E is greater than $c$ , so we can get rid of that. All of the other answers are non-relativistic (even for D, $\beta \approx 0.01$ ) so we can just use $p = mv$ . Solving this, we get $v = \frac{h}{2 d \sin{\theta} m_e}$ Plugging in values gives a result closest to D.

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