Forest ModernPhysics: Difference between revisions

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Another way to calculate
Another way to calculate


<math>p_{X-ray} = \frac{hc}{\lambda c} = \frac{1240\mbox{ eV} \cdot \mbox{nm}}{0.071 \mbox{nm}\cdot \mbox{c}}</math>
<math>p_{X-ray} = \frac{hc}{\lambda c} = \frac{1240\mbox{ eV} \cdot \mbox{nm}}{0.071 \mbox{nm}\cdot \mbox{c}}= 17.5 \frac{\mbox{keV}}{c}</math>


Hit a crystal made of nickel with 54 eV electrons.
Hit a crystal made of nickel with 54 eV electrons.

Revision as of 01:50, 30 September 2009

Matter Waves (Wave Particle Duality)

Special relativity said that

[math]\displaystyle{ E = pc }[/math] if m=0


Plank said he could fit the Black Body radiation data assuming that that

[math]\displaystyle{ E= hf }[/math] where [math]\displaystyle{ h = 6.63 \times 10^{-34} \mbox{ J} \cdot \mbox{s} }[/math] = Plank's constant

Combining the two we have

[math]\displaystyle{ p=\frac{E}{c} = \frac{ h f}{c} }[/math]

[math]\displaystyle{ \Rightarrow }[/math] photons have momentum like a particle (mv)

Do particles reciprocate and behave like photons?

De Broglie's Hypothesis

If photons can behave like particles by having momentum

Then can a particle behave like a wave by having wavelength

[math]\displaystyle{ p=\frac{ h f}{c} = \frac{h}{\lambda} }[/math]

or

[math]\displaystyle{ \lambda_{particle} = \frac{h}{p}= }[/math] de Broglie Hypothesis

Davisson and Germer

We know that X-rays having a wavelength of [math]\displaystyle{ \lambda_{X-rays} = 7.1 \times 10^{-11} \mbox{m} }[/math] make an interference patter on an aluminum foil.

[math]\displaystyle{ p_{X-ray} = \frac{h}{\lambda} = \frac{6.63 \times 10^{-34} \mbox{ J} \cdot \mbox{s}}{7.1 \times 10^{-11}\mbox{m}} \frac{1 \mbox{eV}}{1.6 \times 10^{-19} \mbox{J}} }[/math]

[math]\displaystyle{ = 5.8 \times 10^{-4} \frac{3 \times 10^{8}\mbox{m/s}}{c} }[/math]
[math]\displaystyle{ = 1.75 \times 10^{4} \frac{\mbox{eV}}{c} }[/math]
[math]\displaystyle{ = 17.5 \frac{\mbox{keV}}{c}\Rightarrow \mbox{E}=17.5 \mbox{keV} }[/math]

Another way to calculate

[math]\displaystyle{ p_{X-ray} = \frac{hc}{\lambda c} = \frac{1240\mbox{ eV} \cdot \mbox{nm}}{0.071 \mbox{nm}\cdot \mbox{c}}= 17.5 \frac{\mbox{keV}}{c} }[/math]

Hit a crystal made of nickel with 54 eV electrons.

1.) 54 eV electrons

From hyperphysics:

Bragg Diffraction

[1] Forest_Classes