Forest ModernPhysics
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 a diffraction pattern 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]
- What would be the energy of an electron with the same wavelength as the above X-ray?
relativistic total energy relation
- [math]\displaystyle{ E_{tot} = \sqrt{(pc)^2 + (mc^2)^2} }[/math]
- [math]\displaystyle{ = \sqrt{(17.5 keV)^2 + (511 keV)^2} }[/math]
- = 511.3 keV
relativistic kinetic energy
- [math]\displaystyle{ K = E - mc^2 = (\gamma-1)mc^2 = 0.3 \mbox{keV} = 300 \mbox{eV} }[/math]
- Note [math]\displaystyle{ \beta = \frac{pc}{E_{tot}} = \frac{17.5 \mbox{keV}}{511 \mbox{keV}} = 0.03 \Rightarrow }[/math] classical physics may be used for electrons below 50 keV
Clinton Davisson and Lester Germer in 1927 published conclusive evidence for the diffraction of electron waves using 54 eV electrons impinging a crystal made of nickel.
One problem to overcome for the experiment was that such a low energy electron scatters in air. The had to do the experiment in a vacuum.
From hyperphysics:
Bragg Diffraction
