Wednesday 6 November 2019

quantum mechanics - Proof of de Broglie wavelength for electron


According to de Broglie's wave-particle duality, the relation between electron's wavelength and momentum is $\lambda =h/mv$.


The proof of this is given in my textbook as follows:





  1. De Broglie first used Einstein's famous equation relating matter and energy, $$E=mc^2,$$ where $E=$ energy, $m =$ mass, $c =$ speed of light.




  2. Using Planck's theory which states every quantum of a wave has a discrete amount of energy given by Planck's equation, $$E=h\nu,$$ where $E =$ energy, $h =$ Plank's constant ($6.62607 \times 10^{-34}\:\mathrm{ J\:s}$), $\nu =$ frequency.




  3. Since de Broglie believes particles and wave have the same traits, the two energies would be the same: $$mc^2=h\nu.$$





  4. Because real particles do not travel at the speed of light, de Broglie substituted $v$, velocity, for $c$, the speed of light: $$mv^2=h\nu.$$




I want a direct proof without substituting $v$ for $c$. Is it possible to prove directly $\lambda=h/mv$ without substituting $v$ for $c$ in the equation $\lambda =h/mc$?




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