What is the wave particle duality of electrons?

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What is the wave particle duality of electrons?

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Answer 1 · 2014-10-13T22:53:18

All particles of matter, including electrons, exhibit characteristics of waves. The most obvious of these is the ability of particles to interfere with each other with a characteristic wavelength given by the de Broglie relation

$λ = \frac{h}{p}$ $lambda=h/p$

where $h$ $h$ is Planck's constant, $h = 6.626 \times 10^{- 34} J - s$ $h=6.626times10^(-34)J-s$ , and $p$ $p$ is the momentum of the particle. If $p$ $p$ is expressed in units of $\frac{k g - m}{s}$ $(kg-m)/s$ then the wavelength $λ$ $lambda$ is obtained in units of meters.

Example:

If electrons initially at rest are accelerated through a potential of $10 V$ $10V$ , then the kinetic energy of each electron would be $10 e V = 1.602 \times 10^{- 18} J$ $10eV=1.602times10^(-18) J$ . Using the classical mechanics relation between kinetic energy and momentum, we obtain $p = {(2 m E)}^{\frac{1}{2}} = 1.708 \times 10^{- 24} \frac{k g - m}{s}$ $p=(2mE)^(1/2)=1.708times10^(-24)(kg-m)/s$ .

Using the deBroglie relation, we can calculate the characteristic wavelength of the electrons as $λ = \frac{h}{p} = 3.878 \times 10^{- 10} m = 0.3878 n m$ $lambda=h/p=3.878times10^(-10)m=0.3878 nm$

If such a beam of electrons were diffracted from an ordered crystal, they would form a pattern similar to that for X-rays of the same wavelength ( $0.3878 n m$ $0.3878 nm$ ).

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What is the wave particle duality of electrons?

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