# De Broglie Wavelength

### Deriving the De Broglie Wavelength

De Broglie derived his equation using well established theories through the following series of substitutions:

1. De Broglie first used Einstein's famous equation relating matter and energy:

$E = mc^2$

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:

$E= h \nu$

E = energy, h = Plank's constant(6.62607 x 10-34 J s), υ = 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 subsituted v, velocity, for c, the speed of light.

$mv^2 = h\nu$

5. Through the equation $$\lambda$$, de Broglie substituted $$v/\lambda$$ for $$\nu$$ and arrived at the final expression that relates wavelength and particle with speed.

$mv^2 = \dfrac{hv}{\lambda}$

Hence:

$\lambda = \dfrac{hv}{mv^2} = \dfrac{h}{mv}$

Although De Broglie was credited for his hypothesis, he had no actual experimental evidence for his conjecture. In 1927, Clinton J. Davisson and Lester H. Germer shot electron particles onto onto a nickel crystal. What they see is the diffraction of the electron similar to waves diffractions agaisnt crystals(x-rays). In the same year, an English physicist, George P. Thomson fired electrons towards thin metal foil providing him with the same results as Davisson and Germer.

A majority of Wave-Particle Duality Problems are simple plug and chugs with some variation of canceling out units

 Example

Example Problem 1

Find the de Broglie wavelength for an electron moving at the speed of 6.63 x 106 m/s (mass of an electron = 9.1 x 10-31 kg).

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