De Broglie wavelength for a particle of matter
We can think of waves as particles. They can be thought of as having a mass equivalent and even a momentum  see the photon  now we have to look at matter as waves! Matter has momentum 'p'  the mass multiplied by the velocity but the photon has momentum expressed in terms of wavelength p_{photon}_{} = h/λDe Broglie (pronounced De Broy) proposed that any particle of matter has a wavelength and wave properties as well as particle properties. Just as a light in a photon has particle properties but is wave also, so an electron or proton (or even YOU) would have a wavelength as well as momentum! Rearranging the equation (and specifying that the wavelength is associated with matter waves by using a subscript of 'DB' we have:
Note that only moving particles can act like waves! They must have momentum (a nonzero velocity) to have a wavelength. You will often come across questions in an A level paper that give you the accelerating voltage that a charged particle moves across and nothing else! From this information you have to determine the de Broglie wavelength of the particle. Let us consider this: E = QVEnergy gained by a particle of charge Q when moving across a potential difference V will be the product QV. So the kinetic energy gained by an electron (having a charge 'e') moving across a potential difference of 'V' would be 'eV' You should know how to do this... Now let's try a question!
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