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Is the de Broglie wavelength of a photon equal to the EM ...

May 10, 2013 ... 1) Yes. The photon's matter wave is actually its electromagnetic wave. 2) Photon emission is not such a kind of process when you get some ...

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Electromagnetic radiation - Development of the quantum theory of ...

The Faraday-Maxwell-Hertz theory of electromagnetic radiation seemed to be able to ... It was his idea to use as a good approximation for the ideal blackbody an oven ... The energy hν of a photon of this wavelength is equal to the rest mass energy mc ... with a momentum p has, according to de Broglie, a wavelength λ = h/p.

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Matter wave - Wikipedia

Matter waves are a central part of the theory of quantum mechanics, being an example of ... The de Broglie wavelength is the wavelength, λ, associated with a massive ... to consist of waves of electromagnetic fields which propagated according to ... More recent experiments prove the quantum nature of molecules made of ...

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De Broglie wavelength (video) | Khan Academy

Bohr model energy levels (derivation using physics) ... Since visible light has a wavelength of about 500 nanometers, this means that visible ... couldn't they explain the photoelectric effect by saying that the light is a wave that carries ... it applies only for particles that travel at near-light speed because in the relation you used ...

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Step 1: Wavelength of the electromagnetic radiation

              Wavelength of electromagnetic radiation is inversely proportional to the frequency.

              Equation that relates to wavelength, frequency, and speed of light.

                           \lambda = \frac{c}{\upsilon} ..................(1)

Step 2: de Broglie wavelength of photon

            de Broglie wavelength \lambda = \frac{h}{p} .....................(2)

                    Where, h - Planck constant = 6.63* 10^{-34} J   and p - linear momentum.

                  We know that

                   Where an object has no mass, such as a photon, the energy equation is reduced to

                               E = pc

                              Where, p - momentum and c - speed of light                                      

                                p = \frac{E}{c}

                                p = \frac{h \upsilon}{c}                                    \because Energy of proton E=h\upsilon\

Step 3: Substituting the p(momentum) value in equation (2)

                  de Broglie wavelength  \lambda = \frac{h}{ \frac{h \upsilon}{c}}

                   de Broglie wavelength   \lambda = \frac{c}{\upsilon} ..........................(3)                          

                From equation (1) and (3) we can conclude that

                Wavelength of the electromagnetic radiation = de Broglie wavelength of photon