Show that the wavelength of electromagnetic radiation is equal to the de Broglie wavelength of its quantum (photon).

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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 ...
For more information, see Matter wave - Wikipedia
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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