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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 ...

For more information, see **De Broglie wavelength** (video) | Khan Academy

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