Question

Explain de-Broglie argument to propose his hypothesis. Show that de-

Broglie wavelength of photon equals electromagnetic radiation.​

Answer 1

The momentum of electromagnetic wave of frequency v wavelength

is given by, <br>

, De-broglie wavelength of photon,

<br> Thus wavelength of eletromagnetic radiation is equal to the de-broglie wavelength.

Answer 2

★☆〖Qบęຮτ ı¨ ø nˇ〗☆★

⭐The Dual Nature of Matter⭐

=> de-Broglie's Principle states that "All material particles in motion possess wave characteristics..."

=> de-Broglie's Relationship can be derived by combining the mass and energy relationships proposed by Max Planck, and Albert Einstein...

E = ∫c²dm = Σc²Δm = mc²

E = hν

=> The combination of these two yielded the desired result:

λ = h/mc

=> The above equation is valid for a Photon(γ⁰)

=> The same relation can be extended to every particle of this universe, if the speed of light in vacua(c) is replaced by the ordinary velocity of the particle:

$\: \: \: \: \: \: \: \: \: λ = \frac{h}{ \: mv⃗}$

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