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An electron and a photon have the same wavelength of 109m. If E is the energy of the photon and p is the momentum of the electron, the magnitude of E/P in SI units is?
(A) 1.00×109
(B) 1.50×108
(C) 3.00×108
(D) 1.20×107

Answer
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Hint: Initially you must be aware of the general formula of momentum and photon. De-Broglie wavelength plays an important factor in this question. Deriving a relation between energy and momentum by wavelength will surely help you.

Complete step by step solution:
Given:
λ=109m
Formula:
E=hcλ
Calculation:
 From Planck's equation
E=hcλ
λp=hcE
  From de-Broglie’s equation
λe=hp
 Since,
λp=λe
hcE=hp
  Ep=c
=3×108

Therefore, option C is a correct option.

Note:
DE Broglie wavelength plays an important role in physics. It is said that matter incorporates a dual nature of wave-particles. Nuclear physicist waves, named after the discoverer Louis nuclear physicist, is that the property of a fabric object that varies in time or space while behaving the same as waves. It's also called matter-waves. It holds great similarity to the twin nature of sunshine which behaves as particle and wave, which has been proven experimentally.

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