CONCEPT:
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de Broglie wavelength of electrons: Louis de Broglie theorized that not only light possesses both wave and particle properties, but rather particles with mass - such as electrons - do as well.
- The wavelength of material waves is also known as the de Broglie wavelength.
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de Broglie wavelength of electrons can be calculated from Planks constant h divided by the momentum of the particle.
λ = h/p
where λ is de Broglie wavelength, h is Plank's const, and p is the momentum.
- In terms of Energy, de Broglie wavelength of electrons:
\(λ=\frac{h}{\sqrt{2mE}}\)
where λ is de Broglie wavelength, h is Plank's const, m is the mass of an electron, and E is the energy of the electron.
The equation of energy of photons is given by:
\(E=\frac{hc}{λ}\)
where E is energy, h is the Planck constant, c is the speed of light in a vacuum and λ is the photon's wavelength.
EXPLANATION:
Given that:
\(\lambda_e ={ h\over\sqrt{( 2 mE}) }~and~\lambda_p = {hc \over E}\)
\(\lambda_e^2 ={ h^2\over{( 2 mE}) }\)
\(\lambda_e^2 ={ h^2\over{ 2 m({hc \over \lambda_p}}) }\)
\(\lambda_e^2 ={ h^2\over{ 2 m({hc }}) } \lambda_p\)
\(\lambda_e^2 \propto \lambda_p\)
So the correct answer is option 3.
