CONCEPT:

• de Broglie wavelength of electrons:Louis de Broglietheorized 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 isalso known as thede Broglie wavelength.
• de Broglie wavelength of electronscan be calculated from Planks constanth divided by the momentum of the particle.

λ = h/p

whereλ is de Broglie wavelength,h is Plank's const, and p is the momentum.

• In terms of Energy,de Broglie wavelength of electrons:

\(λ=\frac{h}{\sqrt{2mE}}\)

whereλ is de Broglie wavelength,h is Plank's const, m is the mass of an electron, and E is the energy of the electron.​

Theequation ofenergyof photons is given by:

\(E=\frac{hc}{λ}\)

where E isenergy, h is the Planck constant, c is the speed of light in a vacuum and λ is the photon's wavelength.

EXPLANATION:

Given that:

\(\lambda_e ={ h\over\sqrt{( 2 mE}) }~and~\lambda_p = {hc \over E}\)

\(\lambda_e^2 ={ h^2\over{( 2 mE}) }\)

\(\lambda_e^2 ={ h^2\over{ 2 m({hc \over \lambda_p}}) }\)

\(\lambda_e^2 ={ h^2\over{ 2 m({hc }}) } \lambda_p\)

\(\lambda_e^2 \propto \lambda_p\)

Sothe correct answer isoption 3.