If a photon and an electron have the same De-Broglie wavelength, then?A.Both have same kinetic energyB.Proton has more K.E. than electronC.Electron has more K.E. than protonD.Both have same velocity

The de-Broglie wavelength (λ\lambda) of a particle is given by the formula:
λ=hp\lambda = \frac{h}{p}
where:
• hh is Planck's constant,
• pp is the momentum of the particle.
For a particle with mass mm and velocity vv, momentum is given by:
p=mvp = mv
So, the de-Broglie wavelength becomes:
λ=hmv\lambda = \frac{h}{mv}
Now, if the photon and electron have the same de-Broglie wavelength, this implies that their momentum is the same.
• For the photon: The photon is massless, and its energy is given by E=hνE = h \nu, where ν\nu is the frequency of the photon. Since a photon has no rest mass, it always moves at the speed of light.
• For the electron: The electron has mass mm, and its momentum is given by p=mvp = mv, where vv is its velocity. If the electron has the same de-Broglie wavelength as the photon, it must have a velocity such that its momentum matches the photon's.
However, due to the difference in mass between the electron and the photon, for the same de-Broglie wavelength, the electron must be moving at a much slower speed. This results in different kinetic energies.
• Kinetic energy of electron: K.E.=12mv2K.E. = \frac{1}{2}mv^2
• Kinetic energy of photon: Photons do not have rest mass, but they have energy associated with their momentum.
Since the photon has no rest mass and moves at the speed of light, it will have much more energy than the electron for the same de-Broglie wavelength.
Conclusion:
The correct answer is C. Electron has more K.E. than proton.