Question

which among the electron or proton possess more total energy if they possess same debroglies wave length

Answer 1

Let us consider an electron and a photon of the same de Broglie wavelength λ

Total Energy

Let the electron have a relativistic mass m

Then its momentum =mv = h/ λ, so m= h/ λv… (1)

Momentum of the photon is also h/ λ, so its effective mass mγ =h/λc…(2)

Since v<c, m> mγ

Since, E (the total energy) =Relativistic mass* c2, it follows that the total energy of the electron is greater than that of the photon, both having the same wavelength.

Kinetic Energy (KE)

In the case of the photon, all its energy is kinetic, which is ch/ λ. This can also be written as p2/ =mγ, where E= KE = pc

For the electron, KE = p2/2m

Thus KE (photon)/KE (electron) =2m/ mγ =2c/v (from (1) and (2), which is >1 (since c>v))

Therefore, KE of the photon (which is also the total energy of the photon) is greater than the kinetic energy of the electron, both having the same wavelength.