Question

an electron and proton have same de brogile wavelength which of them has more energy?

Answer 1

Surfing the answers I feel that either I'm having a serious misconception or others do.

As far as I know, the fundamental relation postulated by de Broglie connects Energy of the particle with an associated wavelength.

So E=hν=hcλis the fundamental relation.

In case of a photon, (or for that purpose any massless particle) has energy E=cp,where p is the the momentum of the photon.

Hence, cp=hcλ

Or, λ=hp

This becomes true only for a photon (or a massless particle).

For a massive (non-zero mass) particle, the energy equals E=(cp)2+(m0c2)2−−−−−−−−−−−−−√

Hence, λ=hc(cp)2+(m0c2)2−−−−−−−−−−−−−√

Thus, it is evident that, same de Broglie wavelength implies, total energies of the particles are equal. What makes the difference is the rest mass energies of the particles. Since, proton is more massive and has a rest mass energy greater than that of electron, it must have a low value of “cp” that that of electron. Therefore, it must have a lower momentum than the electron to have the same wavelength.

This is what I've known so far. Do rectify me if I go wrong!!!

Okay, so I had answered this, but I was wrong. My friends rectified me, and in fact de Broglie wavelength λ=hp is the fundamental relation, and is true for all particles, whether massive or not. Others were correct.

Apologies!!!
I hope so it can be help you
please select my answer as brainliest answer