Question

The moving electron and a proton has same de Broglie wavelength. Show that the electron possesses more energy than carried by the proton.

Answer 1

Let us consider an electron and a photon of the same de Broglie wavelength $\lambda$

Total Energy

Let the electron have a relativistic mass m

Then its momentum =mv =$\frac{h}{\lambda }$ … (1)

Momentum of the photon is also $\frac{h}{\lambda }$, so its effective mass m$\gamma$=$\frac{h}{\lambda c}$…(2)

Since v<c, m> m$\gamma$

Since, E (the total energy) =$Relativistic\times massc2$, 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 $\frac{hc}{\lambda }$.

For the electron, KE = $\frac{p^2}{2m}$

Thus$\frac{KE\left(photon\right)}{KE\left(electron\right)}$ = $\frac{2c}{v}$ (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.