Which formula for the de Broglie wavelength of an electron is correct?

Question

So, I have my exams in physics in a week, and upon reviewing I was confused by the explanation of de Broglie wavelength of electrons in my book. Firstly, they stated that the equation was: $\lambda = \frac{h}{p}$, where $h$ is the Planck constant, and $p$ is the momentum of the particle. Later, however, when talking about electron diffraction and finding the angles of the minima, the author gave the formula equivalent to that for light: $\lambda = \frac{hc}{E}$. Now, what I don't understand is if it is simply a mistake made by the author, or whether a different formula have to be used for electron diffraction, as the two formulae are very clearly not equivalent. In the latter case, I don't understand why the formula would be different. I greatly appreciate the help, as the exams are really close, and I would like to make sure I get this right!

Edit: I was told that pictures of text are taking away from the readability of the posts, and thus they were removed. Essentially, the difference between the two cases are that in the first case, the proton did not have any significantly large kinetic energy, while in the second example, the kinetic energy was $400\ MeV$.

Answer 1

Often, when dealing with high-energy (relativistic) particles the rest mass of the particle can be neglected when performing calculations. Use your expression for $p$ from relativistic considerations, plug in the numbers and see the negligible change when you include and neglect to include the mass of the electron.

A good tip for when you enter into higher level physics/astrophysics: approximations are made all the time, where higher order terms will ve neglected.

Answer 2

As the energy of the electrons in that case is much greater than their mass, you can consider the approximation $E \sim pc$. So the formulas are equivalent.

Answer 3

The 4-momentum vector is given by ${\bf p}=(\frac{E}{c},p^{1},p^{2},p^{3})$. Now taking the scalar product with itself we have, \begin{equation} {\bf{p.p}}=E^2-(pc)^2=m_{0}^2c^4 \end{equation} Now for extremely relativistic case , we can use the condition that $E\gg m_0c^2$, thus this yields $p=\frac{E}{c}$.

Answer 4

The momentum of an electron, which is not travelling at very high velocity will not have any relativistic effects. So, its momentum is given by

$$p=m_0v$$

where $m_0$ is the rest mass of electron ($9.1\times 10^{-31}~\rm kg$) and $v$ is the velocity.

But to observe phenomena like diffraction (which observed with radiations like X-rays), the energy of electron should be somewhat in the energy of the radiations. The observed energy level is about $400~\rm MeV$. To achieve that much energy the electron will be moving at relativistic speeds and can be considered as a high energy radiation. In that case, the relativistic momentum of the electron is given by

$$p=\gamma\, m_0v$$

where $$\gamma=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$$

To have an energy of about $400~\rm MeV$, the electron is having a momentum of about $1.33~\rm MeV/c$. From this value, the velocity of electron will be about $0.933~\rm c$ which is about $93\%$ of the speed of light. At that much high speed, the electron behaves like a radiation. It's wave properties start to dominate which is necessary for diffraction to happen. So there is no problem in using the equation for radiation.