 # Which formula for the de Broglie wavelength of an electron is correct? So, I have my exams in phys Jaiden Bowman 2022-05-10 Answered
Which formula for the de Broglie wavelength of an electron is correct?
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
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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.
###### Not exactly what you’re looking for? Osmarq5ltp
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.