The discussion centers on calculating the de Broglie wavelength of a charged electron with a total energy of 1 KeV. The relevant equations include \(E^2 = (mc^2)^2 + (pc)^2\) and \(\lambda = \frac{h}{p}\). A participant encountered an issue when substituting values, resulting in an imaginary momentum. The consensus is that the 1 KeV should be interpreted as the kinetic energy of the electron, not its total energy, which resolves the calculation error.
PREREQUISITESPhysics students, educators, and anyone interested in quantum mechanics and particle physics calculations will benefit from this discussion.
ParrotPete said:Homework Statement
Given an electron with total energy 1 KeV, determine it's deBroigle wavelength.
Homework Equations
E^2 = (mc^2)^2+(pc)^2
[tex]\lambda[/tex] = [tex]\frac{h}{p}[/tex]
The Attempt at a Solution
(pc)^2 = E^2 - (mc^2)^2 <> p = ± [tex]\frac{1}{c}[/tex] [tex]\sqrt{E^2-(mc^2)^2}[/tex]
What am I doing wrong?
When I plug in E = 10^3 eV and (mc^2) = 0.511 * 10^6 eV I get an imaginary result.