This is a beginner question from a self-studying student to give perspective.
In a question it asks to find the de Broglie wavelength of $1$ eV electron by use of relativistic energy formula. The formula given in the book is for this relativistic energy is:
$$E^2=(pc)^2+(m_0c^2)^2 \tag{1} $$
When I put the value $1$ eV for $E$ and $0.511$ MeV for electron rest energy I obtain an imaginary value for the momentum.
My question is how this result should be interpreted? Is there a definition I don't know about this relativistic energy formula? I took $1$ eV as the total energy $E$. Is this a faulty approach?
Best Answer
I would guess the description $1$ eV electron means an electron with a kinetic energy of $1$ eV. In that case the simplest approach is to use the relativistic equation for the kinetic energy:
$$ KE = (\gamma - 1)mc^2 $$
i.e.
$$ \gamma = \frac{KE}{mc^2} + 1 $$
Then you can find the velocity $v$ from:
$$ \gamma = \frac{1}{\sqrt{1 - v^2/c^2}} $$
and use:
$$ p = \gamma m v $$
to find the momentum. In your equation (1) $E$ is the total energy, i.e. the rest mass energy plus the kinetic energy, and this cannot be less than the rest mass energy as the kinetic energy cannot be negative. Hence the total energy cannot be $1$ eV.