An electron is accelerated from rest through a potential difference of 1.0 MV. If the rest energy of the electron is 0.511 MeV, how fast is the electron moving?
A) 0.57 c
B) 0.30 c
C) 0.67 c
D) 0.90 c
E) 0.94 c

Question

Quizplus · Accepted Answer

We can use the equation $K = qV$ to find the kinetic energy gained by the electron, where $q=-e$ is the charge of the electron and $V$ is the potential difference. We can then use the relativistic kinetic energy equation $K = (\gamma-1)mc^2$ to relate the kinetic energy to the speed of the electron, where $\gamma$ is the Lorentz factor and $m$ is the rest mass of the electron. Equating these two expressions for $K$ and solving for $\gamma$, we find:
$$(\gamma-1)mc^2=-eV \implies \gamma = \frac{1+eV/mc^2}{\sqrt{1-(v/c)^2}}$$
Solving for $v/c$, we get:
$$v/c = \sqrt{1-(\gamma^{-2}-1)^2} \approx 0.940$$
Therefore, the electron is moving at 0.94 times the speed of light.

An Electron Is Accelerated from Rest Through a Potential Difference