The kinetic energy of an electron traveling at 0.98c is
A)0.245 MeV
B)0.511 MeV
C)0.756 MeV
D)1.225 MeV
E)2.044 MeV

Question

Quizplus · Accepted Answer

The kinetic energy (KE) of an electron traveling at a significant fraction of the speed of light, such as 0.98c, can be calculated using the relativistic kinetic energy formula: $ KE = (\gamma - 1)mc^2 $, where $ \gamma = \frac{1}{\sqrt{1 - v^2/c^2}} $ is the Lorentz factor, $ m $ is the rest mass of the electron (0.511 MeV/c^2), and $ c $ is the speed of light. Plugging in the values, $ \gamma $ for 0.98c is approximately 5.03. Thus, $ KE = (5.03 - 1) \times 0.511 $ MeV $ \approx 2.064 $ MeV, which is closest to the option E, 2.044 MeV. The slight difference is due to rounding in the calculation.

The Kinetic Energy of an Electron Traveling at 0