Question 26.8: Speed of a High-Energy Electron What s the speed of the elec...
Speed of a High-Energy Electron
What s the speed of the electrons in the 50-GeV linac at SLAC?
Express the answer as a fraction of the speed of light.
ORGANIZE AND PLAN This kinetic energy is much higher than the electron s rest energy (0.511 MeV), so relativity is essential here. Useful relativistic relationships from Chapter 20 are E=K+E_{0} \text { and } E=\gamma m c^{2} \text {, where the relativistic factor is } \gamma=1 / \sqrt{1-v^{2} / c^{2}} .
\text { Known: Electron rest energy } m c^{2}=0.511 MeV ; K=50 GeV .
\text { Because } K \gg E_{0} \text { in this case, } a good approximation is E=K+E_{0} \approx K . \text { Solving } E=\gamma m c^{2} \text { for the relativistic factor, }
\gamma=\frac{E}{m c^{2}}=\frac{5.0 \times 10^{10} eV }{5.11 \times 10^{5} eV }=9.78 \times 10^{4}.
\text { Now solving } \gamma=1 / \sqrt{1-v^{2} / c^{2}} \text { for } v \text {, }\gamma^{2}=1 /\left(1-v^{2} / c^{2}\right).
v^{2} / c^{2}=1-1 / \gamma^{2}=1-\frac{1}{9.56 \times 10^{9}}=0.999999999895.
Thus
v=(0.999999999948) c.
REFLECT That’s amazingly close to the speed of light!