Chapter 37: Problem 58
The mass of an electron is \(9.10938188 \times 10^{-31} \mathrm{~kg}\). To eight significant figures, find the following for the given electron kinetic energy: (a) \(\gamma\) and (b) \(\beta\) for \(K=1.0000000 \mathrm{keV},(\mathrm{c}) \gamma\) and \((\) d \() \beta\) for \(K=\) \(1.0000000 \mathrm{MeV}\), and then (e) \(\gamma\) and (f) \(\beta\) for \(K=1.0000000 \mathrm{GeV}\).
Short Answer
Step by step solution
Convert Energy Units
Calculate Total Energy
Calculate Lorentz Factor \(\gamma\)
Calculate Velocity Ratio \(\beta\)
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Lorentz Factor (γ)
- At rest, \(\gamma = 1\), meaning no relativistic effects.
- As velocity \(v\) increases, \(\gamma\) becomes greater than 1.
- Approaching the speed of light, \(\gamma\) approaches infinity.
Velocity Ratio (β)
- \(\beta = 0\) implies the object is at rest.
- \(\beta\) close to 1 indicates significant relativistic speeds.
- It is impossible for \(\beta\) to equal or exceed 1 as this would require infinite energy.
Electron Kinetic Energy
- \(E\) is the total energy of the electron.
- \(E_0 = mc^2\) is the electron's rest energy.
Energy Conversion
To convert energy from eV to J, use the conversion factor:
- \(1 \text{ eV} = 1.60219 \times 10^{-19} \text{ J}\)