Question
What is $\gamma$ for a proton having a mass energy of 938.3 MeV accelerated through an effective potential of 1.0 TV (teravolt) at Fermilab outside Chicago?
$1070$
Solution Video

# OpenStax College Physics Solution, Chapter 28, Problem 61 (Problems & Exercises) (1:58)

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Video Transcript

This is College Physics Answers with Shaun Dychko. A proton is being accelerated through a potential difference of 1 teravolt which is 1 times 10 to the 12 volts. and it has a mass energy, which is <i>mc</i> squared of 938.3 megaelectron volts and the question is, what is gamma? So we can figure out gamma knowing the kinetic energy and the rest energy of a proton. So kinetic energy is going to be the energy that is given to the proton due to its acceleration through this potential difference and that kinetic energy will be its charge times the number of joules per coulomb, which is, you know, the more base unit of voltage is joules per coulomb so it's energy per charge and then we are multiplying this by charge to get energy. So that's all going to become kinetic energy. So we have a expression for kinetic energy in terms of gamma and then another one, in terms of voltage and so can we equate these two and we do that here on this line. So we have <i>qV</i> equals gamma minus 1 times the rest energy. We'll divide both sides by <i>mc</i> squared and we get gamma minus 1, after we switch the sides around, equals <i>qV</i> over <i>mc</i> squared and then add 1 to both sides and you get gamma is <i>qV</i> over <i>mc</i> squared plus 1 and now we substitute in the numbers. So that's the charge of a proton, which is the elementary charge—1.60 times 10 to the minus 19 coulombs— times 1 teravolt divided by the rest energy in electron volts so it's times 10 to the 6 electron volts here; times 10 to the 6 is the prefix mega. And then we multiply by 1.6 times 10 to the minus 19 joules per electron volt in order to make the units joules in the denominator here. And then add 1 and we get 1070 is the Lorentz factor.