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Question
A Van de Graaff accelerator utilizes a 50.0 MV potential difference to accelerate charged particles such as protons. (a) What is the velocity of a proton accelerated by such a potential? (b) An electron?
Question by OpenStax is licensed under CC BY 4.0.
Final Answer
  1. $0.314c$
  2. $0.99995c$
Solution Video

OpenStax College Physics Solution, Chapter 28, Problem 65 (Problems & Exercises) (4:39)

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Video Transcript

This is College Physics Answers with Shaun Dychko. In part (a) of this question, a proton is being accelerated by a Van der Graff generator through a potential difference of 50 megavolts which is 50 times 10 to the 6 volts. And the question is what will the speed of the proton be after accumulating an amount of kinetic energy which will be equal to the charge on the proton times its potential difference. And we are gonna figure out the speed by figuring out gamma and then we'll solve this formula here for <i>v</i> in terms of gamma. Then part (b) says do it again with electrons instead of protons. So kinetic energy in terms of gamma is gamma minus 1 times the rest energy and we can look up what this rest energy is in the data table for a proton. And we also know that kinetic energy is going to be the charge of the proton multiplied by the joules per coulomb which is this potential difference here and the coulomb's will cancel, leaving us with joules or energy which will become kinetic energy of the proton. So equate the <i>qV</i> with this gamma minus 1 times <i>mc</i> squared here and then divide both sides by <i>mc</i> squared and we have gamma minus 1 is <i>qV</i> over <i>mc</i> squared and then add 1 to both sides. And we now have an expression for gamma in terms of <i>q, V, m</i> and <i>c</i>. So then after we figure out what gamma is, we'll solve this formula for <i>v</i> in terms of gamma. So we'll take both sides to the exponent negative 2 which flips this fraction; this gamma over 1 now becomes 1 over gamma squared and this fraction becomes flipped and we have this square root in the numerator but then it gets squared as well and so it's gonna be 1 minus <i>v</i> squared over <i>c</i> squared and we are switching the sides around to have the unknown on the left. Then add <i>v</i> squared over <i>c</i> squared to both sides and then subtract 1 over gamma squared from both sides and then switch the sides around and we have <i>v</i> squared over <i>c</i> squared is 1 minus 1 over gamma squared. Then multiply both sides by <i>c</i> squared and we have <i>v</i> squared equals <i>c</i> squared times 1 minus 1 over gamma squared and then take the square root of both sides giving us <i>v</i> is <i>c</i> times square root 1 minus 1 over gamma squared. So now we have a formula for the speed of a particle, in terms of gamma. Now we could if we wanted to but we are not going to because it would be really messy, we could substitute this expression for gamma in place here but that'll get too messy so I'm not gonna do that. So let's just figure out what gamma is numerically using this expression here and then we'll substitute that number into this formula here. So gamma is the charge times the voltage of a Van der Graff generator divided by the rest energy of a proton— 938.3 times 10 to the 6 electron volts— converted into joules and then add 1, that is 1.053289 and that number gets squared in the denominator here and we have <i>v</i> is gonna be <i>c</i> times square root 1 minus 1 over that gamma squared and that gives 0.314<i>c</i>. Now in part (b), it's the same type of story but now we have an electron and so the only thing that's changed is this, rest energy has changed, it's a much smaller number and consequently, gamma is larger since the denominator has become much smaller than it was before. So before it was 1.05 for gamma and now gamma is 98.8474. So <i>v</i> for the electron will be <i>c</i> times square root 1 minus 1 over 98.8474 squared which is 0.99995<i>c</i>. And from the number of significant figures in the data we are given, we should not have that many significant figures in our answer but we are doing so anyway just to distinguish this speed from the speed of light.