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A Van de Graaff accelerator utilizes a 50.0 megavolt potential difference to accelerate charged particles such as protons.
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The kinetic energy provided by such a large potential difference is sufficiently great that relativistic effects need to be taken into account when finding the velocity of accelerated particles.
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What is the velocity of a proton accelerated by such a potential?
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What is the velocity of an electron accelerated by such a potential?
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Weβre told in this statement that we have a 50.0 megavolt potential difference that accelerates charged particles.
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We want to solve for the velocity of a proton and the velocity of an electron accelerated by this potential difference.
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Weβll call these values π£ sub π and π£ sub π, respectively.
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We can begin our solution by recalling that potential difference π is equal to the work π€ done on an object per unit charged of that object.
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So the given potential difference in our scenario, which weβll call capital π, is equal to the work it does on the charged particles per charge.
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We can also recall the work-energy theorem, which says that the work done on an object π€ equals that objectβs change in kinetic energy.
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So we can rewrite π€ in our equation as ΞKE.
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Weβre told further in the problem that this potential difference is high enough that we want to consider relativistic effects.
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Relativistic kinetic energy is equal to πΎ minus one quantity times ππ squared, where πΎ equals one divided by the square root of one minus π£ squared over π squared.
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We can use this relationship for relativistic kinetic energy in our problem.
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We substitute this expression in for ΞKE.
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We want to solve this expression for the particle speed π£, which is under the square root sign in our denominator.
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So letβs start algebraically rearranging to solve for π£.
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If we multiply both sides of our equation by π, add ππ squared to both sides, and then divide both sides by ππ squared, we find that capital π times π plus ππ squared all divided by ππ squared equals πΎ one over the square root of one minus π£ squared over π squared.
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If we then cross multiply and square both sides of the resulting equation, we have another intermediate step, which only takes a few more algebraic rearrangements to arrive at an expression for π£, an expression which further simplifies to π times the square root of one minus the quantity squared ππ squared divided by ππ plus ππ squared.
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Weβre now ready to solve for π£ sub π and π£ sub π, the velocities of the proton and electron accelerated through this potential difference capital π.
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Starting with the π£ sub π, the speed of the proton, weβll use a value of 1.67 times 10 to the negative 27th kilograms.
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And for the charge of a proton, weβll use a value of 1.60 times 10 to the negative 19th coulombs for the protonβs charge.
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Weβll assume in this problem that the speed of light π is 3.00 times 10 to the eighth metres per second exactly.
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Weβre now ready to plug in and solve for π£ sub π, the speed of an accelerated proton moving through the potential difference π£.
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After a lot of plugging in, being careful to write our voltage π in units of volts: 50.0 times 10 to the sixth.
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And with all the mass, charge, and speeds plugged in, when we calculate this value on our calculator, we find that to three significant figures itβs 0.314 π; thatβs the speed of proton will acquire starting from rest and accelerated across a potential difference π of 50.0 megavolts.
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Now that weβve solved for π£ sub π, we move on for solving for π£ sub π, the speed an electron would acquire accelerated across this potential difference.
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To do that, weβll change the mass value in this equation as well as the charge which is different for the proton and electron.
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For electron mass, weβll use a value of exactly 9.1 times 10 to the negative 31st kilograms and for electron charge a value of negative 1.60 times 10 to the negative 19th coulombs.
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When we enter these revised values into our equation now for π£ sub π and calculate the result, we find that an electron accelerated across this potential would achieve a very high speed of 0.99995 times the speed of light.