# Video: Relativistic Velocities of Accelerated Particles

A Van de Graaff accelerator utilizes a 50.0 MV potential difference to accelerate charged particles such as protons. 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. What is the velocity of a proton accelerated by such a potential? What is the velocity of an electron accelerated by such a potential?

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### Video Transcript

A Van de Graaff accelerator utilizes a 50.0 megavolt potential difference to accelerate charged particles such as protons. 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. What is the velocity of a proton accelerated by such a potential? What is the velocity of an electron accelerated by such a potential?

Weβre told in this statement that we have a 50.0 megavolt potential difference that accelerates charged particles. We want to solve for the velocity of a proton and the velocity of an electron accelerated by this potential difference. Weβll call these values π£ sub π and π£ sub π, respectively.

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. So the given potential difference in our scenario, which weβll call capital π, is equal to the work it does on the charged particles per charge.

We can also recall the work-energy theorem, which says that the work done on an object π€ equals that objectβs change in kinetic energy. So we can rewrite π€ in our equation as ΞKE. Weβre told further in the problem that this potential difference is high enough that we want to consider relativistic effects.

Relativistic kinetic energy is equal to πΎ minus one quantity times ππ squared, where πΎ equals one divided by the square root of one minus π£ squared over π squared. We can use this relationship for relativistic kinetic energy in our problem. We substitute this expression in for ΞKE. We want to solve this expression for the particle speed π£, which is under the square root sign in our denominator. So letβs start algebraically rearranging to solve for π£.

If we multiply both sides of our equation by π, add ππ squared to both sides, and then divide both sides by ππ squared, we find that capital π times π plus ππ squared all divided by ππ squared equals πΎ one over the square root of one minus π£ squared over π squared. 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 ππ squared divided by ππ plus ππ squared.

Weβre 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. Starting with the π£ sub π, the speed of the proton, weβll use a value of 1.67 times 10 to the negative 27th kilograms. And for the charge of a proton, weβll use a value of 1.60 times 10 to the negative 19th coulombs for the protonβs charge.

Weβll assume in this problem that the speed of light π is 3.00 times 10 to the eighth metres per second exactly. Weβre now ready to plug in and solve for π£ sub π, the speed of an accelerated proton moving through the potential difference π£. After a lot of plugging in, being careful to write our voltage π in units of volts: 50.0 times 10 to the sixth.

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 itβs 0.314 π; thatβs the speed of proton will acquire starting from rest and accelerated across a potential difference π of 50.0 megavolts.

Now that weβve 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. To do that, weβll change the mass value in this equation as well as the charge which is different for the proton and electron.

For electron mass, weβll 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. 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.