A beam of protons is produced by a Van de Graaff generator. The beam produces a 5.00 milliamp current, and the protons in the beam have energies of 1.00 kiloelectron volts. What is the speed of the protons? How many protons are produced each second? We’ll begin with the first part of this question: what is the speed of the protons in the beam. And let’s start that by sketching out what this beam looks like.
We can imagine this beam of protons, where each one has a particular energy given to us in the problem statement, and that overall there’s a current formed of 5.00 milliamps by this beam. Interestingly, we’re given the energy of each of the protons in the beam in units of kiloelectron volts, or more simply electron volts. And this perhaps raises the question what is an electron volt. The definition of an electron volt is that it’s an amount of energy. And that’s correct; it’s an energy rather than a voltage, which is gained by the charge of a single electron when it’s moved across a potential difference of one volt.
A few things to notice then in this definition is that an electron volt is an amount of energy. It has units of joules. And though it’s defined in terms of the charge of a single electron, this amount of energy can be used for any charged object in motion. And finally, there is a correspondence between one electron volt, which is a unit of energy, and one volt, which is a unit of potential difference. Electron volts and volts are very closely related, but they’re not the same thing, and we’ll want to keep that in mind. Now back to our question, what’s the speed of the protons in this proton beam?
Each of the protons in the beam we’re told has an energy of 1.00 kiloelectron volt, and that is a kinetic energy and energy due to motion. If we assume that the protons in this beam are moving slowly enough that their speeds can be considered nonrelativistic, then that means the kinetic energy of each one is equal to one-half the mass of the proton times its speed squared. And it’s that speed we want to solve for. Here’s what we can write, that 𝐸, the energy of each of the protons, is equal to one-half the mass of the proton times its speed squared. Our next step is to rearrange this equation to solve for 𝑉 sub 𝑝, and we’ll do it by multiplying both sides by two over 𝑚 sub 𝑝 and taking the square root of both sides. Once we’ve done that, we see that 𝑉 sub 𝑝 the speed of each of the protons in the beam is equal to the square root of two over the proton mass all multiplied by the energy of each proton.
To find out what 𝑉 sub 𝑝 is, there are two things we’ll want to look up. One, we’ll want to look up the mass of the proton; and secondly, we’ll want to know just how much energy in joules is 1.00 kiloelectron volts. We can start on that by figuring out just how much energy one electron volt is. First things first though, we’ll look up the mass of a proton; it’s approximately equal to 9.1 times 10 to the negative thirty-first kilograms. Now what about the electron volt? One eV, it turns out, is equivalent to 1.6 times 10 to the negative nineteenth joules of energy. If this number looks familiar, you probably have seen it before: 1.6 times 10 to the negative nineteenth coulombs is the charge on a proton, and the negative of that is the charge on an electron. This close correspondence can be traced back to the definition of an electron volt that is defined in terms of the charge of a single electron.
In any case, for our purposes we now have what we want: an expression in units of joules for the energy of each of these protons. Here’s how we’ll write all this into our expression for the speed of the proton. First, when it comes to the energy of each proton, we’ll write that as 1.00 times 10 to the third, because it’s 1.00 kilo, then that multiplied by 1.6 times 10 to the negative nineteenth joules because that’s how much energy is in one electron volt. And then downstairs in the denominator, we have the mass of the proton. When we calculate all this out, here’s what we find: the protons in this beam are moving at a speed of 4.38 times 10 to the fifth meters per second. That’s roughly one one thousandth of the speed of light, so that means our assumption that these protons were moving at nonrelativistic speeds is essentially correct.
Now that we know the proton’s speed, let’s consider part two of the question: how many of these protons are produced each second? In this part of the exercise, we’re less concerned with the amount of energy each proton has and we’re more interested in this current value: 5.00 milliamps. This current has something to say about just how many protons are passing a point every second. And that’s what we want to solve for. To get started, let’s recall the definition of an ampere. One ampere is defined as a coulomb of charge per second. Ampere being the unit of current, that means we’re talking of a coulomb of charge passing through a given point in a circuit every second. As we consider our beam of protons, we know that each one of these as a charged particle has its own charge in coulombs. The charge on an individual proton, we’ll call it 𝑞 sub 𝑝, is equal to 1.6 times 10 to the negative nineteenth coulombs.
Knowing this, let’s see if we can connect this value with our given current value in the problem statement. If we write out the current in this beam, the first thing we can say about it is that 5.00 milliamps is equal to 5.00 times 10 to the negative third amps. Now let’s focus on the units of this expression, amps. We’ve seen that that is equivalent to coulombs per second, so we can write our units according to that expression. Recall that what we want to solve for is how many protons are produced each second; that is, how many move through this beam each second past a given point. We already have units of per second in our expression for the current of this beam. So to find the number of protons produced each second, what we can call in 𝑁 sub 𝑝, all we need to do is divide our current by the charge on a single proton, what we’ve called 𝑞 sub 𝑝.
If we substitute in the value for 𝑞 sub 𝑝, this will start to make a bit more sense. When we replace 𝑞 sub 𝑝 with its value, 1.6 times 10 to the negative nineteenth coulombs, we see that in this expression the units of coulombs cancel out and we’re left with units of per second. That’s good because those are the units we want to end up with, a number which has no units per second. All that’s left for us to do then is to divide 5.00 times 10 to the third by 1.6 times 10 to the negative nineteenth. When we do, we find a result of 3.13 times 10 to the sixteenth. And what is that number? That’s the number of protons per second; that is, the number of protons produced each second by the beam.