# Video: Identifying Particles by How They Move in an Electric Field

A block of an unknown material emits charged particles. These particles are accelerated in a vacuum across an electric potential difference of 6 V. The kinetic energy of each particle is found to be 1.92 × 10⁻¹⁸ J once it has crossed this potential difference. What is the charge of each particle?

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### Video Transcript

A block of an unknown material emits charged particles. These particles are accelerated in a vacuum across an electric potential difference of six volts. The kinetic energy of each particle is found to be 1.92 times 10 to the negative 18th joules once it has crossed this potential difference. What is the charge of each particle?

Okay, let’s say that this is our block of material. And we’re told that this block is emitting particles. And along with that, we’re told that there is a potential difference of six volts, which accelerates these particles through a vacuum. If we follow one of these particles as it moves across this six-volt potential difference, once it reaches the barrier of that six-volt potential difference, we’re told its kinetic energy, we can call it 𝐾𝐸, is 1.92 times 10 to the negative 18th joules. Knowing all this, we first want to answer this question. What’s the charge of each one of these particles?

Now, notice that we don’t know what the particles are. And we don’t know what their mass is. Nonetheless, given what we’re given, it’s still possible to answer this question. We’ll do it by recalling a mathematical relationship between particle kinetic energy, potential difference, and particle charge. According to this relationship, if we have a particle with charge 𝑞 and that particle moves through a potential difference 𝑉, then if we take the product of those two values, we’ll find the kinetic energy that the particle acquires. Now, in this scenario, we’re told that kinetic energy. And we’re also told the potential difference 𝑉. And we want to solve for the charge 𝑞. So let’s do that. Let’s rearrange this equation so that we have 𝑞 by itself on one side.

To do that, we can divide both sides by the potential 𝑉, which means that cancels out on this right-hand side. And we see that particle charge 𝑞 is equal to kinetic energy divided by potential difference. Our next step is to substitute in the given values for kinetic energy and potential. And once we do that and calculate this fraction, we find a result of 3.2 times 10 to the negative 19th coulombs. So that’s the answer we’ll give to this first question of what is the charge of each of the particles emitted from this block of unknown material. Each of these particles has a charge of 3.2 times 10 to the negative 19th coulombs. Now let’s move on to the second part of our question.

Our second question says, using a value of 1.6 times 10 to the negative 19th coulombs for the charge of a proton, what is the relative charge of each particle?

And talking about each particle, we mean the particles emitted from this block of unknown material, the particles whose charge we’ve solved for in the earlier part. To answer this question, we need to know that the relative charge of a proton, we’ll symbolize it with 𝑃, is positive one. So rather than 1.6 times 10 to the negative 19th coulombs, which is a somewhat clumsy number being used for that charge, as a shorthand, we can say that that amount of charge, 1.6 times 10 to the negative 19th coulombs, is equal to one charge unit. And the question is, based on the fact that a proton has a relative charge, we could call it a plus one, we want to know the relative charge of these unknown particles.

Here’s how we can figure that out. We can take the charge of our unknown particles that we solved for earlier, that charge is 𝑞. And then we’ll divide that charge by the charge of a proton, not the plus one but by the true charge of a proton. That is, its charge in units of coulombs. So to figure out the relative charge of these unknown particles, we’ll substitute in the value we found for 𝑞. And then, when we perform this division, we find that the answer is simply positive two. And that this, therefore, is the relative charge of these unknown particles. They have a charge magnitude twice as big as the charge of a proton and the same sign charge, positive. Knowing all this, let’s answer a third and final question about this scenario.

It is found that the charged particles are being emitted by radioactive decay of the material in the block. What kind of particles are being emitted by the block?

Okay, so back to our block, we’re told that the particles emitted by it, which right now we don’t know what they are, are being emitted, thanks to radioactive decay of some material, in the block. Now, based on that, that narrows our options for what kind of particles these particles being emitted by the block are. The particles could be neutrons or beta particles, electrons or positrons. Or, they could also be alpha particles. The way we’ll figure out which one of these particle types is actually being emitted will be based on the charge value of these particles that we’ve solved for previously. We know that whatever these particles are, they have a charge of 3.2 times 10 to the negative 19th coulombs. In other words, a relative charge of plus two, two times the charge of a proton.

Now if these particles have an overall charge of two times the charge of a proton, then it makes sense that we would imagine that two protons are contained within these particles. And it’s at this point that we can recall that an alpha particle, also known as a helium nucleus, is a particle that has two protons as well as two neutrons in it. Now, since these neutrons, the green dots, have no electric charge, they don’t contribute anything to the overall charge of this particle. It still has a charge of plus two. So this is an alpha particle, two protons plus two neutrons. And the charge of an alpha particle, we see, is the same as the charge we solved for, 𝑞. This charge is different from the charge of a beta particle and, of course, different from the charge of a neutron, which has no charge. This then is our answer to this final question. The particles being emitted by this block of unknown material are alpha particles.