### Video Transcript

An electron orbiting in a hydrogen atom is separated from the atom’s nucleus by a distance of 6.32 times 10 to the negative 11th meters. What is the magnitude of the electric field of the nucleus at the point where the electron is located? What is the magnitude of the electric force on the electron from the nucleus at the point where the electron is located?

Okay, let’s start off making a sketch of this hydrogen atom. Hydrogen, we see, if we look it up on the periodic table of elements, comes very first. It’s atomic element number one. We often find that hydrogen also has an atomic mass number of two. This means that the nucleus of the hydrogen atom consists of a single proton and a single neutron. And then, in addition to the nucleus, there’s a single electron that orbits around the center.

Our problem statement tells us that the orbital’s distance of this electron from the center of the nucleus, what we’ll call 𝑑, is equal to 6.32 times 10 to the negative 11th meters. knowing this, we want to answer two questions. The first is, what is the magnitude of the electric field of the nucleus at the point where the electron is located? What this question is saying is, because the nucleus has an overall charge, the neutron doesn’t have a charge but the proton does. Then, it must create an electric field around itself. We can recall that the electric field magnitude, created by a charge 𝑄 a distance 𝑟 away from that charge, is equal to 𝑄 times Coulomb’s constant 𝑘 divided by 𝑟 squared.

In this first question then, we want to apply this relationship where our charge 𝑄 is the charge of the nucleus of our hydrogen atom. And what is the charge of the nucleus? We said that the nucleus is comprised of a neutron, which has no charge, and a proton. The proton does have charge. And it’s equal in magnitude to the charge of an electron. The charge of a proton, we’ll give it the label 𝑞 sub 𝑝, is equal to 1.6 times 10 to the negative 19th coulombs. So then, that electric field created by the nucleus is experienced by the orbiting electron. And that magnitude of electric field is equal to Coulomb’s constant 𝑘 times the charge of a proton divided by the distance 𝑑 between our electron and nucleus.

We’re far enough along now that the only unknown in our equation is Coulomb’s constant 𝑘. When we look up the value of that constant, we find it’s 8.99 times 10 to the ninth newton-meter squared per coulomb squared. With that, we’re now ready to solve for the electric field magnitude 𝐸. When we enter in our values for these terms, notice that, in terms of the units, one factor of coulombs cancels out from our numerator. And both factors of meters cancel out from this expression. What we’ll end up with, this is telling us, is an electric field with units of newtons per coulomb. When we calculate it out, we find a result, to two significant figures, of 3.6 times 10 to the 11th newtons per coulomb. This is the magnitude of the electric field created by the nucleus at the point where the electron is located.

We now move on to our second question. That question says, what is the magnitude of the electric force on the electron from the nucleus at the point where the electron is located? So earlier we solved for the field. And now we want to solve for the force. Considering electric force, let’s write out the equation for that force. And we’ll see an interesting connection. The electric force, 𝐹 sub 𝑒, between two charged particles with charges 𝑄 one and 𝑄 two, is equal to the product of those charges multiplied by Coulomb’s constant divided by the distance between them squared.

Let’s say for a moment that 𝑄 one is the charge of our nucleus, that is, the charge of the proton and that 𝑄 two is the charge of our electron. In that case, this part of our expression for electric force, 𝑘 times 𝑄 one over 𝑟 squared, is equal to the electric field created by the nucleus at the position of the electron. And to see that that’s the case, compare it with our equation for electric field that we used earlier. So this means, when we’re solving for the electric force, we’ll call it 𝐹 sub 𝑒, then we simply need to multiply the electric field created by the nucleus multiplied by the charge of the electron. We’ll call it 𝑞 sub 𝑒.

The charge of an electron is equal in magnitude and opposite in sign to the charge of a proton. So, when we plug in for 𝑞 sub 𝑒, we would use negative 1.6 times 10 to the negative 19th coulombs. But, our question asks us for the magnitude of the electric force. So let’s do this. We’ll use the charge of our electron, which indeed is negative. But, we’ll take the absolute value of that charge. And then, to solve for the electric force, all we need to do is multiply this charge magnitude by the electric field that we solved for earlier. When we multiply these two values together, we find a result of 5.8 times 10 to the negative eighth newtons. That’s the electric force magnitude on the electron from the nucleus.