# Interatomic Forces Example Problem

Okay, for this question we're asked to calculate the Coulomb, big attractive force between a pair of sodium, plus and chlorine, minus ions that are just touching each other. They give us information about how big these ions are. And they tell us that the sodium ion is 0.95 nanometers while the chlorine ion is 0.181 nanometers. So to begin on a question like this since we're asked about the Coulomb Bush attractive force, we should start by remembering what does that attractive force look like and. From physics, you probably saw an equation. Look like this. If the attractive force between ions is equal to q1 times q2 over R squared.

And this is all multiplied by K. So when we talk about inter, attract inter atomic attractive force is it's going to look very, very similar we're going to write it instead as this we're going to write negative, Z 1 times, C 2 multiplied by the charge of these electrons squared divided by 4 PI, epsilon, naught multiplied by the separation distance squared. So all that we've done. Is this term right here came from K. We put a negative out in front because we're going to account for both the positive and negative charges and our ions. And then Q is just the charge of the electrons.

So let's, go ahead. We could go ahead and solve this. But what we need first is to solve for what R is so let's first start by solving for what R is we're told that the radius of the sodium ion is equal to 0.095 nanometers.

Whereas the radius of the chlorine ion is equal to 0.18 1 nanometers. And we. Know that the bond distance let's call it R, naught that's going to be equal to the radius of sodium, plus the radius of the chlorine because we're told in the problem that's, when the ions just touch one another, therefore, when we add these together, 0.095, plus 0.1, 81, I, find that that's equal to 0.276 nanometers. Now we can convert that to meters pretty easily. We just remember that we can multiply this by times 10 let's see to the negative ninth meters per nanometer. If we want to make these. Nanometers cancel out, and that will tell us that we have two point, seven six times 10 to the negative 10 meters is our separation distance.

So now we know our this value here you can plug it in, and we can solve for our attractive force let's do so the force attractive is going to be equal to negative. The first charge of the first ion is positive 1, the charge of our second, our anion Sigma n is negative 1. These are both times the charge of an electron. The fundamental type of electron is one. Point six zero, nine, just at one point, six zero, 10 - 10 to the negative 19, coulombs that's squared divide this by 4 times pi epsilon, naught epsilon naught is the permittivity is the permittivity of free space, which is going to be eight point, eight, five times ten to the negative, twelve that's units of coulombs per Newton meter squared. We multiply this finally, by two point, seven six times ten to the negative 10 meters, and we're going to be squaring that plugging all those values in I find.

That we reach an attractive force of positive three point zero, two times ten to the negative ninth Newtons so that's, our attractive force. This question asked us to solve calculate, the Coulomb look attractive force, so we're done with Part A. Now we can go ahead and move on our B. Part B says, calculate the net potential energy of a simple sodium chloride ion pair by using this following expression, where the net energy is equal to the attractive energy component, plus B over R to the N, where N equals.9 for sodium chloride. So that second term in the energy expression is what's known as the repulsive energy term. Let's recall, a few things first off.

If you have an energy term, II met equals let's just call it EA for a moment, plus B over R to the ninth that was R to the N, but they told us that N equals nine. We can figure out what the forces would be by taking the derivative of this expression. Remember that force equals the derivative with of energy with respect to inter, atomic separation, D R. So if. We take the derivative of this, we can write that the net force is equal to DE let's, just write DE a. Dr. And then this will be this is going to become. It would be minus nine times B over R to the tenth. So now I have an expression for our force. So what were you can next do is we remember that at equilibrium when these ions are touching one another that the net force is equal to zero at equilibrium F net equals zero because the ions aren't moving apart or closer together, if there is a net.

Force, they would be moving either closer together or further apart. But since they're at equilibrium, we can set that equal to one another, and we already solved for the force, the attractive force there for any change colors. This whole term here that is simply equal to our attractive force, which we solved for in the first part of this problem that makes a lot easier let's. Go back. Therefore, we can write the following. We can write that. Zero equals three point zero, two times 10 to the negative ninth.

Newton's minus 9 times B over again. This is our separation distance to point, seven, six times 10 to the negative 10 meters. This is raised to the 10th power right? We can go ahead and bring one of these to the other side. And then multiply both sides by two point, seven six times 10 to the negative 10 meters.

And this will allow us to solve. Well, let's just write it out. This lets us write that 9 B equals C three point zero, two times 10 to the negative, 9 Newtons that will be multiplied by two point. Seven six times 10 to the 10 meters that's, all raised to the tenth power. If we then divide this by 9, we get rid of in there and that's going to be the value for B. So let's, go ahead and plug those values. When I plug those values in I find that B is equal to let's.

See eight point, five nine times ten to the negative, 106 eight point, five nine times ten to the negative, 106 power. And the units are going to be Newton's times meters to the tenth. So now we have an expression for B. So let's, go back up to. What we were originally asked for, they want us to solve for the potential, the net, potential energy using this expression.

Well, we know what B is now we know, R, we know n. We need to figure out what EA is. Well, we were given FA. And we know that this relationship exists that the force is equal to the derivative of energy respect to the inter atomic separation. So now we need to know now you have to go backwards. We know that F an equaled negative Z 1 times, Z 2 times the charge Q squared multiplied by 4. Pi epsilon naught R squared.

If we take the integral of this with respect, excuse me, so if we take and write that the attractive energy component is equal to the integral of F a. Dr, we find that it is equal to positive. Z 1 times, Z 2 times Q squared divided by 4 PI epsilon, naught times R we're, almost there and ready to solve for the final expression. Now, the net energy is equal to Z 1. Z 2, Q squared over 4 PI epsilon, naught over R. This will be plus R, repulsive energy term, which we now know since we. Know, B, this would be plus eight point, five nine times ten to the negative.

One hundred sixth power Newton times meters to the tenth all divided by R to the ninth power. Plugging values in we find that this is then equal to. So the two terms are equal to negative.

Eight point, three four times 10 to the negative 19 joules, plus zero point, nine, two times 10 to the negative 19 joules and combining those together. We finally find that the net energy is a negative. Seven point, four two times 10 to the. Negative 19 joules, or in other words, it's a negative value. This means that when these ions come together, the net energy change actually goes down.

And when things go down in energy, that's, spontaneous and favorable, therefore, we can assess the depth of the potential. Well.

• Times Ten , Newton Times , Times Meters , Net Force , Force Equals , Sodium Ion