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(II) Many chemical reactions release energy. Suppose that at the beginning of a reaction, an electron and proton are separated by $0.110 \mathrm{nm},$ and their final separation is 0.100 $\mathrm{nm}$. How much electric potential energy was lost in this reaction (in units of eV)?

1.3 $\mathrm{eV}$

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Cornell University

Rutgers, The State University of New Jersey

University of Michigan - Ann Arbor

McMaster University

Okay. So in this problem, we are trying to determine the loss of electric potential energy between two situations where a proton electron are near each other. Take this electron proton. The proton is moving towards the electron. So this is a distance here. We'll call this charge one Q one in this charge. Too cute, too. And this is the distance between them are one too. So we can write the change in potential energy as big Delta U, which is just the potential energy at the final position. B minus the potential energy of the original position A. And so we know from the problem statement that are a this our one to situation is your point one one nanometers and our be the final position as your 0.1 nana meters because the proton has moved closer. Okay, So first we need to look up the equation for electric potential energy you, which is just one over four times pi times the constant absolute nut times key one times cute too divided by the distance between them are one too. Okay, so let's see. Eve our equations over here and let's start plugging in. So we're solving for adult to you, Which equals you. Be first. So let's write that one over for pie. Absolutely not. Q. One eyes are electron, so we know that that's the charge of an electron. Q sub. E times The charge of the proton, which is just the opposite of the charge Mint electron. So it's minus. Q t divided by the distance in, um, the final distance are be said so your 0.1. Well, let's keep using symbols for now. Okay, it's right, R b and then we're subtracting the potential energy at position A where we have the same charges involved about it. Bye. Our new distance, our old distance rather are let's simplify. This could take other one over four pipes or not. You could also multiply this times minus q t squared, since that's the same in both situations. And then we have a one over R B minus one over R. A. Okay, so now we can plug in our Constance. You're the one over four times pi, which we can round to 3.14 You look up in the back of our book are absolutely not. That's 8.85 times 10 to the minus 12. Coombs squared over Newton's times. Meters squared. Okay, times we have a minus. Sign here, Charge Mint Electron. We can also look up. That's 1.6 times to minus 19 Coombs Square. What? We have insider parentheses multiplied by the rest of that, says one over. What's changed these two meters Too much. The rest of our units, so are B is your 20.1 animators. So the 0.1 times 10 to the minus nine for nano meters minus one over to your 0.11 This are a I'm still on minus nine meters. Okay, so let's plug in all this math and your calculator we get is 2.9 times 10 to the minus 19. And let's look at our units here. So we have Let's make this clear this all under our denominator. So this cool MB squared gets cancelled out with the cool, um, squared up here so you don't have any qualms. This gets flipped because it's on the bottoms. That's Newton's times meter squared on the top. And then we have a meters to the minus one over here. But we know that this just equals meters to the plus one meters. So this is equal to 2.9 times 10 to the minus 19 Newton meters. But this unit, we know better as Jules. But the question is asking us what the loss of energy is. This is a minus. Sign here. Let's not forget that we have that from the charge of the electron here. Um, the question asked is asking us for the loss of potential energy. That's why we have a negative here. So it's really just asking us for the magnitude of this problem. So the last thing we have to do is convert jewels to electron volts. So you're right to 0.9 Time Stella minus 19. And we need our conversion factor from Jules. Two hoops from Jules to electron volts. So we have our current answer and Jules times or conversion factors. Who No one electron volt is equal to multiply anything by one 1.6 times 10 to the minus 19. Jules, cancel out our jewels and what we get. Minus 19. He's also get canceled. Help turn the minus 19. So it's just 2.9 Divided by 1.6 gives us a minus 1.3 electron volts questions asking us for how much energy is lost. So you really just need the value, which is 1.3 electricals.

Rutgers, The State University of New Jersey