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How much work must be done to pull apart the electron and the proton that make up the hydrogen atom if the atom is initially in (a) its ground state and(b) the stat...

Question

How much work must be done to pull apart the electron and the proton that make up the hydrogen atom if the atom is initially in (a) its ground state and(b) the state with $n=2 ?$

How much work must be done to pull apart the electron and the proton that make up the hydrogen atom if the atom is initially in (a) its ground state and (b) the state with $n=2 ?$



Answers

How much work is needed to assemble an atomic nucleus containing three protons (such as Li)
if we model it as an equilateral triangle of side $2.00 \times 10^{-15}$ m with a proton at each vertex? Assume the protons started from very far away.

Our question says that at first A wants us to calculate how much work is required to accelerate a proton from rest up to a speed which e 0.998 times to see the light and then for V, what would be the momentum of this proton? So also, write down here some other useful constant values that we're gonna use. And that's that M C squared, which is the, uh, rest mass arrest energy of the proton as equal to 938.3 MTV. And then I just went ahead and rearrange that to calculate the mass and units of any of you Percy squared by dividing, um, by dividing that value by C square dance that's 938.3 MTV Percy Square. So first we're gonna calculate the work, and the work is given by the change in the kinetic energy. This is part of a but our initial kinetic energy is zero. So the work here is just the final kinetic energy. We'll just call this K f, which is equal to gamma minus one times m c squared. So we have that m c squared value written above its 938.3 MTV and gamma is one divided by the square root of one minus. V squared over C squared. Well V squared to 0.998 c. And that's got RV A 0.98 c. So v squared is your Where'd so the c squared? They're gonna cancel. You're gonna have 0.9 98 square extended. Include all that notices minus one multiplied by M C squared, which is 938.3 MTV. Okay, so if you, uh, plug this in the calculator, this comes out to be 1.39 times 10 to the fourth. Any of the which is a prevalent two, um 13 g. Okay. Linkenbach send in, Mr Solution. Apart, eh? Now for part B were asked to find Ah mo mental. We can use the equation for calculating relevant relativistic momentum, which is gamma envy course. Gammas one over the square root of one minus V squared over C squared. So this is gonna be one over the square root of one minus 0.998 squared. That's value squared. Oh, and it looks like I forgot toe. Right. That's squared. Above course. It's funny about you squared. Okay, Might and Mrs Gonna be multiplied by the mass, which we found to be 938.3 and maybe her c squared. That's maybe then also multiplied by V, which is 0.998 times this feat of light, that's all. Okay, so if you plug all that into your calculator of they were gonna cancel your feeling, everything else in your calculator and you're gonna have in units of MTV. Percy, it's going to be 1.48 time sent to the four and maybe Percy. Okay, Justus above. We're gonna put that units of G E. V s. We move the decimal place over one to the rights. This is gonna be 14.8 times 10 to the five or 14.8 G. Percy, I'm boxed in his resolution. You're happy

So the three protons woman equilateral triangle. So I'm gonna go ahead and draw a little diagram to help us. Better understand this problem. No, these dots actually represent the protons. And how long? Let me draw third one. And as you can see, these three people protons have formed an equilateral triangle. Let me just have a The radius is between the two to help. You better understand. So has called this distance are. And since it's an equilateral triangle, the radius is the same for all of them. So let me just table the charges you want. You want 2 to 3. What you want you 2 to 3. All right, so now we know the radius is equal to, uh, two times tend to the hole on a me ri, right? You know, at the radius, which is given to us, it's two times 10 to the negative. Negative. 15 years. I obviously this very small distance. But it's also important to note that the distance between all the three charges are, uh, equal to the radius and are constant for all of them. So now, to find, uh, the total amount not to find the month of work for one done by one charge. We know that we know the equation is the Monta work is equal to whom constant times charged one charge to over the radius except rating them. So this is the the work done between the two charges. So if you have three charges, there is a total off, uh, three different exertions of force. So we must add the total weight. I mean, a total work done. So if you were to add the total weight dimensions uh, the charges of the same the radius of the same, you can just I want to play disc on this value by a constant of three. And again, since the charges of the same, we can just make it cute squares since Charge one in charge two are the same thing Que squares three. These air brackets over the whole on the radius is also included in the brackets. You hear me? We could do each going start over right. So three times who is constant times that you charges Q squared over the radius. So this is the total rectum. And now it's here to just plug in concept that we have so cool is constant. She all knows 8.99 times 10 to the power of nine making meters squared over over coons. Weird for the units three times 8.99 Kaine's 10 to the power of nine. And then the other. Charges for electrons are 1.6 times, uh, 10 to the power of 19. Negative 19 These other charges. And we know that this whole thing is square because there are two charges. So it just credit How long? Let me let me get that all of this, right, that a little bit nicer. Negative. 19. All right. And then over the radius, which we already said was two times 10 to the negative. 10. All right, so this is the final equation along that me, I mean, what this little clear. So it's tend to the part of power of Negative 15 over here. Sorry, Mollari here, but yeah. So if you plug these injured help, better get that. The work is equal to three point for Dr to Jules. Now, if you want to convert these into mega electron lost votes, any bees, then you have to just divide this by at one point 1.60 to times 10 to the Sorry. Hold on. This was 3.452 times. So I have for getting a couple digits here. So this is 3.452 times come to the negative. So if you want to convert these into anybody's and we just need to divide by one point there six times 10 to the negative 13 you do that. Okay, so if you put this into a club later, we're going to get that the the the total number, the total work in movies is 2.16 and maybe Yeah, so it's equal to 2.16 megawatt troubles E B. Yeah. So that's it. Thank you for

In this question. Okay. Because we have two electrons Yeah, initially fixed in place to micrometer part. We want to calculate the work done to bring the electron common impunity to this point to form an equally equilateral triangle. Okay, Okay. So to solve this problem. Okay. We'll be using the relationship. Yeah. Um look done because to changing potential energy which is two times their T. V. Yeah. And uh QB F minus uh yes minus V. I. Okay, so the I is equal to zero. Okay. Because the uh electron uh at infinity. Okay. And then yeah. Okay. Yes. Uh this is the final location. Okay. So the BF due to the other to fix electrons. Yeah, at the final point is equal to uh minus two E over four pi X or not. Uh huh. The Okay. Where the is equal to two times 10 to to get these six m to Micro King. So, uh there's two because there are two charges. Okay? And then uh so we can actually calculate w no W C. Good to um, Q B f minus E tends to eat -2 E four Pi Excellent. Not mm. Okay. So to ease where over four pi X or not putting in the numbers, he and we can do this, you're going to get the answer to be 2.30 Times 10 to the -22 Jews. Okay. So this is the work done to bring the electron to form an equilateral triangle. Okay. And that's all for this question.

For about a off the problem. We apply the work and take Innercity room. That is work Easy call Thio Changing can take energy were K for literacy kinetic energies, M E C square game or minus one and M e C square is the rest of energy of electron reaches 0.511 Mega Megan Electron world. Then we can write a chain. Gigantic energy, which will be called to the work, is equal to Ah MPC Square Gamma final minus gamma initial are substituting a gamma in terms off. A better Then we could write Ah, work physical changing kind of energy. This is equal to M p C square one all one word one minus a bit of final square minus one word one minus better initial square. Also substituting your values for the rest. Energy off electron, which is a 511 kill electron volts divided by one minus scuttled off one minus 0.19 square, minus one over square root off one minus 0.18 square and this fuse us the work. Or for you know 0.996 killer like Grown world, which is approximately one kill electron world. I want to be off the problem. Um, the world done for, uh, given speed. Then again, using the same question is w musical to 511 killer that wrong world. Arm one, divided by square. Root off one mine is there a point 99 square minus one work square root off one minus 0.98 square. Excuse us. 1055 Kill electron Walter, which is approximately 1.1 mega electron world. We see ah from the two answers that there is, ah, dramatic increase in difficulty trying to accelerate a particle when it's this initial speed is very close to the speed of light.


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