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Singly ionized carbon atoms are accelerated through $1.00 \times 10^{3} \mathrm{V}$ and passed into a mass spectrometer to determine the isotopes present. (See Topic 19.) The magnetic field strength in the spectrometer is 0.200 T. (a) Determine the orbital radii for the $^{12} \mathrm{C}$ and the $^{13} \mathrm{C}$ isotopes as they pass through the field. (b) Show that the ratio of the radii may be written in the form

$$\frac{r_{1}}{r_{2}}=\sqrt{\frac{m_{1}}{m_{2}}}$$

and verify that your radii in part (a) satisfy this formula.

a. \text { The orbital radius is } 7.9 \mathrm{cm} \text { for } 12 \mathrm{C} \text { and } 8.2 \mathrm{cm} \text { for }^{13} \mathrm{C} \text { . }

b. \begin{array}{l}{\text { The orbital radius and masses are related by the equation } \frac{r_{1}}{r_{2}}=\sqrt{\frac{m_{1}}{m_{2}}} . \text { This }} \\ {\text { may be verified by substituting one of the previous radii in to the equation. }} \\ {\text { For } 13 \mathrm{C}, m_{2} \approx 13 \mathrm{amu} \text { , and } r_{2}=8.2 \mathrm{cm} . \text { Then } r_{1}=r_{2} \sqrt{\frac{m_{1}}{m_{2}}}=8.2 \cdot \sqrt{\frac{12}{13}}=} \\ {7.9 \mathrm{cm}, \text { the same result predicted by the calculation in problem } 6 \mathrm{a} \text { . }}\end{array}

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Numerade Educator

Hope College

University of Sheffield

University of Winnipeg

number eight. I have a carbon Adam that's going to go through a voltage of 1000 volts that's gonna accelerate it. And then it's going to go on a magnetic field with strength point to Tesla, and I wanna know what's the orbit of that radius. So remember, when I have a charged particle in a magnetic field, so, like, this is the velocity of it, the force is that a right angle? Right hand rule tells us that so that this city is kind of going into the page. Um, I should roll. That is an ex, maybe, um so that's called him to go around. So that is the centripetal force. So the centripetal force is calls by the field Force. So some triple forces M v squared R. And this is what I'm trying to find. Or but the radius of the orbit and, uh, the force from a magnetic field iss the charge times of velocity, times the field strength. So I'm just gonna rearrange this equation. Maybe if you're stuck, cancel a factor of the velocity. So now I'm gonna rearrange and get this are by itself. Some brains are over here. Um, so have, um, the on top oven. Q Be on the bottom. Um, so I know what this It's a carbon, Adam. I could figure out the mass in charge of it. I was given the field strength, but 19 other expression for the velocity. So let's come over here and think about that. Um, so it originally got into velocity when it went through this voltage. So the kinetic energy that it got is from the work done by that voltage and work will be Q times V. Um, so I'm gonna just solve this for that. This velocity, um, multiply both sides by two. So now you get that may be squared. Will be to Q, nothing over them. And that was squared t square root of both sides. So that's my expression. Little unflagging over here for this movie. When I do that, I'm gonna bring everything inside this radical and you'll see why. Um so my r he's gonna equal this Mmm. Were taken inside of radical suits, squares at my V, which is this whole thing, which is gonna be a two Q. The on top and I am on the bottom. And then this Cuban I bring in is gonna be accused squared, and this be is gonna be a B squared so you can see why I want to do that. Because now I can cancel some things. I can cancel one factor of the mass, and I can cancel one factor of the charge. So now my expression for the orbital radius will be two, um, really over Q B squared. So right now I want to stop and talk about port. Be a second. Notice that everything in here is constant except the M. So I have, um, things under constant Obviously, number two, the V because that's going to be appear. We were given that that's gonna be 1000 of the queue. It said in the problem that they were Singley charged. So in the queue will be just the elemental charge Q And the magnetic field squared. The B will always be this 0.2. So the only thing in here that varies is the M. So this expression, you know, think of it of this are or is proportional to this, um but the square root of em. So yeah, that's the proportions work I could set two of those and making equation. Now welcome back to the after I solved. Now I'll come over here and come up with my values from using this equation. So for the carbon 12 the radius will be two sums. The manse, which well square root of all this to tempt the mass, which would be 12 of these new plants of 12 times 1.66 times in the 27th. That's in kilograms times uh V, which we were given once I'm sent. The third was going to say 1000. All of that over. Oh, charge. It said they were singly charged. So that's just gonna be the oh, mental charge 1.6 times time that you were 19 times the magnetic field strength squared, so point to squared. And when I do, carbon 13 is gonna be the same thing. Just I'm just changing that 12 or 13 in the mass. So thinking that she's here, just take this, put it down here and then just change this 12. So 13. So now I'm gonna solve appear my arm WAAS point 0789 We'll be in meters and down here My radius was point. Oh, 8 to 1 meters. So now to come back here and finish part B, they want to show that this really works. Um, so my radius one was this So point 0.789 My radius. 2.8 to 1. My masses, I could multiply, but with a top and bottom by this, I'm just gonna recognize that's gonna cancel. So I'm just gonna put 12 over the 13. And sure enough, this side, this 0.961 and this side is 0.961 So I did verify that equation does work.

University of Virginia