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Ch 37: Special Relativity

Chapter 36, Problem 39

(a) The x-coordinate of an electron is measured with an uncertainty of 0.30 mm. What is the x-component of the electron's velocity, vx , if the minimum percent uncertainty in a simultaneous measurement of vx is 1.0%? (b) Repeat part (a) for a proton.

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Hey everyone. Let's go through this problem. Suppose the position and velocity of an alpha particle are measured simultaneously. The position is determined with an accuracy of 0.65 centimeters along the Y axis. While the minimum percent uncertainty on the alpha Y component velocity is 3%. You said y part one, calculate the alpha particle's Y component velocity. Part two asks to suppose that now an electron is observed under the same conditions. Calculate the electrons y component velocity. We have four multiple choice options to choose from option A for V sub Y alpha 4.1 and multiplied by 10 to the power of negative five m per second. And then for part two V sub Y for the electron of 20. m per second. Part B for V sub Y alpha, 8.2 multiplied by 10 to the power of negative five m per second. And then for part two V Suby electron is 20. m per second for option CV sub Y alpha is 2.1 m per second and the sub y electron is 1.5 multiplied applied by 10 to the power of four m per second. And lastly, option D says that V sub Y alpha is equal to 3.8 m per second. And V sub y electron is equal to 2.8 multiplied by 10 to the power of five m per second. Because this problem is having us look for the Y component velocity and in relation to uncertainties and accuracies of position. This is a problem we will want to make use of the Heisenberg uncertainty principle which states that the uncertainty of a position in our case, delta Y because we're looking at the Y axis multiply the uncertainty in momentum. In this case, delta P sub Y is always greater than or equal to the plant constant divided by four pi. So let's use this equation, the Heisenberg uncertainty principle to find the velocity note that the problem says that the minimum percent uncertainty on the velocity is 3%. So that means that the uncertainty in the velocity and the Y component is equal to 3% of the measured velocity. So in in decimal form, that's 0.3 multiplied by the velocity, which we want to find also note that since momentum is what's actually in the Heisenberg equation recall that momentum is equal to mass multiplied by velocity. So the uncertainty in the momentum is going to be equal to the mass. In this case, the mass of the alpha particle multiplied by the uncertainty and the velocity. So if we plug in the uncertainty and the velocity we found that incorporates the 0.3. Then we can see that the uncertainty in the momentum is equal to 0.3 multiplied by M alpha multiplied by V sub Y. Now let's put these into Heisenberg's uncertainty principle to solve. For V sub Y note that even though the uncertainty principle has a greater than or equal to in the equation, we can still use this as a simple equation by realizing that we're given the minimum uncertainty. This means that if we use this uncertainty, then it should be equal to H over four pi or H divided by four pi because that is equal to the minimum possible percentage that we can have. So the uncertainty in Y multiplied by 0. M alpha, V sub Y is equal to the plant constant divided by four pi. And we're solving for V sub Y because we want to find the velocity. So we can find that by dividing both sides of the equation by delta Y multiplied by 0.3 M alpha. So doing that, we find the velocity of the particle V sub Y is equal to H divided by four pi multiplied by 0.3, multiplied by M sub alpha multiplied by delta Y. Now, before we can put anything into our calculator first recall that the plant constant H is equal to 6. multiplied by 10 to the power of negative 34 dual seconds. Also recall that the mass of an alpha particle M sub alpha is equal to about 6. multiplied by 10 to the power of negative 27 kg. Also recall that delta Y the uncertainty in the particle's position is given as 0.65 centimeters. So delta Y is equal to 0.65 centimeters which converting this into units and into meters to make our units consistent also equal to 0.65 multiplied by 10 to the power of negative two m. All that's left for us to do is to plug each of these values into the equation that we wrote on the left of the V sub Y equation to solve for the velocity of the alpha particle. And if we do that, then we find a value for the alpha particle speed of about 4.1 multiplied by 10 to the power of negative five m per second. And that is the answer to the first part of the problem in solving for the speed of the alpha particle. The second part of the problem asks for us to do the exact same thing but with an electron instead of an alpha particle. So for the second part of the problem, we're using the mass of an electron instead, I'm gonna switch colors to indicate that it's otherwise gonna be the exact same formula because it's the same process. So V sub y or an electron, it's equal to the same equations. The plant constant divided by four pi multiplied by 0.3, multiplied by the mass of the electron multiplied by the uncertainty in the electrons position. And this time, so this time, the only thing that's changing is the mass of the particle we're looking at. So recall that the mass of an electron is equal to 9.11, that's 9.11 multiplied by 10 to the power of negative 31 kg. And when we put that into our calculator, instead, we find a value for the electron speed of about 0.30 m per second. And that is the answer then to the second part of the problem. So our answer to the first part of the problem was 4.1 multiplied by 10 to the power of negative five m per second. And our answer to the second part was 0.30 m per second. If we look at our options, we can see that both of these choices agree with option A which states that the volume of the alpha part, the velocity of the alpha particle is 4.1 multiplied by 10 to the power of negative five m per second. And V sub y electron is 50.30 m per second. And that is it then for this problem, I hope this video helped you out if it did, please consider checking out some of our other videos, which will give you more experience with these types of problems, but that's all for now and I hope you all have a good day. Bye bye.
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Textbook Question
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Textbook Question
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