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(III) Estimate the maximum allowable pressure in a $32-\mathrm{cm}-$ long cathode ray tube if 98$\%$ of all electrons must hit the screen without first striking an air molecule.

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$7.3 \times 10^{-3} \mathrm{Pa}$

Physics 101 Mechanics

Chapter 18

Kinetic Theory of Gases

Temperature and Heat

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okay. In this problem, we have a 32 centimeter long tube, specifically a cathode tube. Okay. And we are assuming that we wanna have 98% of the particles hitting the screen. You know, coming from here to here. You want 98% of them to hit the screen on the other side without ever colliding with another particle. And this is pretty practical. This is kind of what you want to accelerate particles in the cathode ray tube. You don't want to deflect it. Besides, you wantto have a clean acceleration the entire way, or at least most of them. So we're asked to find the maximum pressure to make this happen. So this is a problem about mean free pads. And it's a somewhat complicated arguments. We know. So we have to walk through the argument before we even start calculations. So first we start with our argument. If we want 98% of all of the particles to go across the ray tube without any collisions, that's were saying that there's 2% we'll have a collision within 30 centimeters. 2% collision within 32 centimeters. If that's true, it must be the case that 50% that 50% of the electron should have a collision 25 times this 2%. So to get so 50% that 50% of them colliding 50% of I'm not colliding. We take our 2% figure of 32 centimeters multiplied by 25. Okay, and this is about eight meters. So follow this. Stepped from this step to the step for 2% have a collision within 32 centimeters. We want 50% getting a collision. It must be within 25 times 32 which is eight meters. So this is a kind of an operational definition of marine free path. It's the path that it would take. It's the average path, average length for a particle. We have a collision which good idea of an average is 50% client, 50% don't. So this would be a mean free path. It's the length that we would expect at least half of them to have a collision and the other half not to have a collision. So once we have that argument sort of settled, you now have to have two formulas. We need the ideal gas law PV equals and Katie and we need to have the mean free path formula involving some other quantities. It's one over pyre squared and times V and over. Be rather so from there we can say that began over. V is just ah, pee over Katie. You combine that with the mean free path expression to give us Katie over pi r squared p pi r squared okay, and we assumed room temperature. I believe we can assume for this problem. Usually do not operate cathode ray tubes that other temperatures there just in the room in the estimate temperature to be about 300 k. And we know everything else about this problem. Inside of the radius are particles is about 1.5 times 10 the negative 10 meters. That's a good estimate for a molecule, and our mean free path is just eight meters using kind of the argument that we developed in the previous page. Thus, we can isolate P to be Katie over pi r squared keep saying pi squared. Is pi r squared? Of course there's l. We can play in numbers that the bolts been constant. 1.38 10 to the negative 23 jewels for Calvin. I'm not gonna include the units just to keep things a little cleaner. Times 300 kelvin pie The radius 1.5 10 to the negative. 10 meters squared, mean free path used to be eight meters. When we do the number crunch, we end up with a pressure of 7.3 tended the negative three Pascal, which in atmospheres is about seven times 10 the negative six atmospheres. So in other words, if we want to have a nice Catholic ray tube where we have minimal collisions and we get and so each particle has a lot of time, a lot of distance to accelerate in we need to have this tube evacuated. It needs to be roughly vacuum. We're close to vacuum, right, because there is no such thing as a perfect vacuum. But this is, he said, very, very small pressure. That's it

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