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Problem 38 Hard Difficulty

In a nonrelativistic experiment, an electron and a proton are each located along the $x$ -axis to within an uncertainty of 2.50$\mu \mathrm{m}$ . Determine the minimum uncertainty in the $x$ -component of the velocity of (a) the electron, and (b) the proton.

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Top Physics 103 Educators
Elyse G.

Cornell University

Marshall S.

University of Washington

Farnaz M.

Other Schools

Jared E.

University of Winnipeg

Video Transcript

Okay, so we want we We know the electron proton haven't uncertainty in their position of 2.5 my craw craw Mater's. So let's write that Delta X is he cool to 2.50 micrometers. And, um, we want to get the uncertainty and the X component of the velocity. So we want to use the uncertainty principle. Delta X, delta p equal th over for pie. So then delta V is equal toe are sorry. Delta P is equal to am Delta V and then Delta v ous you cool? Is that what we wanted? Um, I guess I don't. Delta V is equal to age over for pie. Um, Delta X um great. And so we just want a value a thought for except for the using. In either case, which is in either case, we're gonna change the mass of the elect the mass to the mass of the electron. Um, let me use for my and I feel like I made that sentence a lot more complicated. We're going to evaluate this for two different masses the mass of the electron and in the mass of the proton. So for the mess throat for the electron. Go ahead and plug in 9.1910 minus 31 for the mass. Okay, let's go to my calculator and do you live, so? Oh, you think I have this exact problem from a different? I think this is very similar to another 10 yeah. Nice. So I've actually done this calculation before. Let me just double check so that I got 23 meters per second and then Delta view for the pro time. Um, I just need to change the mass. If I said earlier Gotham I thinkit's gets 1.67 times seven minus 27. But let me just double Chuck. Yeah. One points. Roll. 672 I guess 673 if I want to add another sigfig 673 Except I don't need all that sick. Thinks so. Okay. And so with that, I got point. Oh, point. How many six pigs were in the problem. 36 figs, 360.1 to 6 meters per second.

University of Washington
Top Physics 103 Educators
Elyse G.

Cornell University

Marshall S.

University of Washington

Farnaz M.

Other Schools

Jared E.

University of Winnipeg