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Problem 40 Hard Difficulty

(a) Show that the kinetic energy of a nonrelativistic particle can be written in terms of its momentum as $K E=$ $p^{2} / 2 m .$ (b) Use the results of part (a) to find the minimum kinetic energy of a proton confined within a nucleus having a diameter of $1.0 \times 10^{-15} \mathrm{m} .$

Answer

a. K E=\frac{p^{2}}{2 m}
b. 5.3 \times 10^{6} e V

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Top Physics 103 Educators
Christina K.

Rutgers, The State University of New Jersey

Andy C.

University of Michigan - Ann Arbor

Marshall S.

University of Washington

Jared E.

University of Winnipeg

Video Transcript

Okay, so we want to get the the We wantto show that the kinetic energy you could be written his p squared over to, um so generally kinetic energy is written as one happened B squared, But let's go ahead and try to write of this p squared over to him and make sure we get back. The one happened B squared. So I'll just write p squared over two m and then peace where that's m v squared over to, um and then wearing both of these, you get m squared, you squared, and then you have ah, and am down here. That Castle's willing those EMS and so that could be 1/2 on B squared. Um, so yeah, that works. And we wanna find the minimum kinetic energy if the of a proton, if the it's confined to a diameter 1 10 of the minus 15 meters. So Delta Axe is equal to one times 10 to the minus 15 years, and we want to get what was that? The minimum kinetic energy. And so Okay, so we want to use start off with the relationship that, um, see Delta X times Delta P is equal to h over for pie for trying to find sort of minimum uncertainties. So from that we can get Delta P. Um, and so whoa to remember how uncertainties legate I don't actually remember. So if you're squaring in a certainty, actually, sure you possibly think about it. Okay, so I was thinking about the problem. Why? I think I don't have to worry about propagation. Zay want certainty? I think it's worded a little weirdly so. I mean, first you can evaluate Delta X Delta P S H over to pie. And then if you saw that for Delta P I'II defined by Delta X and calculate, um, I ended up getting delta up. Iwas 5.27 times 10 to the minus 20. Um, kilogram, You know, for a second. And then I mean, I just really did the only thing you can do with this problem. I playground in for P here and then used the mass of the electron. So then I got that k e men, I guess, would be equal. Thio, um of five point. Let's see Hominy sig figs to sig figs. 5.2 times 10 to the sixth, 5 to 10 of the sexy V again. I just plugged in this number into this formula, um, and then converted TV. But I mean, in principle by the swaddle showed the minimum. I mean, the minimum momentum could in principle, be zero. Because if you go back and forth right, like, it's not like specifically, um, well, it's in a specific direction. I mean, I guess you could say that the momentum in a certain direction, um, the like, the absolute minimum value if it's going in the negative direction or excite the absolute absolute minimum value would be the value of if it were going in the negative direction. And so the biggest it could be would be, I suppose, um, given in this model, this number. So I guess in that sense, it kind of makes sense. But, um, yeah, it's sort of have to make that jump in logic to make that true

University of Washington
Top Physics 103 Educators
Christina K.

Rutgers, The State University of New Jersey

Andy C.

University of Michigan - Ann Arbor

Marshall S.

University of Washington

Jared E.

University of Winnipeg