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Problem 7 Medium Difficulty

An alpha particle $\left(Z=2, \text { mass }=6.64 \times 10^{-27} \mathrm{kg}\right)$ approaches to within $1.00 \times 10^{-14} \mathrm{m}$ of a carbon nucleus (Z = 6). What are (a) the maximum Coulomb force on the alpha particle, (b) the acceleration of the alpha particle at this time, and (c) the potential energy of the alpha particle at the same time?

Answer

a. 27.6 \mathrm{N}
b. 4.16 \times 10^{27} \mathrm{m} / \mathrm{s}^{2}
c. 1.73 \mathrm{MeV}

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ME

Manail E.

October 14, 2020

Top Physics 103 Educators
Marshall S.

University of Washington

Zachary M.

Hope College

Jared E.

University of Winnipeg

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McMaster University

Video Transcript

for number seven. There's an Alfa particle that has so a mass of 6.64 times tending 27 kilograms and has two protons. It's approaching a carbon Adam. It has six pro tones, and the closest distance they get is one Timpson negative, 14 meters apart. First to ask, What is there Colin Force? So they're getting repelled from each other. We want to know Use coolers Law here. So that is one over for pie. Absolutely not. Times the charge of the one, the charge of the other divide by their distant support squared. No, I usually this I use Kay. It's over in, since that's all constant. You just calculate all that and always used this. Kay. So Kay is 8.99 times 10 to the ninth, the charge of the one with my alpha particle. So it's two times the elementary charge 1.6 times 10 to the negative, mating the charge of the other thing, which was the carbon and that has six pretends or six times 1.6 times 10 negative 19 and then the distance between them is this one times tends and negative 14 and that gets squared. Calculate. I get 27 0.6 Newton's import. Be there were no, but see acceleration. So it's just net force is called acceleration. The figures are way I just found my force was 27.6. I'm given my mass of the alpha particles I want to acceleration of the alpha particle 27. And I'm so sorry. Divide both sides and I give acceleration is 4.16 times 10 to the 27 meters per second squared in the report. See, they want to know what's the potential energy of the outfit Particle. At that point, essential energy is okay. That same K, um, charge of the one the charge of the other divided by just the distance, not squared. So that is the only difference between this and this is that, um, extra power or so if I just take what I found here and multiplied by our figure, um, bring that up multiplied by another. Effective are that will be the potential energy to be a lot easier. So I'm gonna take this 27.6 and just multiply it by the factor of are so it wasn't squared. Mr. Starr and I get to point 76 times. Tell negative 13. Jules

University of Virginia
Top Physics 103 Educators
Marshall S.

University of Washington

Zachary M.

Hope College

Jared E.

University of Winnipeg

Meghan M.

McMaster University