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Problem 6 Easy Difficulty

In a Rutherford scattering experiment, an a - particle (charge $=+2 e )$ heads directly toward a gold nucleus (charge $=+79 e ) .$ The $\alpha$ -particle had a kinetic energy of 5.0 $\mathrm{MeV}$ when very far $(r \rightarrow \infty)$ from the nucleus. Assuming the gold
nucleus to be fixed in space, determine the distance of closest approach. Hint: Use conservation of energy with $P E=k_{e} q_{1} q_{2} / r$.

Answer

$r^{\prime}=4.5 \times 10^{-14} \mathrm{~m}$

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Top Physics 103 Educators
Elyse G.

Cornell University

LB
Liev B.

Numerade Educator

Zachary M.

Hope College

Jared E.

University of Winnipeg

Video Transcript

In this exercise, we have an awful particle that has a charge. Q equals two plus two e that hats towards ah, gold nucleus That has a charge Capital que It was him plus 79 e. And we also have the information that very far away from the the nucleus, that is when our tends to infinity the kinetic energy ke off the of the offer Particle is five mega like from votes, Okay. And we need to find what is the distance of closest approach that I'm gonna call or prime between the alpha particle and the gold nucleus. And in order to do that, I'm gonna plant conservation of energy. So I'm gonna play the I'm gonna ride the energy when our tends to infinity K is equal to the energy in the distance of closest approach our prime. Okay, so the total kinetic energy of the of the alpha particle. I'm sorry to total mechanic. Energy of the system is even by the kinetic energy of the awful particle. So this is the kinetic energy of the local particle. When Argos Community plus the potential the electric electric potential energy you when our ghost infinity and this is equal to the kinetic energy at our prime. Plus the potential energy the Lakers stuck electrostatic potential energy when r equals R prime. Okay, now, when we know that gonna write it down here, you know that you the electrostatic potential energy, is given by K times que capital Q over our. So in our case, this is 158 which is us 79 times two okay, e squared, divided by are so notice that this means that the energy the the petitioner you went Argo Symphony tends to zero okay. And also noticed that the kinetic energy in the business of closest approach which in the figure is right here Okay, the kinetic energy is zero because the alpha particle is neither, ah, approaching nor going already from the gold nucleus at that specific time. Ah, it's still okay that that specific a specific time There's no kinetic energy of the offer particle. And for that reason, our formula, our conservation of energy equation here that high just pointed to in red becomes the kinetic energy. When Argos infinity, which is five Meg electoral votes, is equal to this term here, uh, vanishes And so this the Easter is equal to 158. Okay, e squared over our prime. And we want to find our prime, which is 158 times K E squared, divided by five times 10 to the sixth Electron volts. Okay, One of the electrons here, one of the electric charges cancels out. So we have 158 times K, which is nine times 10 to the ninth. Mutants meters squared. Coolum squared, divided by five times. 10 to the fifth, the sixth votes. Let me just write five little better here. Okay? Eso What we have is that are trying is equal to four point five 4.5 tires, too to the minus 14 meters. And this is that This is of closest approach, okay?

Universidade de Sao Paulo
Top Physics 103 Educators
Elyse G.

Cornell University

LB
Liev B.

Numerade Educator

Zachary M.

Hope College

Jared E.

University of Winnipeg