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The structure of the atom. During $1910-1911,$ Sir Ernest Rutherford performed a series of experiments to determine the structure of the atom. He aimed a beam of alpha particles (helium nuclei, of mass $6.65 \times 10^{-27} \mathrm{kg}$ ) at an extremely thin sheet of gold foil. Most of the alphas went right through with little deflection, but a small percentage bounced directly back.These results told him that the atom must be mostly empty space with an extremely small nucleus. The alpha particles that bounced back must have made a head-on collision with this nucleus. A typical speed for the alpha particles before the collision was $1.25 \times 10^{7} \mathrm{m} / \mathrm{s},$ and the gold atom has a mass of $3.27 \times 10^{-25} \mathrm{kg}$ . Assuming (quite reasonably) elastic collisions, what would be the speed after the collision of a gold atom if an alpha particle makes a direct hit on the nucleus?

$4.98 \times 10^{5} \mathrm{m} / \mathrm{s}$

Physics 101 Mechanics

Chapter 8

Momentum

Physics Basics

Kinetic Energy

Potential Energy

Energy Conservation

Moment, Impulse, and Collisions

Esperanza C.

November 8, 2020

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University of Michigan - Ann Arbor

Simon Fraser University

McMaster University

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so here in this question, were given that the moss off alpha particle is 6.65 times 10 to the power minus 27 k g on the mass off gold atom were given us 3.27 times 10 to the power minus 25 kg. On the initial velocity off the alpha particles, we are given us 1.25 times 10 to the power 7 m per second on. We have to calculate the speed after the collision off a gold atom on if the alpha particle makes a direct hit on the nucleus. And we're also given that the initially gold atom is at rest or the gold foil is at rest so we can take the initial velocity off gold atom Toby zero on were also given that we have to assume the collision Toby Elastic collision. So this think we are given in the question Now it's a special case where the collision is elastic collision on the second object, that is the gold foil is it rest. So in this case, we can say that the final everlasting off the alpha particle will be given by the expression ml FA minus the mask off Gold atom Divide by M. L fa plus m g times we initial velocity off alpha particle on the final velocity off gold particles is given by the expression two times ml fa over m l Fabulous m g James we off in the Alfa Particles Initiative lost your alpha particles. So these are the two formulas that we're going to use toe Calculate the velocity. Now let's substitute values in these formula. So the final velocity off alpha particle will be closed toe ml for we have 6.65 times 10 to the power minus 27 k g minus. This one is mg. We have 3.27 times 10 to the power, minus 25 kg times the velocity. So velocity we have 1.25 times 10 to the power 7 m per second divide by M l far. We have 6.65 times 10 to the power minus 27 plus 3.27 times 10 to the power minus 25 kg. So from her will get a final velocity as minus 1.20 times 10 to the power 7 m per second. So this is the final velocity off alpha particles. And here negative sign indicates that the direction is reversed. Advance. If initial alphabetical is traveling in the right side, the finally we lost it after the collision would be in the left side. So this is a solution for the velocity for the alpha particles. Now, next will be calculating this finally, the last year off gold at him. So we're going to pluck the village here two times m l for we have 6.665 times 10 to the power minus 27 k g divide by M. L. For we have 6.65 times 10 to the power minus 27 k g plus M. G. We have 3.27 times 10 to the power minus 25 kg times the velocity. So velocity we have they need to lost If alpha particles, we have 1.25 times 10 to the power 7 m per second. So from here we'll get the value as 4.98 times 10 to the power five media per second. So this is the final velocity off gold particle. So I hope you have understood the problem. Thank you

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