Rutherford's atom experiment problem

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In 1911, Rutherford's experiment revealed the atomic nucleus through helium nuclei scattering off gold nuclei. A problem arises when calculating how close a helium nucleus, with a mass of 6.68 x 10^-27 kg and a charge of +2e, approaches a gold nucleus with a charge of +79e, given its initial velocity of 1.30 x 10^7 m/s. An incorrect formula was presented, leading to confusion about its derivation and dimensional accuracy. The discussion emphasizes the importance of understanding the underlying physics rather than relying on peer solutions. Ultimately, grasping the concepts is crucial for mastering the material.
Dana Fishel
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Homework Statement


In 1911, Ernest Rutherford discovered the nucleus of the atom by observing the scattering of helium nuclei from gold nuclei. If a helium nucleus with a mass of 6.68
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10^-27 kg, a charge of +2e, and an initial velocity of 1.30
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10^7 m/s is projected head-on toward a gold nucleus with a charge of +79e, how close will the helium atom come to the gold nucleus before it stops and turns around? (Assume the gold nucleus is held in place by other gold atoms and does not move.)

Homework Equations


electric field equations involving k constant, charge, mass, and velocity

The Attempt at a Solution


I was told by a classmate that this should work:

d = (9e9)4(79)(1.602e-19)^2 / (6.68e-27)(1.3e7)
= 8.4e-7 meters
(where e means x10^#)

But it's wrong, and I'm not sure which equation this solution came from in the first place.
 
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Dana Fishel said:
d = (9e9)4(79)(1.602e-19)^2 / (6.68e-27)(1.3e7)
It's not reasonable to expect others to decode that by reverse engineering the numbers.
Please post it as an equation using symbols, preferably defining them.
 
Think energy.
 
Haru is absolutely right. But you yourself can easily do the reverse engineering, isn't it ?

d = (9e9)4(79)(1.602e-19)^2 / (6.68e-27)(1.3e7)

I see a kcoulomb, q1, q2, m, and a v. In short: all you need.
But dimensionally it doesn't fit. Andrev's hint should be enough to discover the little thingy that's still missing from your friendly classmate's spoiler !

[warning]: make sure you learn from this. Getting the answer from a mate means that in fact you miss the exercise. You can't have someone else train for the marathon in your place and still expect to be able to run it yourself!
 
It's very similar to throwing a ball upwards, except in this case the force increases drastically "upwards".
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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