# Rutherford Experiment, energy and acceleration.

• skg94
In summary: Nucleus with an initial potential energy. At that point, the alpha particle will experience the maximum possible acceleration. If the net force is zero, then there would be no acceleration.
skg94

## Homework Equations

If i can make that out its Ep=kq1q2/r where k is coloumbs constant

## The Attempt at a Solution

Im a little confused how do i know the radius of the alpha particles?

I was thinking its conservation of energy, Ep+Ek=Eka then try to solve for a that way, but would Fnet be 0 on this alpha particle?

skg94 said:
If i can make that out its Ep=kq1q2/r where k is coloumbs constant
Right.
Im a little confused how do i know the radius of the alpha particles?
Assume it's a point.
I was thinking its conservation of energy, Ep+Ek=Eka then try to solve for a that way,
That's the way to start, but it's a two-step process; you don't get $a$ directly.
but would Fnet be 0 on this alpha particle?
Of course not. If the net force were 0 then there would be no acceleration.

skg94 said:

## Homework Equations

If i can make that out its Ep=kq1q2/r where k is coloumbs constant

## The Attempt at a Solution

Im a little confused how do i know the radius of the alpha particles?
Well, you don't know and don't need their radius. You will need to look up their mass and charge though. The radius being referred to in the potential energy equation is the radial distance of the alpha particle from the center of the gold atom nucleus.

Oh yes, you'll need to know the charge on the gold atom nucleus, too!
I was thinking its conservation of energy, Ep+Ek=Eka then try to solve for a that way, but would Fnet be 0 on this alpha particle?
Conservation of energy will certainly play a role here. Assuming that the alpha particle heads directly towards a gold nucleus, at what point in its trajectory do you think it'll experience maximum acceleration?

Is the conservation set properly? IF it is, than in the Ep+Ek, what do i use for r?, or is Ep ignorable? Mass of an alpha particle and the charge are, 6.65*10^-27kg and 2e, and gold is what 79 protons, how do i know the charge of that?

It would be at its maximum right before it hits, just as the question suggest wouldn't it be?

What's the charge on one proton? What's the charge due to 79 of them?

The alpha particle is unlikely to have enough energy to enter (hit) the gold nucleus. It'll approach and rebound due to electrostatic repulsion. The thing you'll have to determine is, how close will it get with its given initial KE? What'll be the net force acting on it when it reaches that closest approach?

Oh, since its a positive, that meets a positive atom?, but isn't the nucleus neutral, of any atom? It would be 79+ i would assume. Wouldnt net force just be Force electric, i suppose gravity is too but it is negligible, there was no magnets what so whatever from what i remember from the rutherford experiment.

The nuclei of all atoms (except for antimatter atoms, of course) is positively charged. It's the electrons surrounding the atom that give the atom a net neutral charge. The electric force is the only one you need be concerned with here.

how do i calculate, Fe, i don't see a formula i can use, V=E/d and E=V/d only work for parallel plates right. and i don't know r, the standard Fe=Eq, applies, but both of you said to use conservation of energy, at the moment i can't see to factor it in

The problem statement tells you how to determine potential energy for two charges separated by distance r, and as for force, Coulomb's Law comes to mind...

But you said you don't need to know r, and coulomb's law involves r as well Fe=kq1q2/r^2?

skg94 said:
But you said you don't need to know r, and coulomb's law involves r as well Fe=kq1q2/r^2?
I said you don't need to know the radius of an alpha particle. The radial distance of the alpha particle from the center of the gold nucleus is a different matter -- you will want to calculate a particular value.

Can you give me some sort of a hint?, its not centripetal since it bounces straight back, besides no mag fields, also the is r in the Ep formula, the radius of an alpha particle, or the radius between the two particles?

skg94 said:
Can you give me some sort of a hint?, its not centripetal since it bounces straight back, besides no mag fields, also the is r in the Ep formula, the radius of an alpha particle, or the radius between the two particles?

There is NO alpha particle radius ANYWHERE in this problem. All r's are the distance between charges (alpha particle and nucleus).

Use conservation of energy to determine the closest distance that the alpha particle can approach to the nucleus.

You start with what you know. You know the initial kinetic energy, and the initial potential energy can be set to zero. You know the potential energy due to the electric field at the point of closest approach, and you know what the kinetic energy will be there, too (assume the alpha goes straight in and straight out). Write down the expressions for the initial and final energy, and equate them.

Ek+Ep = Ekf +Epf
1*10^-12 + 0 = 1/2mv^2 + Kq1q2/r ?

Don't use numbers at this stage (except for 0); do everything symbolically. It will be easier to understand.

What is the kinetic energy at the point of closest approach? (This time give a number).

I don't know, 0?

skg94 said:
I don't know, 0?
You surely do know. Throw a ball straight up. What is its kinetic energy at the highest point of its trajectory?

zero right, since at the top as it comes back down in that instant the velocity is zero?

Ek+ 0 = 0 + Ep ? find r through that and use that r in Fe=kq1q2/r^2 ? using charge of alpha particle and charge of gold nucleus, as the other poster stated?

Right. Although you should be stating these things, not asking them.

## 1. What was the purpose of the Rutherford Experiment?

The Rutherford Experiment, also known as the Gold Foil Experiment, was conducted in order to understand the structure of the atom and the distribution of its positive charge.

## 2. How did the Rutherford Experiment contribute to our understanding of energy?

The Rutherford Experiment showed that atoms have a dense, positively charged nucleus at its center, which contains almost all of the atom's mass and positive charge. This discovery led to the development of the nuclear model of the atom and the understanding of nuclear energy.

## 3. What is the role of energy in the acceleration of particles in the Rutherford Experiment?

In the Rutherford Experiment, alpha particles were accelerated towards a thin gold foil. The kinetic energy of these particles allowed them to penetrate the foil and interact with the atoms in the foil, resulting in a scattered pattern. This showed that the alpha particles were being accelerated and their energy was being changed by the positively charged nucleus of the atom.

## 4. How did the results of the Rutherford Experiment impact our understanding of atomic structure?

The Rutherford Experiment disproved the previous "plum pudding" model of the atom, which proposed that the positive charge and mass of the atom were evenly distributed throughout. Instead, it showed that the atom has a small, dense nucleus and a mostly empty space surrounding it. This led to further research and understanding of the structure of the atom.

## 5. What are the implications of the Rutherford Experiment for our understanding of energy and matter?

The Rutherford Experiment had significant implications for our understanding of energy and matter. It demonstrated that the atom is made up of smaller, subatomic particles and that energy plays a crucial role in the interactions between these particles. It also paved the way for the development of nuclear energy and other applications of atomic structure in various fields of science and technology.

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