Rutherford scattering experiment question

In summary, the student is trying to calculate the value of the electrical force between an alpha particle and a gold nucleus at the point of closest approach, using the Coulomb potential equation. They have correctly identified the charge on the atom and the closest approach distance, but are unsure about the identity of the nucleus. They have also correctly identified that the kinetic energy at the point of closest approach is 0 and the potential energy is equal to the initial kinetic energy. They are struggling to find the correct answer and are seeking urgent help.
  • #1
stony
4
0
Rutherford scattering experiment question (urgent)

Homework Statement


I need to find the value of the electrical force between the alpha particle and the nucleus of the atom at the point of closest approach.

Closest approach = 7.27 x 10^-14 m
Charge on atom = (79 * 1.60 x 10 ^-19) C
initial kinetic energy of alpha particle = 5.00 x 10^-13 J

Homework Equations



I think you're supposed to use [Epe = (kq1q2)/r] or [Fe = (kq1q2)/r^2]

The Attempt at a Solution



(8.99x10^9)(2*1.60x10^-19)(79*1.60x10^-19)
________________________________________
7.27x10^-14

= 5.00175x10^-13 (wrong answer)

Am I trying to solve this right? Need an answer ASAP please!

Thank you.
 
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  • #2


stony said:

Homework Statement


I need to find the value of the electrical force between the alpha particle and the nucleus of the atom at the point of closest approach.

Closest approach = 7.27 x 10^-14 m
Charge on atom = (79 * 1.60 x 10 ^-19) C
initial kinetic energy of alpha particle = 5.00 x 10^-13 J

Homework Equations



I think you're supposed to use [Epe = (kq1q2)/r] or [Fe = (kq1q2)/r^2]

The Attempt at a Solution



(8.99x10^9)(2*1.60x10^-19)(79*1.60x10^-19)
________________________________________
7.27x10^-14

= 5.00175x10^-13 (wrong answer)

Am I trying to solve this right? Need an answer ASAP please!

Thank you.
Your are correct to use the Coulomb potential. What is the atomic nucleus that is scattering the alpha particle (how many protons and neutrons)?

What is the kinetic energy of the alpha particle at the point of closest approach? What is its potential energy? You should work out an algebraic solution first and then plug in the numbers. Otherwise it is difficult to see your reasoning.

AM
 
  • #3


Andrew Mason said:
Your are correct to use the Coulomb potential. What is the atomic nucleus that is scattering the alpha particle (how many protons and neutrons)?

What is the kinetic energy of the alpha particle at the point of closest approach? What is its potential energy? You should work out an algebraic solution first and then plug in the numbers. Otherwise it is difficult to see your reasoning.

AM

It doesn't outright say what the nucleus is, but since it has 79 protons I guess that means it's gold.

Kinetic energy at point of closest approach = 0
Potential energy at the point of closest approach = the kinetic energy that it started with

Sorry, what do you mean it's difficult to see my reasoning?

I just tried it using this equation [Fe = (kq1q2)/r^2] but the answer I got is 6.8799, which is also wrong. I don't understand why this equation won't give me the right answer, am I not using it right?
 

1. What is the Rutherford scattering experiment?

The Rutherford scattering experiment, also known as the gold foil experiment, was conducted by Ernest Rutherford in 1911 to study the structure of the atom. It involved firing alpha particles at a thin sheet of gold foil and observing the pattern of scattering to understand the distribution of positive charge within the atom.

2. What were the key findings of the Rutherford scattering experiment?

The key findings of the Rutherford scattering experiment were that most of the alpha particles passed straight through the gold foil, but a small percentage were deflected at large angles, indicating the presence of a concentrated positive charge within the atom. This led to the discovery of the atomic nucleus and the understanding that most of the atom is empty space.

3. How did the Rutherford scattering experiment impact our understanding of the atom?

The Rutherford scattering experiment revolutionized our understanding of the atom by disproving the previously accepted plum pudding model, which proposed that the positive charge and electrons were evenly distributed throughout the atom. It also laid the foundation for the development of the modern atomic model, with a small, dense nucleus surrounded by orbiting electrons.

4. What were the limitations of the Rutherford scattering experiment?

One of the limitations of the Rutherford scattering experiment was that it only provided information about the distribution of positive charge within the atom, not the distribution of electrons. It also did not account for the phenomenon of atomic orbitals, which were later discovered and explained by quantum mechanics.

5. How is the Rutherford scattering experiment still relevant today?

The Rutherford scattering experiment is still relevant today as it remains one of the most important experiments in the history of physics and has greatly influenced our understanding of the atom and its structure. It also paved the way for further research and discoveries in the field of nuclear physics and led to the development of technologies such as nuclear power and medical imaging.

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