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
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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?
 

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