Rutherford scattering experiment question

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SUMMARY

The discussion revolves around calculating the electrical force between an alpha particle and a gold nucleus during the Rutherford scattering experiment. The closest approach is given as 7.27 x 10^-14 m, with the charge of the gold nucleus calculated as (79 * 1.60 x 10^-19) C. The user attempts to apply Coulomb's law using the equations Epe = (kq1q2)/r and Fe = (kq1q2)/r^2, but encounters errors in their calculations, yielding incorrect results. The correct approach involves recognizing that the kinetic energy of the alpha particle at the point of closest approach is zero, and its potential energy equals the initial kinetic energy.

PREREQUISITES
  • Coulomb's law for electric force calculations
  • Understanding of potential energy in electrostatics
  • Basic knowledge of atomic structure, particularly gold's atomic composition
  • Familiarity with algebraic manipulation of equations
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  • Review Coulomb's law and its applications in electrostatics
  • Study the relationship between kinetic and potential energy in particle interactions
  • Learn about the properties of gold as an atomic nucleus
  • Practice algebraic problem-solving with physics equations
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Students studying physics, particularly those focusing on atomic interactions and electrostatics, as well as educators seeking to clarify concepts related to the Rutherford scattering experiment.

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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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
 


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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