Alpha particle approaching a gold nucleus

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SUMMARY

An alpha particle approaching a gold nucleus with a charge of 79e experiences an electric force calculated using Coulomb's Law, represented by the equation FE=k(q1q2)/d². The correct calculation yields a force of 9.1e7 N, but the textbook states 9.1e-12 N, indicating a potential error in charge units. The charge of an electron, e, is 1.602e-19 Coulombs, and since it appears twice in the calculation, the correct expression must account for this factor. The discussion concludes that the distance provided in the problem may be a misprint, as it is significantly smaller than the typical size of a nucleus.

PREREQUISITES
  • Coulomb's Law
  • Understanding of electric charge in Coulombs
  • Basic algebra for solving equations
  • Familiarity with atomic structure and nucleus size
NEXT STEPS
  • Review Coulomb's Law and its applications in electrostatics
  • Learn about the significance of atomic units and conversions
  • Explore the concept of nuclear sizes and their implications in physics
  • Investigate common errors in physics problem statements and how to identify them
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Randell Julius
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Homework Statement


An alpha particle approaches at high speeds a gold nucleus with a charge of 79e. What is the electric force acting on the alpha particle when it is 2.0e-17 m from the gold nucleus?

Homework Equations


FE=k(q1q2)/d2

The Attempt at a Solution


FE=k(q1q2)/d2
FE=9e9((2e)(79e))/(2.0e-17)2
FE=9.1e7 N

The answer in the back of the book says that it is 9.1 e -12 N.

Not sure where I went wrong.
 
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If you want the force to be in units of Newtons, what should be the units for electric charge?
 
TSny said:
If you want the force to be in units of Newtons, what should be the units for electric charge?
It should be in Coulombs.
 
Randell Julius said:
It should be in Coulombs.
Yes. It looks like you made a minor error in expressing the two charges in terms of Coulombs. Check to see if you missed a factor of 10-19.
 
TSny said:
Yes. It looks like you made a minor error in expressing the two charges in terms of Coulombs. Check to see if you missed a factor of 10-19.
The value we used for e was 1.602e-19 Coulombs
 
Randell Julius said:
The value we used for e was 1.602e-19 Coulombs
Yes. Note that e occurs twice, once for each charge. So, how many factors of 1.602 x 10-19 occur in the numerator of the force calculation?
 
TSny said:
Yes. Note that e occurs twice, once for each charge. So, how many factors of 1.602 x 10-19 occur in the numerator of the force calculation?
2 factors of 1.602 * 10-19.
 
OK. My mistake. Since your answer was off by exactly a factor of 10-19, I assumed (wrongly) that you had missed a factor of 10-19 in substituting for the charges. Now that I have grabbed my calculator and checked your work, I'm getting the same answer as you. The distance for d given in the problem looks way too small to be reasonable. (It is much less than the size of a nucleus!) So, I think there must have been a misprint in the problem statement for the distance d.
 
TSny said:
OK. My mistake. Since your answer was off by exactly a factor of 10-19, I assumed (wrongly) that you had missed a factor of 10-19 in substituting for the charges. Now that I have grabbed my calculator and checked your work, I'm getting the same answer as you. The distance for d given in the problem looks way too small to be reasonable. (It is much less than the size of a nucleus!) So, I think there must have been a misprint in the problem statement for the distance d.
Okay thank you very much.
 

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