Alpha particle approaching a gold nucleus

[Randell Julius]

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

F_E=k(q₁q₂)/d²

The Attempt at a Solution

F_E=k(q₁q₂)/d²
F_E=9e9((2e)(79e))/(2.0e-17)²
F_E=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.

 

[TSny]

If you want the force to be in units of Newtons, what should be the units for electric charge?

 

[Randell Julius]

If you want the force to be in units of Newtons, what should be the units for electric charge?

It should be in Coulombs.

 

[TSny]

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.

 

[Randell Julius]

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

 

[TSny]

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?

 

[Randell Julius]

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.

 

[TSny]

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.

 

[Randell Julius]

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.