Total kinetic energy of two protons

  • Thread starter Zvaigzdute
  • Start date
  • #1
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Homework Statement



Two protons are held fixed at a distance of 3 angstroms (3 x 10-10 m) from
one another. The protons have a charge of +1.6 x 10-19 C. After they are released they
repel each other and fly apart and each acquires some kinetic energy. What is the final
total kinetic energy of the two particles when they are at a large distance from each other?

Homework Equations



Potential Electric Energy= (1/4pie)*q1q2/r
Ki+Ui=Kf+Uf

The Attempt at a Solution



3A * (3e-10m/1 A) = 9e-10m
Ue=(9e9 Nm^2/C^2)(1.6e-19C)(1.6e-19C)/(9e-10m)= 2.56e-19 Nm

And I know that as r goes to infinity, potential energy goes to 0
thus
0+Ui=Kf+0
2.56e-19J=Kf
But the answer is 7.7e-19, what i am doing wrong, what step am I missing thank you!
 

Answers and Replies

  • #2
Redbelly98
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Welcome to PF :smile:

The Attempt at a Solution



3A * (3e-10m/1 A) = 9e-10m
When you multiply by (3e-10m/1 A) to convert units, that is saying that
1 A = 3e-10m​
and this is incorrect.

Actually, you do not need to calculate the distance in m; the problem statement tells you that the protons are ____m apart.

Hope that helps.
 
  • #3
10
0
Yes thank you very much! Wow that was really stupid of me
 

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