How to calculate the Coulomb barrier between a proton and a lithium nucleus

  • Thread starter eigenmax
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  • #1
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Homework Statement


How to calculate the Coulomb barrier between a proton and a lithium nucleus. The variable is r (interaction radius). The known data is Z1, Z2, k (Coulomb constant), and e (elementary charge).

Homework Equations


The equation is UCoul = kZ1Z2e2/r

The Attempt at a Solution


I used the sum of lithium's atomic radius and the radius of a proton as the interaction radius. I got 1.5200084x10-10m. I plugged in all the numbers and got 4.552437x10-18 as my Coulomb barrier, which I think is incorrect.
 

Answers and Replies

  • #2
kuruman
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You need to consider the nuclear not the atomic radius of lithium.
 
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  • #3
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Ok, thanks. The nuclear radius of lithium is 4.8fm and the radius of a proton is 0.85fm. That gives an interaction radius of 5.65fm, or 5.65x10-15. Plugging in the new numbers gives 1.224733x10-13, which is still incorrect.
 
  • #4
kuruman
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What do you believe to be the correct answer? Please include units.
 
  • #5
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I would think the answer to be around 350keV
 
  • #6
kuruman
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And in what units is 1.224733x10-13?
 
  • #8
kuruman
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eV
Nope. If you use the equation and the numbers that you provided, the answer comes out in Joules.
 
  • #9
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Oh, ok. Sorry. That comes out to be 7.66eV. That seems awfully low. One can't run a proton accelerator on 7.66 volts.
 
  • #10
kuruman
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Oh, ok. Sorry. That comes out to be 7.66eV. That seems awfully low. One can't run a proton accelerator on 7.66 volts.
It's more than that. Recheck your powers of 10.
 

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