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

AI Thread Summary
To calculate the Coulomb barrier between a proton and a lithium nucleus, the relevant equation is UCoul = kZ1Z2e2/r, where r is the interaction radius. Initial calculations using the atomic radius of lithium yielded a Coulomb barrier of 4.552437x10-18, which was deemed incorrect. The correct approach involves using the nuclear radius of lithium (4.8 fm) and the proton radius (0.85 fm) to find an interaction radius of 5.65 fm. After recalculating, the result was 1.224733x10-13, which was initially misidentified as low; however, it should be noted that this value is in Joules, equating to 7.66 eV. The discussion highlights the importance of using the correct radii and units in such calculations.
eigenmax
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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.
 
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You need to consider the nuclear not the atomic radius of lithium.
 
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Likes Dr. Courtney
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.
 
What do you believe to be the correct answer? Please include units.
 
I would think the answer to be around 350keV
 
And in what units is 1.224733x10-13?
 
eV
 
eigenmax said:
eV
Nope. If you use the equation and the numbers that you provided, the answer comes out in Joules.
 
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
eigenmax said:
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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