Calculating Speed to Reach 1.11 fm from Oxygen Nucleus

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SUMMARY

The problem involves calculating the speed required for a proton to reach a turning point of 1.11 femtometers (fm) from the surface of an oxygen nucleus (16O) with a radius of 2.91 fm. The correct approach combines kinetic energy and potential energy equations, specifically using the formula for potential energy as 1/(4πE₀) * (q₁ * q₂)/r. The final calculated speed of the proton, after correcting for the mass of the proton and ensuring accurate values, is 23,448,253.12 m/s, confirming the solution's validity.

PREREQUISITES
  • Understanding of kinetic energy and potential energy equations.
  • Familiarity with the concept of electric charge and Coulomb's law.
  • Knowledge of the mass of a proton (9.109 x 10^-31 kg).
  • Basic understanding of nuclear physics, specifically regarding atomic nuclei and their radii.
NEXT STEPS
  • Study Coulomb's law and its application in nuclear physics.
  • Learn about the properties of atomic nuclei, including radii and binding energy.
  • Explore advanced topics in kinetic and potential energy calculations in particle physics.
  • Review the principles of electric fields and forces between charged particles.
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Students in physics, particularly those studying nuclear physics or electromagnetism, as well as educators looking for problem-solving strategies in particle dynamics.

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Homework Statement



The oxygen nucleus 16O has a radius of 2.91 fm. With what speed must a proton be fired toward an oxygen nucleus to have a turning point 1.11 fm from the surface?

Homework Equations



Kinetic energy = 1/2 mv^2
Potential energy = 1/4(pi)Enot * q/radius min

The Attempt at a Solution



I combined the two above equations as follows:

1/4(pi)Enot * q/radius min = 1/2 mv^2
9x10^9 (1.602x10^-19)/(1.11x10^-15 m) = 1/2 (9.109x10^-31)v^2
1.688x10^18 m/s = v

The above answer was incorrect. I have roughly 31 hours to solve this question until it is due. Can someone please tell me if the equations I am using are wrong and which equations I should be looking at?
 
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Potential energy = 1/4(pi)Enot * q/radius min
Can't be right - you need the charge on the proton multiplied by the charge on the nucleus. The 1/4(pi) doesn't look right either - you must mean 1/(4*pi*Enot).
 


Sorry for not including the brackets, I did mean 1/(4*pi*Enot).

So I changed the way I approached this question...below is my revised attempt which still ended up being wrong:


1/(4(pi)Enot) * (q oxygen * q proton)/radius min = 1/2 mv^2
9x10^9 (8*1.602x10^-19 * 1.602x10^-19)/(2.91x10^-15m +1.11x10^-15m)=1/2 (9.109x10^-31)v^2
9x10^9 ((8*1.602x10^-19 * 1.602x10^-19)/4.02x10^-15) = 1/2 (9.109x10^-31)v^2
4.5965x10^-13 = 1/2mv^2
1.00923x10^18 = v^2
1,004,605,104 m/s = v

This answer is incorrect. I decided to add the radius of the oxygen nucleus with the turning point radius as the radius between the two point charges. Maybe my radius should be something else but the way I did it seems right to me...unfortunately it isn't right :(
 


1/2 (9.109x10^-31)v^2
Looks like you used the mass of an electron rather than the mass of a proton.
 


I recalculated by answer and used the protons mass this time. I had a final answer of 2344825.312 m/s which is still wrong. I don't know why the answer is not correct.

Everything else that has been inputted into the equations seems logical, I still must be missing something.
 


Ok, all is fine :)

I double checked everything, I must have missed a digit in one of my numbers the first time!

Final answer is 23448253.12 m/s and it is correct! Thank you for all your help Delphi51!
 


Great; most welcome!
 

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