Calculating Speed to Reach 1.11 fm from Oxygen Nucleus

AI Thread Summary
To determine the speed a proton must have to reach a turning point of 1.11 fm from the oxygen nucleus, the potential and kinetic energy equations were used. Initial calculations were incorrect due to misapplication of constants and the mass of the proton instead of the electron. After revising the calculations and ensuring the correct mass and charge values were used, the final speed calculated was approximately 23,448,253.12 m/s. The discussion highlighted the importance of careful attention to detail in calculations and the correct application of physical constants. The problem was ultimately resolved with collaborative input and verification.
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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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