Determine the radius of the electron's orbit.

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In one model of the hydrogen atom, the electron revolves in a circular orbit around the proton with a speed of 4.6 106 m/s

f = mv^2/r

F is the force of attraction between the electron and proton
m is the mass of the electron
v is the velocity it spins at
r is the radius

(9.11)(1.67*10^-27)K/r^2=[(1.67*10^-27)((4.6*10^6)^2)]/r

when i solve for r, it is not correct, can you tell me if it is setup right

thanks
 
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Frogger Man said:
(9.11)(1.67*10^-27)K/r^2=[(1.67*10^-27)((4.6*10^6)^2)]/r
I don't understand what those numbers are. What's the mass of the electron? The charge on the electron and proton?
 
The 1.67*10^6 should be 9.109 × 10-31

On the left i am trying to find the force of attraction, is this the wrong way to go about this problem?
 
Your approach is probably correct, but you're messing up some of the numbers. Start by writing your equation just in terms of symbols. Once we agree that the equation is correct, then you can plug in numbers.
 
q1q2k/r^2 = mv^2/r
 
Frogger Man said:
q1q2k/r^2 = mv^2/r
Good. Now solve that for r before plugging numbers in.
 
r = q1q2k/(mv^2)
 
Frogger Man said:
r = q1q2k/(mv^2)
Good. Now plug away!
 
9.11x10^-31*1.6x10^-19*8.99x10^9/(9.11x10^-31*(4.6x10^6)^2)
 
  • #10
Frogger Man said:
9.11x10^-31*1.6x10^-19*8.99x10^9/(9.11x10^-31*(4.6x10^6)^2)
Why do you have the electron mass here?
 
  • #11
I must have a number wrong, when I plug it
 
  • #12
1.6x10^-19*1.6x10^-19*8.99x10^9/(9.11x10^-31*(4.6x10^6)^2)

That should be better, I used the mass instead of the charge.
 
  • #13
1.194e-11

Got it, again, thanks for your help. Wish I had you as my teacher in school.
 
  • #14
Much better.
 

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