Determine the radius of the electron's orbit.

In summary, the conversation discusses the model of the hydrogen atom where the electron revolves around the proton in a circular orbit with a speed of 4.6 * 10^6 m/s. The force of attraction between the two particles is represented by the equation F = mv^2/r, where m is the mass of the electron, v is the velocity it spins at, and r is the radius. The conversation also includes a discussion on the correct setup of the equation and the correct values to use for calculations. After some clarification and corrections, the correct value for r is found to be 1.194 * 10^-11.
  • #1
Frogger Man
14
0
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
 
Physics news on Phys.org
  • #2
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?
 
  • #3
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?
 
  • #4
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.
 
  • #5
q1q2k/r^2 = mv^2/r
 
  • #6
Frogger Man said:
q1q2k/r^2 = mv^2/r
Good. Now solve that for r before plugging numbers in.
 
  • #7
r = q1q2k/(mv^2)
 
  • #8
Frogger Man said:
r = q1q2k/(mv^2)
Good. Now plug away!
 
  • #9
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.
 

1. What is the radius of the electron's orbit?

The radius of the electron's orbit is approximately 0.53 angstroms or 0.053 nanometers.

2. How is the radius of the electron's orbit determined?

The radius of the electron's orbit is determined by the balance between the centripetal force of the electron's motion and the electrostatic attraction between the electron and the nucleus of the atom.

3. Does the radius of the electron's orbit vary in different atoms?

Yes, the radius of the electron's orbit varies depending on the atomic number of the element. As the atomic number increases, the radius of the electron's orbit also increases.

4. Can the radius of the electron's orbit be measured?

Yes, the radius of the electron's orbit can be measured using various experimental techniques such as X-ray diffraction, electron diffraction, and spectroscopy.

5. Does the radius of the electron's orbit remain constant?

No, the radius of the electron's orbit is not a fixed value. It can change due to factors such as the electron's energy level, the presence of other electrons, and external influences such as electric and magnetic fields.

Similar threads

  • Introductory Physics Homework Help
Replies
9
Views
1K
  • Introductory Physics Homework Help
Replies
26
Views
2K
Replies
13
Views
2K
  • Introductory Physics Homework Help
Replies
7
Views
1K
  • Introductory Physics Homework Help
Replies
4
Views
1K
  • Introductory Physics Homework Help
Replies
3
Views
1K
  • Introductory Physics Homework Help
Replies
7
Views
3K
Replies
2
Views
928
  • Introductory Physics Homework Help
Replies
5
Views
1K
  • Introductory Physics Homework Help
Replies
6
Views
2K
Back
Top