Centripedal acceleration of an electron

  • Thread starter azurken
  • Start date
  • #1
azurken
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Homework Statement

Multiple-Concept Example 3 provides some pertinent background for this problem. Suppose a single electron orbits about a nucleus containing two protons (2e), as would be the case for a helium atom from which one of the two naturally occurring electrons is removed. The radius of the orbit is 2.65 x 10^11 m. Determine the magnitude of the electron’s centripetal acceleration.

From the example on that page
Given:
Electron charge -1.60 x 10^-19 C
Electron mass 9.11 x 10^-31 kg
Proton charge +1.60 x 10^-19 C
Radius of orbit 5.29 x 10^-11 m

Homework Equations


F = (k)(q1)(q2) /r^2
Fc=Ac x M
Ac=Fc/M

The Attempt at a Solution


I don't even know where to begin with this problem. Yes it's a chug and plug problem but the book is being annoyingly vague. The example it was referring me to was for a hydrogen atom. I'm not sure if I was supposed to use the values from the example but otherwise they really didn't give me ANYTHING besides the radius to find the force of the electron.

From using the Charges from the example (1.6x10^-19 ), I get 3.27x10^-7 .

Plugging that number into find the acceleration, using the mass of the electron from the example I get 3.60x10^23

The answer in the back says 7.19x10^23 m/s^2
 

Answers and Replies

  • #2
kushan
256
0
Hello there ,

Why don't you make a force diagram on electron revolving around nucleus , and try to find the acceleration from there ?
 
  • #3
azurken
15
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You mean like a free diagram?
 
  • #4
kushan
256
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Yea , a FBD ( free body diagram )
That might get the juices flowing in the right direction :cool:
 
  • #5
azurken
15
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Alright.. I still don't get it though, how do I know what the mass is for the e-?

And also I don't know how to draw an fbd for this.. Theres 2 protons, (I assume not at the center), so where do I draw them?

Sorry it's been awhile I've done a fbd... well tbh I've never done one for an electron.

Is it basically just a proton (nucleus) in the middle with an electron going circles around it?
 
  • #6
kushan
256
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Ok , first of all mass of electron should be known . They assume that you know the mass of electron which is 9.1X10^-31 kg
and for FBD , simply imagine a satellite orbiting a planet , you don't need to draw two protons , just one big thing of charge .As the electron will revolve around the nucleus ,it will experience a force of k2e^2/r^2 . Equate this to centripetal force and divide by mass of electron . You will be left with centripetal acceleration on one side and that is what you want :cool:
 
  • #7
azurken
15
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Ahh I got it! I didn't know I had to multiply by 2 for the electrons.
 
  • #8
azurken
15
0
Ok , first of all mass of electron should be known . They assume that you know the mass of electron which is 9.1X10^-31 kg
and for FBD , simply imagine a satellite orbiting a planet , you don't need to draw two protons , just one big thing of charge .As the electron will revolve around the nucleus ,it will experience a force of k2e^2/r^2 . Equate this to centripetal force and divide by mass of electron . You will be left with centripetal acceleration on one side and that is what you want :cool:

Wait.. am I supposed to double the charge for the electron since it's a 2e- substance? The hydrogen example given in the book didn't have the 2 in it and just went k(e^2)/r^2
 

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