anonymous
  • anonymous
A single electron orbits a lithium nucleus that contains three protons (+3e). The radius of the orbit is 1.76E-11m. Determine the kinetic energy of the electron.
Physics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
Is this Intro Physics E&M or Modern Physics+ ?
anonymous
  • anonymous
I'm not sure what either of those mean. This is college physics 2
anonymous
  • anonymous
Then it is intro.. I ask that so that if I need to look up or do any other things.. But since this is intro, we only have to worry about Orbit Radius, Coulomb Force and momentum to come up with a number in Joules. Which is where we start..

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anonymous
  • anonymous
Well coulumb's law is what most of this chapter is about. i looked up physics e&m and i suppose this would fall under that category
anonymous
  • anonymous
I'm just not sure how to apply it to KE
anonymous
  • anonymous
Ok.. I will work you though it.. I was curious what Carlos was writing.. but I might as well start.. First get is the mass of the electron.. You can treat it as if it was a planet moving around the sun.. In that case, your Coulomb force is your "gravity" with sign convention that can basicly be ignored.
anonymous
  • anonymous
Ok, 9.11E-31
anonymous
  • anonymous
should i figure out coulomb force?
anonymous
  • anonymous
Yeah. Go a head and calculate that up..
anonymous
  • anonymous
Centrifugal force must equal electrostatic attraction:\[F_c=m_e·\frac{ v^2 }{ r }\]\[F_E=\frac{ 1 }{ 4 \pi \epsilon_0 }\frac{ 3e^2 }{ r^2 }\] \[F_E=F_C \rightarrow m_e \frac{ v^2 }{ r }=\frac{ 1 }{ 4 \pi \epsilon_0 }\frac{ 3e^2 }{ r^2}\rightarrow E_K=\frac{ 1 }{ 2 } m_ev^2=\frac{ 3e^2 }{ 8 \pi \epsilon_0r}\]
anonymous
  • anonymous
can you follow that?
anonymous
  • anonymous
Not really
anonymous
  • anonymous
thank you though carlos
anonymous
  • anonymous
He basicly took 2 Equations and substituted them together.
anonymous
  • anonymous
welcome. what is it that you dont understand?
anonymous
  • anonymous
well is Me mass of the electron?
anonymous
  • anonymous
yup
anonymous
  • anonymous
why are you only counting the proton charges and not the electron charges?
anonymous
  • anonymous
Well.. How much is the charge of an electron?
anonymous
  • anonymous
1e=1.6E-19 correct?
anonymous
  • anonymous
why is that negated?
anonymous
  • anonymous
which is the same as a proton.. Just with a flipped charge sign.
anonymous
  • anonymous
right, there are 4 charges in total. -e and +3e why is the -e dropped?
anonymous
  • anonymous
I am counting both. Coulomb force between q1 and q2 is:\[F_E=\frac{ 1 }{ 4 \pi \epsilon_0 }\frac{ q_1q_2 }{ r^2 }\]Now see that q1=3e (three protons) and q2=e (one electron) whwre e=1.6·10^(-19) coulombs and you get my formula
anonymous
  • anonymous
na.. his is right. (sorry thought you were saying something else, my fault)
anonymous
  • anonymous
yeah but up there it goes from (3e)(e)/r^2. why is the q2 dropped?
anonymous
  • anonymous
ohhhhh
anonymous
  • anonymous
If Earth is orbiting around the sun.. the mass of the earth is multiplied into the mass of the sun... not added... That's where you are confused.
anonymous
  • anonymous
nothing dropped, the formula says e^2
anonymous
  • anonymous
yeah it just clicked
anonymous
  • anonymous
got it now?
anonymous
  • anonymous
yeah, i do actually just had to go step by step in that formula on paper
anonymous
  • anonymous
thanks a lot for being so patient guys
anonymous
  • anonymous
awesome.. np.. I used to be a tutor for my school. :)
anonymous
  • anonymous
awesome, you guys are good. thanks again i really appreciate it
anonymous
  • anonymous
anyway.. what is the next thing sticking you?
anonymous
  • anonymous
By the way, this is the Rutherford model for the atom, which is no longer valid and was replaced by Bohr´s model
anonymous
  • anonymous
well that I will have to see. I have been sitting on this problem for hours. I know ill run into more speed bumps. mind if i message you later if youre on?
anonymous
  • anonymous
the bohr is the cloud model?
anonymous
  • anonymous
yep, based on probability
anonymous
  • anonymous
the Ruth model was basicly saying atoms travel in descrete orbits.. like planets.. which is all you end up doing in intro phy 2.
anonymous
  • anonymous
ok, i feel this is merely a stepping stone. bohr would complicate things further
anonymous
  • anonymous
atoms... pshh.,.. I mean electrons orbit
anonymous
  • anonymous
i got what you were saying
anonymous
  • anonymous
So, back to it.. Are you seeing the Kinetic Energy from Carlos' last line?
anonymous
  • anonymous
yes
anonymous
  • anonymous
KE=coulombs force
anonymous
  • anonymous
sort of i mean
anonymous
  • anonymous
:( Force is not energy...
anonymous
  • anonymous
I see the math, the understanding of it all is lost. However I think if i stick with it knowing the solution i will figure it out
anonymous
  • anonymous
"Work" is energy. You have to find out how much energy there is by Work done by the force on the object in question.
anonymous
  • anonymous
I move from centrifugal forces to Kinetic Energy multiplying by 1/2r and do the same with Coulomb force
anonymous
  • anonymous
I urge you not to use the solution as a means of understanding... (just saying)
anonymous
  • anonymous
yeah, you put it better. Working backwards is my best ally when it comes to understanding concepts.
anonymous
  • anonymous
@QuantumQQ I concur
anonymous
  • anonymous
well guys, I have to go. @scottman you are in good hands, bye
anonymous
  • anonymous
ok thanks guys i appreciate it.
anonymous
  • anonymous
thanks @CarlosGP
anonymous
  • anonymous
actually I found that really the directions to be the hardest part.. It is also easy to make up interpretations.. Up as Carlos's equations are put.. and with the question... You want to make the Fe = FE <------ important key (take care carlos)
anonymous
  • anonymous
Fe meaning force of electron and FE meaning electric field correct?
anonymous
  • anonymous
And I will say.. Scroll up and look at the equations.. tell me what the KE equation is...
anonymous
  • anonymous
1/2MeV^2
anonymous
  • anonymous
Electrostatic <-- static meaning not moving, and electro meaning.. :)
anonymous
  • anonymous
yup.. now.. tell me.. what are your unknowns in the KE equation?
anonymous
  • anonymous
velocity
anonymous
  • anonymous
yup.. and from the equations given.. where can we dig up a velocity?
anonymous
  • anonymous
coulombs law?
anonymous
  • anonymous
wait so if we get Force out of coulombs law and we know F=ma we know coulombs law=Me(a)
anonymous
  • anonymous
no.. there is a v in the Fe Since, the Fe needs to be equal to the FE.. this would imply that orbit (r) is fixed and not changing (orbit collapsing or going out.)
anonymous
  • anonymous
yeah... there is a little identity that says a = v^2/r
anonymous
  • anonymous
(for rotational motion only that is)
anonymous
  • anonymous
yeah so since we now would know r and a we just need to square root to find v?
anonymous
  • anonymous
yeah.. write the equality in terms of the velocity squared, since there is no reason to unsquare it.... we can just directly insert the rewritten equation directly into 1/2 m_e v^2
anonymous
  • anonymous
so then we have our asnwer?
anonymous
  • anonymous
Yup. 1/2 m_e v^2 gives you the kinetic energy of the system.. We don't have to worry about any potential energies at all.. yay!
anonymous
  • anonymous
oh wow, you made me conceptually understand this problem. You truly are awesome.
anonymous
  • anonymous
Just tell me if the energy is positive or neg..... and you should be home free.
anonymous
  • anonymous
positive
anonymous
  • anonymous
awesome... nailed it. :)
anonymous
  • anonymous
thank you. you have a gift. are you a teacher or prof of some sort?
anonymous
  • anonymous
i know you said tutor but.. wow
anonymous
  • anonymous
remember that if you are having trouble with a problem.. Chances are, the problem is really in the conceptual understanding... No.. I'm a tutor.. a Physics major.. doing my thing :) Thank you for the awesome compliment though :D
anonymous
  • anonymous
Yeah sometimes it's hard bridging the gap without some assistance. Again thank you, would you mind if i messaged you if I run into any future snags?
anonymous
  • anonymous
sure... I infrequently come here but I will try to get back to you if I get the message.
anonymous
  • anonymous
Cool thanks again.
anonymous
  • anonymous
(that's why I only have 53 as a score) Good luck with your class and take care :)
anonymous
  • anonymous
you too

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