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.

- anonymous

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- anonymous

Is this Intro Physics E&M or Modern Physics+ ?

- anonymous

I'm not sure what either of those mean. This is college physics 2

- 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

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

I'm just not sure how to apply it to KE

- 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

Ok, 9.11E-31

- anonymous

should i figure out coulomb force?

- anonymous

Yeah. Go a head and calculate that up..

- 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

can you follow that?

- anonymous

Not really

- anonymous

thank you though carlos

- anonymous

He basicly took 2 Equations and substituted them together.

- anonymous

welcome. what is it that you dont understand?

- anonymous

well is Me mass of the electron?

- anonymous

yup

- anonymous

why are you only counting the proton charges and not the electron charges?

- anonymous

Well.. How much is the charge of an electron?

- anonymous

1e=1.6E-19 correct?

- anonymous

why is that negated?

- anonymous

which is the same as a proton.. Just with a flipped charge sign.

- anonymous

right, there are 4 charges in total. -e and +3e why is the -e dropped?

- 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

na.. his is right. (sorry thought you were saying something else, my fault)

- anonymous

yeah but up there it goes from (3e)(e)/r^2. why is the q2 dropped?

- anonymous

ohhhhh

- 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

nothing dropped, the formula says e^2

- anonymous

yeah it just clicked

- anonymous

got it now?

- anonymous

yeah, i do actually just had to go step by step in that formula on paper

- anonymous

thanks a lot for being so patient guys

- anonymous

awesome.. np.. I used to be a tutor for my school. :)

- anonymous

awesome, you guys are good. thanks again i really appreciate it

- anonymous

anyway.. what is the next thing sticking you?

- 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

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

the bohr is the cloud model?

- anonymous

yep, based on probability

- 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

ok, i feel this is merely a stepping stone. bohr would complicate things further

- anonymous

atoms... pshh.,.. I mean electrons orbit

- anonymous

i got what you were saying

- anonymous

So, back to it.. Are you seeing the Kinetic Energy from Carlos' last line?

- anonymous

yes

- anonymous

KE=coulombs force

- anonymous

sort of i mean

- anonymous

:(
Force is not energy...

- 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

"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

I move from centrifugal forces to Kinetic Energy multiplying by 1/2r and do the same with Coulomb force

- anonymous

I urge you not to use the solution as a means of understanding... (just saying)

- anonymous

yeah, you put it better. Working backwards is my best ally when it comes to understanding concepts.

- anonymous

@QuantumQQ I concur

- anonymous

well guys, I have to go. @scottman you are in good hands, bye

- anonymous

ok thanks guys i appreciate it.

- anonymous

thanks @CarlosGP

- 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

Fe meaning force of electron and FE meaning electric field correct?

- anonymous

And I will say.. Scroll up and look at the equations.. tell me what the KE equation is...

- anonymous

1/2MeV^2

- anonymous

Electrostatic <-- static meaning not moving, and electro meaning.. :)

- anonymous

yup.. now.. tell me.. what are your unknowns in the KE equation?

- anonymous

velocity

- anonymous

yup.. and from the equations given.. where can we dig up a velocity?

- anonymous

coulombs law?

- anonymous

wait so if we get Force out of coulombs law and we know F=ma we know coulombs law=Me(a)

- 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

yeah... there is a little identity that says a = v^2/r

- anonymous

(for rotational motion only that is)

- anonymous

yeah so since we now would know r and a we just need to square root to find v?

- 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

so then we have our asnwer?

- 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

oh wow, you made me conceptually understand this problem. You truly are awesome.

- anonymous

Just tell me if the energy is positive or neg..... and you should be home free.

- anonymous

positive

- anonymous

awesome... nailed it. :)

- anonymous

thank you. you have a gift. are you a teacher or prof of some sort?

- anonymous

i know you said tutor but.. wow

- 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

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

sure... I infrequently come here but I will try to get back to you if I get the message.

- anonymous

Cool thanks again.

- anonymous

(that's why I only have 53 as a score)
Good luck with your class and take care :)

- anonymous

you too

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