Question

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.

Answer 1

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

Answer 2

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

Answer 3

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..

Answer 4

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

Answer 5

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

Answer 6

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.

Answer 7

Ok, 9.11E-31

Answer 8

should i figure out coulomb force?

Answer 9

Yeah. Go a head and calculate that up..

Answer 10

Centrifugal force must equal electrostatic attraction:\[F_c=m_e·\frac{ v^2 }{ r }\]\[F_E=\frac{ 1 }{ 4 \pi \epsilon_0 }\frac{ 3e^2 }{ r^2 }\]
\[F_E=F_C \rightarrow m_e \frac{ v^2 }{ r }=\frac{ 1 }{ 4 \pi \epsilon_0 }\frac{ 3e^2 }{ r^2}\rightarrow E_K=\frac{ 1 }{ 2 } m_ev^2=\frac{ 3e^2 }{ 8 \pi \epsilon_0r}\]

Answer 11

can you follow that?

Answer 12

Not really

Answer 13

thank you though carlos

Answer 14

He basicly took 2 Equations and substituted them together.

Answer 15

welcome. what is it that you dont understand?

Answer 16

well is Me mass of the electron?

Answer 17

yup

Answer 18

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

Answer 19

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

Answer 20

1e=1.6E-19 correct?

Answer 21

why is that negated?

Answer 22

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

Answer 23

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

Answer 24

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

Answer 25

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

Answer 26

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

Answer 27

ohhhhh

Answer 28

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.

Answer 29

nothing dropped, the formula says e^2

Answer 30

yeah it just clicked

Answer 31

got it now?

Answer 32

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

Answer 33

thanks a lot for being so patient guys

Answer 34

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

Answer 35

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

Answer 36

anyway.. what is the next thing sticking you?

Answer 37

By the way, this is the Rutherford model for the atom, which is no longer valid and was replaced by Bohr´s model

Answer 38

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?

Answer 39

the bohr is the cloud model?

Answer 40

yep, based on probability

Answer 41

the Ruth model was basicly saying atoms travel in descrete orbits.. like planets.. which is all you end up doing in intro phy 2.

Answer 42

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

Answer 43

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

Answer 44

i got what you were saying

Answer 45

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

Answer 46

yes

Answer 47

KE=coulombs force

Answer 48

sort of i mean

Answer 49

:(
Force is not energy...

Answer 50

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

Answer 51

"Work" is energy.
You have to find out how much energy there is by Work done by the force on the object in question.

Answer 52

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

Answer 53

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

Answer 54

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

Answer 55

@QuantumQQ I concur

Answer 56

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

Answer 57

ok thanks guys i appreciate it.

Answer 58

thanks @CarlosGP

Answer 59

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)

Answer 60

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

Answer 61

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

Answer 62

1/2MeV^2

Answer 63

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

Answer 64

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

Answer 65

velocity

Answer 66

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

Answer 67

coulombs law?

Answer 68

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

Answer 69

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.)

Answer 70

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

Answer 71

(for rotational motion only that is)

Answer 72

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

Answer 73

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

Answer 74

so then we have our asnwer?

Answer 75

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!

Answer 76

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

Answer 77

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

Answer 78

positive

Answer 79

awesome... nailed it. :)

Answer 80

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

Answer 81

i know you said tutor but.. wow

Answer 82

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

Answer 83

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?

Answer 84

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

Answer 85

Cool thanks again.

Answer 86

(that's why I only have 53 as a score)

Good luck with your class and take care :)

Answer 87

you too