Bohr Model in Polar Co-ordinates

  • Thread starter dirtyhippy
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  • #1
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The angular momentum of the Bohr atom is ⃗L = ⃗r× ⃗p

Let the Oz axis of the unit vector ⃗uz be determined by the relation L = L⃗uz, in which L stands for the magnitude of the angular momentum vector ⃗L .
(Basically L is in Positive z axis)

We specify the position of the electron within its plane of movement
by the polar coordinates r = OM and θ = ∢(⃗ux,OM), where ⃗ux is the unit vector of the Ox axis. The unit vectors in polar coordinates are to be denoted ⃗ur and ⃗uθ. Calculate the absolute value L of the angular momentum as a function of m, r and dθ/dt.

What ive tried

Ive tried little things like dp=rdθ, to get ⃗v= ⃗w x ⃗r to ⃗v =dθ/dt x ⃗r
but really i have no idea how to bring that back ⃗L= ⃗r x ⃗p

Thanks for any help
 
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Answers and Replies

  • #2
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I'm not sure I understand your notation or work but here's my 2 cents.

[tex] \vec{r} \times \vec{p} = \abs{\vec{r}}\abs{\vec{p}}sin\theta [/tex] and what is [tex] \theta [/tex]?
 
  • #3
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I'm not sure I understand your notation or work
?

Sorry I should elaborate,

when there is a ⃗ in front of a letter it means vector, if not its a scalar
 
  • #4
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[tex] \vec{r} \times \vec{p} = \abs{\vec{r}}\abs{\vec{p}}sin\theta [/tex] and what is [tex] \theta [/tex]?

Theta is the angle between the x-axis and r (as the electron goes round).

Am I right in saying that it should be the magnitudes on the right hand side of your suggestion as opposed to vectors i.e.
[tex] \vec{r} \times \vec{p} = \abs{{r}}\abs{{p}}sin\theta [/tex]
as then that may help me.
 
  • #5
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yes about the magnitudes, couldn't get it in latex. Theta is the angle between the two vectors r and p.
 
  • #6
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What is the question? To find the magnitude of L in terms of m, r and do/dt right? (o=theta)
 
  • #7
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aha got it, thankyou,

[tex] \vec{r} \times \vec{p} = \abs{{r}}\abs{{p}}sin\theta [/tex]

using dx/dt = sin (x) dθ/dt and p =mv

we arrive at

L = r mv sin θ
L = r mv dθ/dt sin x
where x is actually ⃗ux

so we get L = r mv dθ/dt sin ( ⃗ux)


I think this is correct, thankyou zachzach
 
  • #8
258
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aha got it, thankyou,

[tex] \vec{r} \times \vec{p} = \abs{{r}}\abs{{p}}sin\theta [/tex]

using dx/dt = sin (x) dθ/dt and p =mv

we arrive at

L = r mv sin θ
L = r mv dθ/dt sin x
where x is actually ⃗ux

so we get L = r mv dθ/dt sin ( ⃗ux)


I think this is correct, thankyou zachzach

I don't think so. Theta is a constant. It is the angle between r and p so the angle between r and v. Your answer is not in terms of what it asks for because you have a v in there. In fact your answer even has the wrong units. Also, is x an angle? Otherwise how why are you taking the sine of it?
 
  • #9
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ah balls your right, ive been looking at this question too long. I had to translate it from french as well which probably isnt helping. I'll take a nap and come back to it later.

but upto L=r m v sin theta is correct and v = r dθ/dt right, and theta being a constant means sin theta is also, so would it be

L=r^2 m dθ/dt sin theta, could i say theta is 90 ie sin theta is 1?
L=r^2 m dθ/dt ?
 
  • #10
258
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That is what I got. Theta does equal 90 degrees always since it is traveling in a circle. V is tangential to the circle and r is the radial direction.
 
  • #11
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well in that case thankyou once again, much appreciated.
 
  • #12
258
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no prob good luck.
 

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