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MIT Quantum Physics Problem Set 1

  • Thread starter lekh2003
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lekh2003

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1. The problem statement, all variables and given/known data
upload_2019-3-8_12-11-22.png


2. Relevant equations
I've used:
##mv^2/r## = Centripetal
##q^2/r^2## = Force pulling the electron in
A bunch of other ones which I really can't be bothered listing.
3. The attempt at a solution
I managed to get Part A of the question using pretty simple methods.

I set ##mv^2/r = q^2/r^2##
That gave me ##v = sqrt(q^2/mr)##

And I plugged that into the equation for KE and got ##q^2/2r##

Then I need to transform the change in energy to the energy lost per orbital, so I used the time it takes to take each orbit using some orbital formula: ##t=2πr/v##

Keep going, and keep going, and eventually I found an equation for the change in energy due to radiation of energy, which I compared as a ratio to the kinetic energy, and it was basically negligible as an effect.

But Part B is beating me up. I really don't understand what to do. I was suspicious I must find a correlation between the energy and the radius, so I checked the solution and I got this:
upload_2019-3-8_12-19-39.png

However, I simply don't understand how they seemed to get these magical values.

After this, its some differential equations which I might have trouble understanding later, but right now I dont understand this step and what they're doing.

Thanks.
 

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DrClaude

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But Part B is beating me up. I really don't understand what to do. I was suspicious I must find a correlation between the energy and the radius, so I checked the solution and I got this:
View attachment 239917
However, I simply don't understand how they seemed to get these magical values.
Could you clarify where you think the magic appears? Considering that you have found yourself the equation for KE, the derivation in the picture is quite straightforward.
 

lekh2003

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I understad where K comes from. But why do we take away from K what is essentially 2K. What is ##q^2/r##?
 
I understad where K comes from. But why do we take away from K what is essentially 2K. What is ##q^2/r##?
Can you tell us what exactly your doubt is? The ##q^2/r## is potential energy of electron in system where k=1
 

lekh2003

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I apologize, I don't think I have my thoughts compiled very well.

Why is Energy = KE - PE, like E = ##q^2/2r - q^2/r##?

Also another issue I see, which might be separate is, why do they ignore constants here? Coulomb^2/metres is obviously not energy, but why are the constants that maintain dimensional accuracy eliminated?
 

DrClaude

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Why is Energy = KE - PE, like E = ##q^2/2r - q^2/r##?
As always, Energy is KE + PE. KE is ##q^2/r##, as you calculated yourself. As for the potential energy, it is the Coulomb interaction between two charges ##+q## (the nucleus) and ##-q## (the electron), hence ##U = -q^2/r##. (##q## is here the elementary charge).

Also another issue I see, which might be separate is, why do they ignore constants here? Coulomb^2/metres is obviously not energy, but why are the constants that maintain dimensional accuracy eliminated?
They are not ignoring constants here, but rather are using the theoreticians favourite units for electromagnetism, Gaussian units. So the charge ##q## is in statcoulomb, ##r## in centimetre, and ##q^2/r## will be an energy in ergs.
 

lekh2003

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As always, Energy is KE + PE. KE is q2/rq2/rq^2/r, as you calculated yourself. As for the potential energy, it is the Coulomb interaction between two charges +q+q+q (the nucleus) and −q−q-q (the electron), hence U=−q2/rU=−q2/rU = -q^2/r. (qqq is here the elementary charge).
Ohhh. That makes sense. I don't seem to understand why I didn't realize that. Thanks.
They are not ignoring constants here, but rather are using the theoreticians favourite units for electromagnetism, Gaussian units. So the charge qqq is in statcoulomb, rrr in centimetre, and q2/rq2/rq^2/r will be an energy in ergs.
Ohhh, this is one of those things which would've been helpful knowing prior to jumping into the problem. I guess it all makes sense now. If I had done the prereq courses with MIT maybe it would be fine, but thanks for the help.

I'll keep working.
 

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