# How Does Energy Transformation Occur in Quantum Orbital Mechanics?

• lekh2003
In summary, the student is trying to solve a problem for part B of a homework assignment but is having difficulty understanding what to do. He found an equation for KE but is unsure how to apply it to the problem. He also doubts whether or not he understood correctly what the magical values were.
lekh2003
Gold Member

## Homework 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.

## 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:

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 don't understand this step and what they're doing.

Thanks.

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lekh2003 said:
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.

I understad where K comes from. But why do we take away from K what is essentially 2K. What is ##q^2/r##?

lekh2003 said:
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

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?

lekh2003 said:
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).

lekh2003 said:
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.

DrClaude said:
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.
DrClaude said:
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.

## 1. What is the purpose of "MIT Quantum Physics Problem Set 1"?

The purpose of "MIT Quantum Physics Problem Set 1" is to test your understanding and application of basic concepts in quantum physics. It is designed to challenge your problem-solving skills and prepare you for more complex problems in the field.

## 2. How many problems are included in "MIT Quantum Physics Problem Set 1"?

There are typically 5-10 problems included in "MIT Quantum Physics Problem Set 1". However, the number may vary depending on the specific version of the problem set.

## 3. What level of difficulty can I expect from "MIT Quantum Physics Problem Set 1"?

"MIT Quantum Physics Problem Set 1" is typically designed for students at an introductory or intermediate level in quantum physics. The problems may range from easy to challenging, but they are all meant to test your understanding of the basic principles.

## 4. How much time should I allocate to complete "MIT Quantum Physics Problem Set 1"?

The amount of time needed to complete "MIT Quantum Physics Problem Set 1" may vary depending on your level of understanding and problem-solving skills. However, on average, it may take 1-2 hours to complete the set.

## 5. Are there any resources available to help me with "MIT Quantum Physics Problem Set 1"?

Yes, there are resources available to help you with "MIT Quantum Physics Problem Set 1". These may include lecture notes, textbooks, online tutorials, and study groups. It is always recommended to seek help and clarification whenever needed to better understand the material.

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