Hey Khashishi,
Very interesting reply. Good, I shall look into natural line broadening. Hummm, I knew molecules have rotational and vibrational states but are photons released from these states and if so what is the mechanism of their release/absorption? Is it that a change in the associated...
Define: Contiguous
Source:
https://www.merriam-webster.com/dictionary/contiguous
touching or connected throughout in an unbroken sequence
Define: Continuous
https://www.merriam-webster.com/dictionary/continuous
having the property that the absolute value of the numerical difference between the...
Great Info, I did not know this. Hummm, during the bonding process is that when the photon is released? This bonding is between two or more atoms and in some cases when an electron is shared, is that when the photon it released? If so, is the change in the electron energy what releases the photon?
I understand the EM spectrum is made up of photons at different energy levels (wavelengths); where, the higher the energy level the shorter the wavelength. What I do not understand is how collectively photons are created at a range of energy levels that produce a contiguous EM spectrum. (When I...
Did not know about factoring out the constants; however, it presents a problem.
It doesn’t seem right that the derivative can have 2 solutions.
This:
F_C=-\left(-\frac {e^2 }{4\pi \epsilon_0 }\right) \frac{d}{dr} \frac{-1}{r}
=-\frac{e^2}{4\pi \epsilon_0 r^2}
Or this:
F_C=-...
Thanks bpatrick, your tip to factor out the unaffected variables will be quite useful.
I checked some other references also and everything is starting to come back and clear up; however, I have one persistent question about equation (5.36)
Equation (5.36)
F_C=-\frac{d}{dr}\left(-\frac...
What a silly mistake. All I can say is that it was late, I was in a rush and didn’t double check my work which I always try to do. Anyway, what you got does make sense and does balance out.
From you answer I see you used h and hbar. Thus, I am still working with a handicap because I cannot...
Wow, it has been a long time since I have performed a derivative.
Yep, I think I understand.
Does any constant goes to 1. Then what ever power r is raised to becomes a constant and then the r is raised to the next higher (lower) power.
Is this example correct?
\frac {d}{dr} ( \frac {1}...
In equation (5.35) the constant drops out when the derivative with respect to r is taken. However, in equation (5.36) the constant does not drop out. Does anybody know why?
Equation (5.35)
F_L=-\frac {d}{dr}\left(\frac {n^2h^2}{2\mu r^2}\right)=\frac{n^2h^2}{\mu r^3}
Equation (5.36)...
Hey Deveno, I went back and looked at both formulas with my laptop and still didn’t see the hbar. Could be the resolution of the laptop is not so great. When first working this problems I wondered if those were Planck’s constants or Dirac’s constants. Subsequent to your note I posted some...
There was a typo in the denominator in the first term of the last equation in Bluestar's last post. That equation should look like the following:
E_n=\frac {\mu e^4}{32\pi^2 \epsilon_0^2 n^2h^2} - \frac{\mu e^4 }{16 \pi^2 \epsilon_0^2 n^2h^2}= -\frac {e^4\mu}{8\epsilon_0^2 n^2h^2}...