An electron of mass me is moving around a nucleus along a...

An electron of mass $m_e$ is moving around a nucleus along a ring with a fixed radius $r$ and zero potential energy $V(\phi) = 0$. The following is a general solution to the Schrodinger equation for this particle-on-a-ring system: $\Phi(\phi) = Acos(m_\ell\phi)$ where $m_\ell^2 = \frac{2IE}{\hbar^2}$ and $I = m_er^2$ a) Apply the appropriate boundary conditions to obtain allowed values for $m_\ell$. b) Can $m_\ell$ be zero? Why or why not? c) Can we include negative integers for $m_\ell$? Why or why not?

Transcript

00:01 So for a particle in a box, we have a restriction on the wavelengths that the particle can have because it, of course, has a wave -like nature.

00:07 And the wavelengths we get are something like 2l over n.

00:13 And so then our momentum from this, using the debroy wavelength, is going to be h times n over 2l.

00:21 And so the energy, which is p squared over 2m, is going to be h squared, n squared, over 8m .l squared.

00:35 And so we can see this is quantized, of course, because it depends on in or n squared in this case.

00:39 So we're going to have a variety of energy levels, but they're going to be discrete.

00:44 So that would be like one fact you could state for the first part.

00:49 For part b, so that was part a.

00:52 For part b, we want to know why.

00:54 Is there no possibility of finding the electron outside the wells? and that's because the potential is infinite by you know, in the problem of design outside of, you know, the box that we're describing.

01:06 And so, you know, if you wanted to get technical, you could look at, like, the transmission coefficients for an electron, like the probability of the electron moving through.

01:15 And it looks like e to the minus, you know, let's see, what is it? it's a square to 2m over h bar squared.

01:23 The energy of the electron minus the potential energy, something like that.

01:27 Well, if the potential energy is infinite, then this is like e to the minus infinity.

01:34 Actually, i think it's v minus e.

01:37 But we can ignore the sign of it for now and times like the width of the box or the length of the box...

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