Physics Help Forum MIT 8.04 Quantum Physics Problem

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Nov 8th 2017, 10:57 PM   #1
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MIT 8.04 Quantum Physics Problem

I've decided to work through the assignments on https://ocw.mit.edu/courses/physics/...i-spring-2013/ where they have problems and posted solutions. Of course I am not enrolled in any course, just doing this for my own edification.

I am looking at:

 Suppose the world was actually governed by classical mechanics. In such a classical universe, we might try to build a Hydrogen atom by placing an electron in a circular orbit around a proton. However, we know from 8.03 that a non-relativistic, accelerating electric charge radiates energy at a rate given by the Larmor formula, $\displaystyle \frac{dE}{dt} = -\frac{2}{3}\frac{q^2 a^2}{c^3}$ (in cgs units) where q is the electric charge and a is the magnitude of the acceleration. So the classical atom has a stability problem. How big is this effect? (a) Show that the energy lost per revolution is small compared to the electron’s kinetic energy. Hence, it is an excellent approximation to regard the orbit as circular at any instant, even though the electron eventually spirals into the proton. ◦ (b) Using the typical size of an atom (1A) and a nucleus (1 fm), calculate how long it would take for the electron to spiral into the proton. ◦ (c) Compare the velocity of the electron (assuming an orbital radius of 0.5 A) to the speed of light – will relativistic corrections materially alter your conclusions? (d) As the electron approaches the proton, what happens to its energy? Is there a minimum value of the energy the electron can have?
At the moment I am working on (a). Now I'm not quite sure how to proceed.

I started off with trying to compute the energy lost per revolution and started thusly:

$\displaystyle \frac{m_e v^2}{r} = \frac{k}{r^2} = m_e a\: \textrm{where} \: k=\frac{1}{4 \pi \epsilon_0} \: \textrm{and} \: m_e \: \textrm{is the mass of the electron}$

If I solve for "a" (as a function of r) and substitute into the above differential equation I am no better off as r is not independent of time.

Initially I thought I could use conservation of angular momentum to my advantage only to finally realise that probably doesn't hold due to the radiation of energy.

What am I missing for the second step? I kind of figure I must be missing something obvious. Is there some other law I can substitute in there to make this a straightforward differential equation I can solve by separation of variables?
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Last edited by kiwiheretic; Nov 8th 2017 at 11:05 PM.

 Nov 9th 2017, 03:29 AM #2 Senior Member   Join Date: Apr 2015 Location: Somerset, England Posts: 926 Both the kinetic energy and the acceleration are functions of the electron velocity. You have to show that the ratio of the energy loss to the KE is small. Try writing both as functions of electron velocity and taking the ratio with the hope of cancelling the unknown velocity. topsquark likes this.
 Nov 12th 2017, 08:55 PM #3 Senior Member   Join Date: Nov 2013 Location: New Zealand Posts: 534 Ok I have been messing around with this again and lets see if I am on the right track. I substituted $\displaystyle a = \frac{k}{m_e r^2} \:\textrm{and}\:k=\frac{1}{4 \pi \epsilon_0}$ where $\displaystyle m_e$ is the mass of the electron $\displaystyle \frac{dE}{dt} = -\frac{2}{3}\frac{q^2 k^2}{m_e^2 r^6 c^3}$ Now because we are talking about a circular orbit we have: $\displaystyle m_e a = m_e \frac{v^2}{r} = \frac{k}{r^2}$ and so $\displaystyle E_k = \frac{1}{2}m_e v^2 = \frac{k}{r}$ If we now divide kinetic energy via the change in energy we get: $\displaystyle (dE/dt) / E_k= -\frac{2}{3}\frac{q^2 k}{m_e^2 r^5 c^3}$ Does that look like the right approach? It doesn't actually give the energy lost per revolution tho. __________________ Burn those raisin muffins. Burn 'em all I say.

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