- #1
buttersrocks
- 29
- 0
Hi,
I originally posted this question in the homework section, but I really don't need any help calculating anything, my answers are right. I'm having conceptual trouble, so I figured that this question belongs here.
So, let's say there is a field driving a single atomic oscillator (hydrogen for arguments sake). As a further simplification, let's say the oscillator has the same frequency as the driving field. I found the time average power absorbed by the oscillator and the time average number of electrons per unit time absorbed. (ie these quantities:)
[tex] \bigl \langle P \bigr \rangle &=& \bigl\langle\frac{dW}{dt}\bigr\rangle\\ &=& \frac{1}{2}eE_0\omega\nu[/tex]
(nu is just a constant that is frequency dependent... this equation is correct)
[tex]\bigl\langle\frac{dN}{dt}\bigr\rangle=\frac{eE_0}{\hbar}\nu[/tex] (this equation should be right but I can't grasp it conceptually.
What I can't understand is the factor of 2 that is associated with the number of electrons absorbed. Why isn't the number of electrons absorbed per unit time simply the time average power divided by [tex]\hbar\omega[/tex]?
Does it have something to do with the fact that the oscillator makes two displacements from and back to equilibrium in one period? Or, does it have something to do with emission? Where is the extra energy going, heat? Any help getting my mind around this one would be VERY much appreciated.
I originally posted this question in the homework section, but I really don't need any help calculating anything, my answers are right. I'm having conceptual trouble, so I figured that this question belongs here.
So, let's say there is a field driving a single atomic oscillator (hydrogen for arguments sake). As a further simplification, let's say the oscillator has the same frequency as the driving field. I found the time average power absorbed by the oscillator and the time average number of electrons per unit time absorbed. (ie these quantities:)
[tex] \bigl \langle P \bigr \rangle &=& \bigl\langle\frac{dW}{dt}\bigr\rangle\\ &=& \frac{1}{2}eE_0\omega\nu[/tex]
(nu is just a constant that is frequency dependent... this equation is correct)
[tex]\bigl\langle\frac{dN}{dt}\bigr\rangle=\frac{eE_0}{\hbar}\nu[/tex] (this equation should be right but I can't grasp it conceptually.
What I can't understand is the factor of 2 that is associated with the number of electrons absorbed. Why isn't the number of electrons absorbed per unit time simply the time average power divided by [tex]\hbar\omega[/tex]?
Does it have something to do with the fact that the oscillator makes two displacements from and back to equilibrium in one period? Or, does it have something to do with emission? Where is the extra energy going, heat? Any help getting my mind around this one would be VERY much appreciated.