1
$\begingroup$

How can the power (in Watts) absorbed by the electron be calculated, knowing the incident electric field amplitude $ E_0 $, wavelength $ \lambda $, and the electron momentum relaxation time $ \tau $ in the medium ?

The units seem to check out for $ \frac {|q_e| E_0 \lambda }\tau $ (result is in Watts), but the correct answer is off by several orders of magnitude. What is missing?

$\endgroup$
0
$\begingroup$

I guess you're also doing the Plasmonics course of edX. (Great course by the way, although I find the exercises quite difficult.)

I solved the task using the formula for power that was shown in the lecture video:
$$ P = <\cfrac{\vec{F}+\vec{F}^*}{2}\cdot \cfrac{\vec{v}+\vec{v}^*}{2}> = \cfrac{1}{2}\Re (\vec{F}^* \cdot \vec{v}) $$

The velocity is time derivative of position $\vec{r}$. To get $\vec{r}$ let's solve the equation of motion that is accounting for collisions (this was done in the edX course):
$$ m_e\ddot{\vec{r}}=-\gamma m_e \dot{\vec{r}} - |e|\vec{E_0}e^{-i\omega t}$$ which yields the solution:
$$\vec{r}(t)=\cfrac{|e|}{m_e\, \omega \,(\omega + i\gamma)}\vec{E_0}e^{-i\omega t}$$ $\vec{F}$ is Lorentz force due to electric field:
$$\vec{F} = |e| \vec{E_0}e^{-i\omega t}$$

So that gives me: $$ P = \cfrac{1}{2}\Re (\vec{F}^* \cdot \vec{v}) = \cfrac{|e|^2}{2} \Re \left(E_0^2 \cfrac{-i\omega}{m_e \omega(\omega + i\gamma)} \right) = \cfrac{E_0^2\,e^2 \gamma}{2 m_e \,(\omega^2 + \gamma^2)} $$

Now we now that $\lambda = \frac{2 \pi c}{\omega}$, from this we can calculate $\omega$ knowing the wavelength. And following again the defitions from edX course $\gamma = \frac{1}{\tau}$.

As you can see, the unit is still OK although the expression changed a lot.

I would like to know if there is a simpler way but I cannot think of any right now.

$\endgroup$
  • $\begingroup$ Incredibly clear and straightforward explanation. Thank you so much. $\endgroup$ – WantsToLearn Feb 4 at 21:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.