# Power absorbed by electron in plane electromagnetic wave

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?

## 1 Answer

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.

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