Calculate electric polarizability

[DannyJ108]

Homework Statement

Considering electric cloud charge distribution and obtaining the value for non linear polarizability coefficients

Relevant Equations

$\rho(r) = \frac {Ze} {\pi a^3} e^{-\frac {2r} a$

Hello everyone,

I've got my non-uniform electric cloud distribution formula given by:

$\rho(r) = \frac {-Ze} {\pi a^3} e^{-{2r}/a}$

Where $Z$ is the atomic number of the atom in question and $a$ Bohr's radius and $E_L$ the local electric field.
Considering the previus expression , we obtain a value for non linear polarizability ( $\alpha \equiv f(\vec {E_L}$ ). If we write this as:

$\alpha(E_L) =\alpha_0 +\alpha_1 E_L + ···$

Determine the values of $\alpha_0$ and $\alpha_1$

That is the homework statement. How I've proceeded:

I determined the electronic cloud's electric field ($E_e$) ussing Gauss law. And expanded in powers for the $e^{-2r/a}$ term (and not writing terms in higher order than 4) and got:

$E_e=\frac {Ze} {3 \pi \epsilon_0 a^3} r$

I then matched $E_e = E_L$ at a certain distance $d$, and knowing that

$p=Ze·d=\alpha E_L$

I obtain that $\alpha=3\pi \epsilon_0a^3$

How should I proceed from here? Do I just substitute my $\alpha$ in the expression given by the homework statement? If that's so, to obtain $\alpha_0$ I assumed d=0 (right on the atom's nucleus) and I get that $\alpha_0=\alpha$, which doesn't seem right. And then I've got a nuisance to obtain $\alpha_1$

Also the homework statement says we obtain $\alpha$ in terms of $E_L$, though I don't know if it means once I found $\alpha_0$ and $\alpha_1$ it will be in terms of the local field, or if this $\alpha=3\pi \epsilon_0a^3$ I obtained is wrong and it should be another in terms of $E_L$.

Thanks for the help!

 

[mark726]

Hello there,

First of all, great job in determining the values of $\alpha_0$ and $\alpha_1$ using the given formula and your calculations. However, I would like to suggest a few things to consider.

Firstly, it is important to note that the expression for $\alpha(E_L)$ is a Taylor series expansion, which means that it is valid for small values of $E_L$. In your calculation, you have assumed that $E_e=E_L$ at a certain distance $d$. However, this may not always be the case and it is possible that the value of $E_L$ may be much larger than the value of $E_e$ at that distance. Therefore, it would be more accurate to use a more general expression for $\alpha(E_L)$, such as:

$\alpha(E_L)=\alpha_0+\alpha_1E_L+\alpha_2E_L^2+...$

where $\alpha_2$ and higher order terms take into account the non-linearity of the polarizability.

Secondly, it is important to consider the units of the values you have obtained for $\alpha_0$ and $\alpha_1$. The units of polarizability are usually in $C^2m^2J^{-1}$, so make sure your values also have the correct units.

Finally, in terms of the statement about obtaining $\alpha$ in terms of $E_L$, I believe it means that the final expression for $\alpha(E_L)$ should be in terms of the local electric field $E_L$. This means that your final answer should not have any other variables, such as $a$ or $Z$.

I hope this helps and good luck with your calculations!