这是一份GLA格拉斯哥大学MATHS3021_1作业代写的成功案例

We consider an applied field $E_{0}$ that is small compared to the internal fields of the atoms. that is,
$$
E_{0} \sim \frac{e}{a_{B}^{2}} \sim \frac{26 \mathrm{Volts}}{a_{B}} \sim 5 \times 10^{\circ} \mathrm{Volts} / \mathrm{cm}
$$
where $e$ is the electron’s charge and $a_{B}$ the Bohr radius:
$$
a_{N}=\frac{\hbar^{2}}{m e^{2}} \sim \frac{1}{2} \times 10^{-8} \mathrm{~cm}
$$
Here, $\hbar$ is Planck’s constant divided by $2 \pi$ and $m$ is the electron mass.

MATHS5054_1 COURSE NOTES ：

$$
Q_{u j}=\alpha_{Q}\left(\frac{\partial E_{1}}{\partial x_{j}}+\frac{\partial E_{i}}{\partial x_{j}}\right)
$$
The quadrupole polarizability $\alpha_{O}$ in has the dimensionality $L^{S}$. so we expect $\alpha_{Q}$ for an atom to be $\sim a_{m}^{5}$. The field generated by the induced moment, in analogy with, will be
$$
\delta E_{Q} \sim \alpha_{\circlearrowright} \frac{\partial E_{i 1}}{\partial x} \sum_{i} \frac{1}{\left|\mathbf{r}-\mathbf{r}{i}\right|^{+}} $$ or $$ \delta E{\circlearrowright}-\frac{E_{11}}{R} \times \frac{N \alpha_{Q}}{R^{4}}
$$