According to quantum mechanics, the electron cloud for a hydrogen atom in the ground state has a charge density
\rho(r)=\frac{q}{\pi a^{3}} e^{-2 r / a} ,
where q is the charge of the electron and a is the Bohr radius. Find the atomic polarizability of such an atom. [Hint: First calculate the electric field of the electron cloud E_{e}(r) ; then expand the exponential, assuming r \ll a \cdot^{1}
First find the field, at radius r, using Gauss’ law: \int E \cdot d a =\frac{1}{\epsilon_{0}} Q_{ enc }, \text { or } E=\frac{1}{4 \pi \epsilon_{0}} \frac{1}{r^{2}} Q_{\text {enc }} .
Q_{ enc }=\int_{0}^{r} \rho d \tau=\frac{4 \pi q}{\pi a^{3}} \int_{0}^{r} e^{-2 \bar{r} / a} \bar{r}^{2} d \bar{r}=\left.\frac{4 q}{a^{3}}\left[-\frac{a}{2} e^{-2 \bar{r} / a}\left(\bar{r}^{2}+a \bar{r}+\frac{a^{2}}{2}\right)\right]\right|_{0} ^{r}
=-\frac{2 q}{a^{2}}\left[e^{-2 r / a}\left(r^{2}+a r+\frac{a^{2}}{2}\right)-\frac{a^{2}}{2}\right]=q\left[1-e^{-2 r / a}\left(1+2 \frac{r}{a}+2 \frac{r^{2}}{a^{2}}\right)\right] .
\left[\text { Note: } Q_{\text {enc }}(r \rightarrow \infty)=q .\right] So the field of the electron cloud is E_{ e }=\frac{1}{4 \pi \epsilon_{0}} \frac{q}{r^{2}}\left[1-e^{-2 r / a}\left(1+2 \frac{r}{a}+2 \frac{r^{2}}{a^{2}}\right)\right] . The proton will be shifted from r = 0 to the point d where E_{ e }=E (the external field):
E=\frac{1}{4 \pi \epsilon_{0}} \frac{q}{d^{2}}\left[1-e^{-2 d / a}\left(1+2 \frac{d}{a}+2 \frac{d^{2}}{a^{2}}\right)\right] .
Expanding in powers of (d/a):
e^{-2 d / a}=1-\left(\frac{2 d}{a}\right)+\frac{1}{2}\left(\frac{2 d}{a}\right)^{2}-\frac{1}{3 !}\left(\frac{2 d}{a}\right)^{3}+\cdots=1-2 \frac{d}{a}+2\left(\frac{d}{a}\right)^{2}-\frac{4}{3}\left(\frac{d}{a}\right)^{3}+\cdots
1-e^{-2 d / a}\left(1+2 \frac{d}{a}+2 \frac{d^{2}}{a^{2}}\right)=1-\left(1-2 \frac{d}{a}+2\left(\frac{d}{a}\right)^{2}-\frac{4}{3}\left(\frac{d}{a}\right)^{3}+\cdots\right)\left(1+2 \frac{d}{a}+2 \frac{d^{2}}{a^{2}}\right)
=1-1-2 \frac{d}{a}-2 \frac{d^{2}}{d^{2}}+2 \frac{d}{a}+4 \frac{d^{2}}{d^{2}}+4 \frac{d^{3}}{d^{3}}-2 \frac{d^{2}}{d^{2}}-4 \frac{d_{1}^{3}}{a^{3}}+\frac{4}{3} \frac{d^{3}}{a^{3}}+\cdots
=\frac{4}{3}\left(\frac{d}{a}\right)^{3}+ higher order terms.
E=\frac{1}{4 \pi \epsilon_{0}} \frac{q}{d^{2}}\left(\frac{4}{3} \frac{d^{3}}{a^{3}}\right)=\frac{1}{4 \pi \epsilon_{0}} \frac{4}{3 a^{3}}(q d)=\frac{1}{3 \pi \epsilon_{0} a^{3}} p . \quad \alpha=3 \pi \epsilon_{0} a^{3} .
[Not so di↵erent from the uniform sphere model of Ex. 4.1 (see Eq. 4.2).
\alpha=4 \pi \epsilon_{0} a^{3}=3 \epsilon_{0} v (4.2)
Note that this result predicts \frac{1}{4 \pi \epsilon_{0}} \alpha=\frac{3}{4} a^{3}=\frac{3}{4}\left(0.5 \times 10^{-10}\right)^{3}=0.09 \times 10^{-30} m ^{3} , compared with an experimental value (Table 4.1) of 0.66 \times 10^{-30} m ^{3} . Ironically the “classical” formula (Eq. 4.2) is slightly closer to the empirical value.]
| He | Li | Be | C | Ne | Na | Ar | K | Cs |
| 0.205 | 24.3 | 5.60 | 1.67 | 0.396 | 24.1 | 1.64 | 43.4 | 59.4 |