Atom-ion interplay

To start off, we note that the electrostatic force $F$ between the ion and the polarized atom is given as $F = \frac{Qq}{4\pi\epsilon_0}\left(\frac{1}{(r-d)^2} - \frac{1}{(r+d)^2}\right) \ ,$ where $r$ is the distance between the ion and the centre of the atom, $Q$ is the charge of the ion, and $q$ is the charge of the polarized atom.

The charges on the atom that are attracted to the ion will naturally be closer to the ion, hence the $r-d$ portion of the equation, and the charges that are repelled by the ion will be on the opposite side of the atom, hence the $r+d$ portion of the equation.

Now we note that $p = \alpha E_{external} = 2qd\ .$ Rearranging for $q$ , we obtain $q = \frac{\alpha E_{external}}{2d} \ .$ We now substitute this result into our first equation, divide $F$ by $\alpha$ , and factor out an $\frac{1}{r^2}$ to obtain $\frac{F}{\alpha} = \frac{Q}{4\pi\epsilon_0}\frac{E_{external}}{2dr^2}\left(\frac{1}{\left(1-\frac{d}{r}\right)^2} - \frac{1}{\left(1-\frac{d}{r}\right)^2}\right) \ .$ Now using the binomial series, which can be generally written as $(1+x)^n = 1 + nx + \frac{n(n-1)}{2!}x^2\, + \,\, ... \ ,$ we can rewrite our equation to be $\frac{F}{\alpha} = \frac{Q}{4\pi\epsilon_0}\frac{E_{external}}{2dr^2}\left(\left(1 + \frac{2d}{r} + ...\right) - \left(1 - \frac{2d}{r}+...\right)\right) \ .$ We only take the first two terms in our binomial series, since $d << r$ . Thus we get the equation $\frac{F}{\alpha} = \frac{Q}{4\pi\epsilon_0}\frac{2E_{external}}{r^3} \ .$ $E_{external}$ is simply the electric field of the ion, so we get $\frac{F}{\alpha} = 2\left(\frac{Q}{4\pi\epsilon_0}\right)^2\frac{1}{r^5} = 161.6 \ .$

We can know that the attraction force field from the ion is taken as $E_{external} = \frac{1}{4\pi\epsilon_0}\frac{q_{ion}}{r^2}$ . It is also of common knowledge that a dipole charge can make fields such like $F_{attraction} = \left(\frac{1}{4\pi\epsilon_0}\left(\frac{2p}{r^3}\right)-\frac{1}{3\epsilon_0}p\delta^3(r)\right) q_{ion}$ . The second delta term being 0.

We also have knowledge that $F_{attraction}/\alpha = F_{attraction}\frac{E_{external}}{p}$ . The term on the left is what we like to solve for, the attraction force over the polarizability. Working through the mess, the right piece turns into $2\left(1/4\pi\epsilon_0\right)q_{ion}^2/r^5$ . If we now simply plug into the values for the ionic charge and the distance of 1 meter, we find out that that ratio is ~162.

the field the ion causes on the atom: E=\frac{kq}{r^2}

the dipole induced by that field: p= /alpha E=/alpha \frac{kq}{r^2}

the attractive force: F=2\frac{kqp}{r^3}=2\alpha*\frac{k^2q^2}{d^5}

\alpha=2\frac{k^2q^2}{d^5}=162

the formula of attractive form below can be deduced form forces between individual pairs of charge.(the two charges Q of dipole are separated by d) F=\frac{kqQ}{r^2}-\frac{kqQ}{(d+r)^2}. Since d<<r, we obtain that F=2\frac{kqQd}{r^3}=2\frac{kqp}{r^3}

E = k Q/R^2
F = k
q Q/(R-d)^2-k q Q/(R+d)^2 ~= k q Q((R+d)^2-(R-d)^2)/R^4 ~= 4 k q Q d/R^3
a = 2
q d/E
F/a = F
E/(2 q d) = 2 k^2 Q^2/R^5 ~= 2 * 8.98755e9^2 * 1e-9^2 (N^2/C^2/m) ~= 161.552
But why 161.869? For that we should have k=8.99636e9 instead of 8.98755e9, that is, round pi to 3.14 and e0 to 8.85e-12 from 8.8541878e-12, that is quite too rough for intermediate calculations, so getting last 3 digits completely wrong. To get pi=3.1416, e0=8.8542e-12 would be much better.

Let the distance from the ion to the center of the atom be $r$ , the charge of the ion be $Q$ . Let the Coulomb constant be $k=\frac{1}{4 \pi \epsilon_0} \approx 8.987552 \times 10^9$

The attractive force a the atom acts on the ion

$F=\frac{kQq}{(r-d)^2} - \frac{kQq}{(r+d)^2}$ $=kQq \times [\frac{1}{(r-d)^2} - \frac{1}{(r+d)^2}]$ $=kQq \times \frac{2d \times 2r}{((r-d)(r+d))^2}$

Because $r>>d$ , then $(r-d)(r+d) \approx r^2$

Thus, $F=kQq \times \frac{2d \times 2r}{r^4} = \frac {4kQqd} {r^3} = \frac {2kQp} {r^3}$

On the other hand, the atom's polarizability $\alpha = \frac {p} {E_{external}} = \frac {p} {\frac {kQ} {r^2}} = \frac {pr^2} {kQ}$

Hence, the ratio is $\frac {F}{p} = \frac {2kQp} {r^3} \times \frac {kQ} {pr^2} = \frac {2(kQ)^2} {r^5} = 161.552$

$r=1 \ \mathrm{m}$ is the distance between the atom and the ion. $Q=1 \ \mathrm{nC}$ is the charge on the ion.

The magnitude of the force $F$ between the atom and the ion is the sum of the forces by each of the two charges in the dipole. The orientation of the dipole is such that the negative end is closer to the ion than the positive end, so the net force on the ion will be attractive. The distance between the negative end and the ion is $r-d$ , and the distance between the positive end and the ion is $r+d$ . So, by Coulomb's Law, the magnitude of the force is

$$F=\frac{kqQ}{(r-d)^2}-\frac{kqQ}{(r+d)^2}.$$

We are asked for the ratio of the force and the polarizability, so we must manipulate the equations given for dipole moment.

$$p=\alpha E, \ p=2qd$$

$$\alpha=2qd/E.$$

The electric field $E$ is simply given by the equation for electric field at a distance $r$ due to a charge $Q$ , which is

$$E=\frac{kQ}{r^2}.$$

So our equation for $\alpha$ becomes

$$\alpha = \frac{2r^2 qd}{kQ}.$$

So the ratio between the force and polarizability is

$$\frac{F}{\alpha}=\frac{k^2 Q^2}{2r^2d(r-d)^2}-\frac{k^2 Q^2}{2r^2d(r+d)^2}.$$

Algebraically manipulating this (getting a common denominator and cancelling) yields

$$\frac{F}{\alpha}=\frac{2k^2 Q^2}{r(r^2-d^2)^2}.$$

We are told to assume the separation between the charges in the atomic dipole is much smaller than the atom-ion distance, so we will take a look at what happens for very small $d$ . Formally, we take the limit as $d \to 0$ .

$$\frac{F}{\alpha}=\lim_{d \to 0} {\frac{2k^2 Q^2}{r(r^2-d^2)^2}=\frac{2k^2 Q^2}{r^5}}$$

$$=\frac{2(8.99 \times 10^9 \ \mathrm{\frac{N\cdot m^2}{C^2}})^2 (1 \times 10^{-9} \ \mathrm{C})^2}{(1 \ \mathrm{m})^5}=161.6 \ \mathrm{\frac{N^2}{m \cdot C^2}}. $$

The electrical dipole moment is

$p = \alpha E = 2qd$

And $E = k \frac{Q}{r^{2}}$

Q is the charge of the ion r is the atom-ion distance Through algebraic manipulation we can obtain

$\alpha = \frac{2r^{2}qd}{kQ}$

The net force acting on the ion is

$F= \frac{kQq}{(r-d)^{2}} - \frac{kQq}{(r+d)^{2}} = kQq\frac{4rd}{(r-d)^{2}(r+d)^{2}}$

Therefore, the ratio is

$\frac{F}{\alpha} = kQq\frac{4rd}{(r-d)^{2}(r+d)^{2}} \times \frac{kQ}{2r^{2}qd} = \frac{2k^{2}Q^{2}}{r(r-d)^{2}(r+d)^{2}}$

And since we can assume d = 0, the value of this ratio will be 162 (to 3s.f.)

When the positively charged ion will pass by the spherical atom , it will induce an equal but opposite negative charge - q on the hemispherical surface facing the ion , which will further induce an equal but opposite positive charge + q on the other hemispherical surface of the atom . Since the atom-ion distance is very large as compared to the size of the atom , we can assume that the induced negative and positive charges lie at points on the surface of the atom directly opposite to ion and diametrically opposite to the first point . After this assumption , both the points of induced charge and the position of the ion will be collinear .

Let some symbols used here be defined as follows:

r = distance of the ion from the centre of the atom

2d = diameter of the atom

E_ {external} = electric field acting on the polarized atom due to the ion

F_ {+} = force on the ion due to the induced positive charge

F_ {-} = force on the ion due to the induced negative charge

F_ {net} = net force exerted on the ion due to the dipole of the polarized atom

k = \frac{1}{4πε_{₀}}

Again , since the atom-ion distance is very large as compared to the size of the atom , electric field for the atom due to the ion can be written as

E_ {external} = \frac{ kq }{ r ^2}

Now ,

F_ {+} = \frac{ kq ^2}{{ r + d }^2}

F_ {-} = \frac{ kq ^2}{{ r - d }^2}

As , F *{-}*>F {+}

therefore , F *{net}* = F {-} - F_ {+}

= \frac{ kq ^2}{{ r - d }^2} - \frac{ kq ^2}{{ r + d }^2}

= kq ^2[\frac{1}{{ r - d }^2} - \frac{1}{{ r + d }^2}

= kq ^2[\frac{{ r + d }^2 - { r - d }^2}{{ r ^2- d ^2}^2}

Since d << r , d ^2 is negligible .

\Rightarrow F_ {net} = \frac{4 kq ^2 d }{ r ^3}

Now , p = α E_ {external} = 2 qd

Hence , α = \frac{2 qd }{E_ {external} }

= \frac{2 r ^2 d }{ k }

Thus , the required ratio = \frac{F_ {net} }{ α }

= \frac{2 k ^2 q ^2}{ r ^5}

Substituting the values : q = {10}^{-9} C , r = 1 m , k = 9\times {10}^9 Nm^2C^{-2} , we get the required ratio as 162 .

In order to find the force on the ion, $F = Q E$ , we have to find the electric field with which the polarized atom acts on it. That electric field comes from the atomic dipole with dipole moment $p$ induced by the ion, $p = \alpha E_i = \alpha \frac{Q}{4 \pi \epsilon_0 r^2}$ .

The electric field from a general dipole with a dipole moment $p$ can be written as $E(r) = \frac{q}{4 \pi \epsilon_0} (\frac{1}{(r-a)^2} - \frac{1}{(r+a)^2})$ , where $p = 2qa$ . This can be further simplified using the fact that $\frac{1}{(r-a)^2} = \frac{1}{r^2 (1-\frac{a}{r})^2)} = \frac{1}{r^2} \times (1 + 2\frac{a}{r})$ when $a$ is much smaller than $r$ . We can use the same approximation for $r+a$ , and then we get that $E(r) = \frac{\alpha E_{external}}{2 \pi \epsilon_0 r^3}$ .

Plugging in the electric field from the ion into the electric field from the dipole, we get $E = \frac{Q \alpha}{8 \pi^2 \epsilon_0^2 r^5}$ , and $\frac{F}{\alpha} = \frac{Q E}{\alpha} = \frac {Q^2}{8 \pi^2 \epsilon_0^2 r^5} = 161.869 \frac{\mbox{N}^2}{\mbox{m C}^2}$ .