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IPhO 2011, Теория — Задача 3. To Commemorate the Centenary of Rutherford’s Atomic Nucleus: the Scattering of an Ion by a Neutral Atom

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Всички материали от IPhO 2011

Theoretical Competition: 12 July 2011 — Question 3 · 12 юли 2011 г.

Условие

Trajectory of an ion (labelled Ion, m, Q) coming from the left with speed v_0 along a dashed horizontal line at impact parameter b above a dotted line through the atom (Atom, M). The ion's path bends down towards the atom; at an intermediate point the ion has velocity v and position vector r from the atom at angle φ; the closest-approach distance r_min is marked perpendicular to the path; the path then leaves downwards to the right. — FIGURE 1

An ion of mass $m$ , charge $Q$ , is moving with an initial non-relativistic speed $v_0$ from a great distance towards the vicinity of a neutral atom of mass $M \gg m$ and of electrical polarisability $\alpha$ . The impact parameter is $b$ as shown in Figure 1.

The atom is instantaneously polarised by the electric field $\vec{E}$ of the in-coming (approaching) ion.

The resulting electric dipole moment of the atom is $\vec{p} = \alpha\vec{E}$ . Ignore any radiative losses in this problem.

3.1 Calculate the electric field intensity $\vec{E}_p$ at a distance $r$ from an ideal electric dipole $\vec{p}$ at the origin O along the direction of $\vec{p}$ in Figure 2. [1,2 т.]

An electric dipole: charge −q and charge +q at distance a on each side of the origin O along a straight line, with the dipole moment arrow p pointing from −q to +q; a point on the extended axis at distance r from O is marked. The label reads p = 2aq, r ≫ a. — FIGURE 2

3.2 Find the expression for the force $\vec{f}$ acting on the ion due to the polarised atom. Show that this force is attractive regardless of the sign of the charge of the ion. [3 т.]

3.3 What is the electric potential energy of the ion-atom interaction in terms of $\alpha, Q$ and $r$ ? [0,9 т.]

3.4 Find the expression for $r_{\mathrm{min}}$ , the distance of the closest approach, as shown in Figure 1. [2,4 т.]

3.5 If the impact parameter $b$ is less than a critical value $b_0$ , the ion will descend along a spiral to the atom. In such a case, the ion will be neutralized, and the atom is, in turn, charged. This process is known as the “charge exchange” interaction. What is the cross sectional area $A = \pi b_0^2$ of this “charge exchange” collision of the atom as seen by the ion? [2,5 т.]

Отговори

Решение

QUESTION 3: SOLUTION

Using Coulomb’s Law, we write the electric field at a distance $r$ is given by

E_p = \frac{q}{4\pi\varepsilon_0 (r-a)^2} - \frac{q}{4\pi\varepsilon_0 (r+a)^2}

E_p = \frac{q}{4\pi\varepsilon_0 r^2}\left(\frac{1}{\left(1-\frac{a}{r}\right)^2} - \frac{1}{\left(1+\frac{a}{r}\right)^2}\right) \qquad \ldots\ldots(1)

Using binomial expansion for small $a$ ,

E_p = \frac{q}{4\pi\varepsilon_0 r^2}\left(1+\frac{2a}{r}-1+\frac{2a}{r}\right)

= +\frac{4qa}{4\pi\varepsilon_0 r^3} = +\frac{qa}{\pi\varepsilon_0 r^3} \qquad \ldots\ldots(2)

= \frac{2p}{4\pi\varepsilon_0 r^3}

The electric field seen by the atom from the ion is

\vec{E}_{ion} = -\frac{Q}{4\pi\varepsilon_0 r^2}\hat{r} \qquad \ldots\ldots(3)

The induced dipole moment is then simply

\vec{p} = \alpha\vec{E}_{ion} = -\frac{\alpha Q}{4\pi\varepsilon_0 r^2}\hat{r} \qquad \ldots\ldots(4)

From eq. (2)

\vec{E}_p = \frac{2p}{4\pi\varepsilon_0 r^3}\hat{r}

The electric field intensity $\vec{E}_p$ at the position of an ion at that instant is, using eq. (4),

\vec{E}_p = \frac{1}{4\pi\varepsilon_0 r^3}\left[-\frac{2\alpha Q}{4\pi\varepsilon_0 r^2}\hat{r}\right] = -\frac{\alpha Q}{8\pi^2\varepsilon_0^2 r^5}\hat{r}

The force acting on the ion is

\vec{f} = Q\vec{E}_p = -\frac{\alpha Q^2}{8\pi^2\varepsilon_0^2 r^5}\hat{r} \qquad \ldots\ldots(5)

The “-’’ sign implies that this force is attractive and $Q^2$ implies that the force is attractive regardless of the sign of $Q$ .

The potential energy of the ion-atom is given by $U = \int_r^{\infty} \vec{f}.d\vec{r}$ $\ldots\ldots(6)$

Using this, $U = \int_r^{\infty} \vec{f}.d\vec{r} = -\frac{\alpha Q^2}{32\pi^2\varepsilon_0^2 r^4}$ $\ldots\ldots(7)$

[Remark: Students might use the term $-\vec{p}\cdot\vec{E}$ which changes only the factor in front.]

At the position $r_{\mathrm{min}}$ we have, according to the Principle of Conservation of Angular Momentum,

m v_{\mathrm{max}} r_{\mathrm{min}} = m v_0 b

v_{\mathrm{max}} = v_0 \frac{b}{r_{\mathrm{min}}} \qquad \ldots\ldots(8)

And according to the Principle of Conservation of Energy:

\frac{1}{2} m v_{\mathrm{max}}^2 + \frac{-\alpha Q^2}{32\pi^2\varepsilon_0^2 r^4} = \frac{1}{2} m v_0^2 \qquad \ldots\ldots(9)

Eqs.(12) & (13):

\left(\frac{b}{r_{\mathrm{min}}}\right)^2 - \frac{\alpha Q^2 \Big/ \frac{1}{2} m v_0^2}{32\pi^2\varepsilon_0^2 b^4}\left(\frac{b}{r_{\mathrm{min}}}\right)^4 = 1

\left(\frac{r_{\mathrm{min}}}{b}\right)^4 - \left(\frac{r_{\mathrm{min}}}{b}\right)^2 + \frac{\alpha Q^2}{16\pi^2\varepsilon_0^2 m v_0^2 b^4} = 0 \qquad \ldots\ldots(10)

The roots of eq. (14) are:

r_{\mathrm{min}} = \frac{b}{\sqrt{2}}\left[1 \pm \sqrt{1-\frac{\alpha Q^2}{4\pi^2\varepsilon_0^2 m v_0^2 b^4}}\right]^{\frac{1}{2}} \qquad \ldots\ldots(11)

[Note that the equation (14) implies that $r_{\mathrm{min}}$ cannot be zero, unless $b$ is itself zero.]

Since the expression has to be valid at $Q = 0$ , which gives

r_{\mathrm{min}} = \frac{b}{\sqrt{2}}\left[1 \pm 1\right]^{\frac{1}{2}}

We have to choose “+” sign to make $r_{\mathrm{min}} = b$

Hence,

r_{\mathrm{min}} = \frac{b}{\sqrt{2}}\left[1 + \sqrt{1-\frac{\alpha Q^2}{4\pi^2\varepsilon_0^2 m v_0^2 b^4}}\right]^{\frac{1}{2}} \qquad \ldots\ldots(12)

A spiral trajectory occurs when (16) is imaginary (because there is no minimum distance of approach).

$r_{\mathrm{min}}$ is real under the condition:

1 \ge \frac{\alpha Q^2}{4\pi^2\varepsilon_0^2 m v_0^2 b^4}

b \ge b_0 = \left(\frac{\alpha Q^2}{4\pi^2\varepsilon_0^2 m v_0^2}\right)^{\frac{1}{4}} \qquad \ldots\ldots(13)

For $b < b_0 = \left(\frac{\alpha Q^2}{4\pi^2\varepsilon_0^2 m v_0^2}\right)^{\frac{1}{4}}$ the ion will collide with the atom.

Hence the atom, as seen by the ion, has a cross-sectional area $A$ ,

A = \pi b_0^2 = \pi\left(\frac{\alpha Q^2}{4\pi^2\varepsilon_0^2 m v_0^2}\right)^{\frac{1}{2}} \qquad \ldots\ldots(14)

Оригинал в Архива: 3.pdf · официални решения: 3_sol.pdf