IPhO 2011 · Theory problem 3

To Commemorate the Centenary of Rutherford's Atomic Nucleus: the Scattering of an Ion by a Neutral Atom

Download PDFIPhO 2011
Figure 1

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 points]

Figure 2

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. [3.0 points]

3.3 What is the electric potential energy of the ion-atom interaction in terms of \(\alpha\), \(Q\) and \(r\)? [0.9 points]

3.4 Find the expression for \(r_{\min}\), the distance of the closest approach, as shown in Figure 1. [2.4 points]

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 points]

Official English text of the problem.

Answers

Let \(k=1/(4\pi\varepsilon_0)\), and let \(\hat{\mathbf r}\) point from the atom towards the ion. Set \(U(\infty)=0\).

  • 3.1 \[ \boxed{\vec E_p=\frac{2k}{r^3}\vec p.} \]

  • 3.2 \[ \boxed{\vec f=-\frac{2\alpha k^2Q^2}{r^5}\hat{\mathbf r}.} \] The force is attractive for either sign of \(Q\).

  • 3.3 \[ \boxed{U(r)=-\frac{\alpha k^2Q^2}{2r^4} =-\frac{\alpha Q^2}{32\pi^2\varepsilon_0^2r^4}.} \]

  • 3.4 For a scattered trajectory, \(b>b_0\), \[ \boxed{ r_{\min}= \left[ \frac{b^2+\sqrt{b^4-\dfrac{4\alpha k^2Q^2}{mv_0^2}}}{2} \right]^{1/2}.} \] At \(b=b_0\), the ion asymptotically approaches \(r=b_0/\sqrt2\). For \(b<b_0\), there is no positive radial turning point.

  • 3.5 \[ \boxed{b_0=\left(\frac{4\alpha k^2Q^2}{mv_0^2}\right)^{1/4},} \qquad \boxed{ A=\frac{2\pi k|Q|}{v_0}\sqrt{\frac{\alpha}{m}} =\frac{|Q|}{2\varepsilon_0v_0}\sqrt{\frac{\alpha}{m}}.} \]

Solution

Since \(M\gg m\), we neglect the recoil of the atom and treat its position as fixed. We use the ideal induced-dipole model specified in the question, with \(\alpha>0\).

3.1

Represent the dipole by charges \(+q\) and \(-q\), separated by \(2a\), so that \(p=2aq\). At a point on the axis in the direction of \(\vec p\), the distances from the two charges are \(r-a\) and \(r+a\).

The fields of the two charges point in opposite directions, giving \[ \vec E_p =kq\left[\frac{1}{(r-a)^2}-\frac{1}{(r+a)^2}\right]\hat{\mathbf p}, \] where \(\hat{\mathbf p}\) is the unit vector along the dipole moment.

Combining the fractions, \[ \vec E_p =\frac{4kqar}{(r^2-a^2)^2}\hat{\mathbf p}. \] In the ideal-dipole limit \(a/r\to0\), with \(p=2aq\) fixed, \[ \boxed{\vec E_p=\frac{2kp}{r^3}\hat{\mathbf p} =\frac{2k}{r^3}\vec p.} \]

The field on the opposite side of the dipole axis also points along \(\vec p\) and has the same magnitude at the same distance.

3.2

Let the ion be at \(\vec r=r\hat{\mathbf r}\) relative to the atom. The electric field produced by the ion at the atom is \[ \vec E_{\mathrm{ion}} =-\frac{kQ}{r^2}\hat{\mathbf r}. \] Consequently, the induced dipole moment is \[ \vec p=\alpha\vec E_{\mathrm{ion}} =-\frac{\alpha kQ}{r^2}\hat{\mathbf r}. \]

The ion lies on the axis of this dipole. Using the result of 3.1, \[ \vec E_p =\frac{2k}{r^3}\vec p =-\frac{2\alpha k^2Q}{r^5}\hat{\mathbf r}. \] Thus the force on the ion is \[ \boxed{ \vec f=Q\vec E_p =-\frac{2\alpha k^2Q^2}{r^5}\hat{\mathbf r} =-\frac{\alpha Q^2}{8\pi^2\varepsilon_0^2r^5}\hat{\mathbf r}.} \]

Because \(\alpha>0\) and the charge enters as \(Q^2\), the force always points towards the atom, regardless of the sign of \(Q\).

3.3

For a central conservative force, \[ f_r=-\frac{dU}{dr}. \] Choosing zero potential energy at infinity, \[ U(r) =-\int_\infty^r f_r(s)\,ds =-\int_\infty^r\left(-\frac{2\alpha k^2Q^2}{s^5}\right)ds. \] Therefore, \[ \boxed{U(r)=-\frac{\alpha k^2Q^2}{2r^4}.} \]

Equivalently, \[ U=-\frac12\alpha E_{\mathrm{ion}}^2. \] The factor \(1/2\) accounts for the energy required to induce the dipole. Indeed, the polarization energy \(p^2/(2\alpha)\), together with the coupling energy \(-pE_{\mathrm{ion}}\), gives \(-\alpha E_{\mathrm{ion}}^2/2\) when \(p=\alpha E_{\mathrm{ion}}\).

3.4

The force is central, so angular momentum about the atom is conserved. Its magnitude is determined by the incoming trajectory: \[ L=mv_0b. \]

Mechanical energy is also conserved. At infinity, the potential energy vanishes, so \[ \mathcal E=\frac12mv_0^2. \] Separating the kinetic energy into radial and transverse contributions gives \[ \frac12m\dot r^2+\frac{L^2}{2mr^2} -\frac{\alpha k^2Q^2}{2r^4} =\frac12mv_0^2. \]

At the closest approach of a scattered trajectory, \(\dot r=0\). Substituting \(L=mv_0b\), \[ \frac{mv_0^2b^2}{2r_{\min}^2} -\frac{\alpha k^2Q^2}{2r_{\min}^4} =\frac12mv_0^2. \] Multiplication by \(2r_{\min}^4/(mv_0^2)\) gives \[ r_{\min}^4-b^2r_{\min}^2 +\frac{\alpha k^2Q^2}{mv_0^2}=0. \]

This is a quadratic equation in \(r_{\min}^2\), whose roots are \[ r_\pm^2 =\frac{b^2\pm\sqrt{b^4-\dfrac{4\alpha k^2Q^2}{mv_0^2}}}{2}. \] An ion arriving from infinity encounters the larger root first. Hence \[ \boxed{ r_{\min}= \left[ \frac{b^2+\sqrt{b^4-\dfrac{4\alpha k^2Q^2}{mv_0^2}}}{2} \right]^{1/2}.} \]

The two roots are distinct for \(b>b_0\), with \(b_0\) determined below. As a check, when the interaction tends to zero, this expression gives \(r_{\min}\to b\), as expected for straight-line motion.

3.5

The radial motion is governed by the effective potential \[ U_{\mathrm{eff}}(r) =\frac{L^2}{2mr^2}-\frac{\alpha k^2Q^2}{2r^4}. \] For nonzero angular momentum, the positive centrifugal term dominates at large distances, while the attractive term dominates at short distances. The resulting potential barrier can prevent the ion from reaching the atom.

The critical trajectory occurs when the two radial turning points merge. Thus the discriminant found in 3.4 must vanish: \[ b_0^4-\frac{4\alpha k^2Q^2}{mv_0^2}=0. \] It follows that \[ \boxed{b_0=\left(\frac{4\alpha k^2Q^2}{mv_0^2}\right)^{1/4}.} \]

For \(b<b_0\), the radial-energy equation \[ \dot r^2 =v_0^2\left( 1-\frac{b^2}{r^2} +\frac{\alpha k^2Q^2}{mv_0^2r^4} \right) \] has no positive zero. The incoming ion therefore continues inward and, under the assumption in the question, undergoes charge exchange.

At exactly \(b=b_0\), \[ \dot r^2 =v_0^2\left(1-\frac{b_0^2}{2r^2}\right)^2. \] The incoming ion approaches the unstable circular-orbit radius \[ r_c=\frac{b_0}{\sqrt2} \] asymptotically. This boundary has zero area in the impact-parameter plane and does not affect the cross section.

The captured trajectories occupy a disk of radius \(b_0\) in that plane. Therefore, \[ A=\pi b_0^2 =\pi\sqrt{\frac{4\alpha k^2Q^2}{mv_0^2}}, \] or \[ \boxed{ A=\frac{2\pi k|Q|}{v_0}\sqrt{\frac{\alpha}{m}} =\frac{|Q|}{2\varepsilon_0v_0}\sqrt{\frac{\alpha}{m}}.} \] With SI inputs, \(A\) is in \(\mathrm{m^2}\). No numerical parameter values are supplied.

The solution was checked against the official solution.