IPhO 2024 · Theory problem 2

Trapping Ions and Cooling Atoms

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In recent decades, trapping and cooling atoms and ions has been a fascinating topic for physicists, with several Nobel prizes awarded for work in this area. In the first part of this question, we will explore a technique for trapping ions, known as the "Paul trap". Wolfgang Paul and Hans Dehmelt received one half of the 1989 Nobel Prize in Physics for this work. Next, we investigate the Doppler cooling technique, one of the works cited in the press release for the 1997 Nobel Prize in Physics awarded to Steven Chu, Claude Cohen-Tannoudji, and William Daniel Phillips "for developments of methods to cool and trap atoms with laser light".

A. The Paul Trap

It is known that with electrostatic fields, it is not possible to create a stable equilibrium for a charged particle. Therefore, creating a stable equilibrium point for ions requires more sophisticated techniques. The Paul trap is one of these techniques.

Consider a ring of charge with a radius \(R\) and a uniform positive linear charge density \(\lambda\). A positive point charge \(Q\) with mass \(m\) is placed at the center of the ring.

Figure 1

Figure 1 - A positively charged ring with a uniform linear charge density \(\lambda\) and radius \(R\): the origin of the coordinate system is at the center of the ring.

A-1 a) In cartesian coordinates \((x,y,z)\), obtain the electric field due to the charged ring in the vicinity of the ring’s center to the first order in \(x/R\), \(y/R\), and \(z/R\).

b) Find the angular frequency of small oscillations of the charged particle around the center of the ring in the directions for which a stable equilibrium exists.		 1.5 pt

In order to trap the charge \(Q\) fully, we would like to apply alternating fields to produce a dynamic equilibrium. Assume that the charge density is \(\lambda=\lambda_0+u\cos\Omega t\) in which \(\lambda_0\), \(u\), and \(\Omega\) are adjustable. We shall ignore radiative effects. Then the equation of motion for small displacements from the center of the ring, along the direction perpendicular to the plane of the ring will turn out to be:

\[ \ddot z=(+k^2+a\Omega^2\cos\Omega t)z \tag{1} \]

A-2 Write \(a\) and \(k\) in terms of the known parameters. 0.4 pt

We would like to obtain an approximate solution to Equation (1) by making the following simplifying assumptions: \(a\ll1\), \(\Omega\gg k\), and \(a\Omega^2\gg k^2\). With these assumptions, it can be shown that the solution of this equation can be split into two parts: \(z(t)=p(t)+q(t)\), where \(p(t)\) is a slowly varying component and \(q(t)\) is a small-amplitude rapidly-varying component with a mean value of zero. In other words, \(p(t)\) may be assumed constant over a few oscillations of \(q(t)\) (see Figure 2).

Figure 2

Figure 2 - A typical solution for the equation of motion of the charged particle: \(p(t)\) gives the overall motion, and \(q(t)\) represents small oscillations around this trajectory. The ellipse on the right is a magnification of a part of this trajectory.

A-3 a) Using the approximations stated above, find the equation of motion for \(q(t)\) in terms of \(a\), \(\Omega\), and \(p\).

b) Find the solution of this equation by considering appropriate initial conditions corresponding to the required properties of this function.		 1.8 pt
A-4 a) Using the mean effect of the rapidly varying component and obtain an effective equation of motion for \(p(t)\).

b) Investigate the stability of the equilibrium point and find the condition for a stable equilibrium.		 1.5 pt

Assume that \(\lambda_0=8\times10^{-9}\ \mathrm{C/m}\) and \(R=10\ \mathrm{cm}\). We would like to use this device to trap a singly ionized atom 100 times heavier than a hydrogen atom.

A-5 Calculate \(k\). Assume \(a=0.04\) and estimate the smallest frequency required to stabilize the motion of this ion. Use the data given at the end of the question. 0.4 pt

B. Doppler Cooling

It may be necessary to cool a trapped atom or ion. Assume that a trapped atom of mass \(m\), has two energy levels with an energy difference of \(E_0=\hbar\omega_{\mathrm A}\). Electrons in the lower level may absorb a photon and jump to the higher level, but after a period \(\tau\) they will return to the lower level and emit a photon with a frequency predominantly within \([\omega_{\mathrm A}-\Gamma,\omega_{\mathrm A}+\Gamma]\).

B-1 Use the Heisenberg’s uncertainty principle to find \(\Gamma\). 0.5 pt

With a similar reasoning, when we shine a laser light on the trapped atom, if the angular frequency of the laser, \(\omega_{\mathrm L}\), falls in the interval \([\omega_{\mathrm A}-\Gamma,\omega_{\mathrm A}+\Gamma]\), the atom may absorb the photon. Assume that the frequency \(\omega_{\mathrm L}\) of the laser light is slightly lower than \(\omega_{\mathrm A}\). For a particular device, the rate of photon absorption by an atom in the reference frame of the atom is given in Figure 3. The absorbed photon is then re-emitted in a random direction. To make things simple, we consider the problem in one dimension, i.e. we assume that the atoms can only move in the \(x\)-direction and the laser light shines on them both from the left and from the right. In the atom’s reference frame, the light has a higher or lower frequency due to the motion of the atoms. Since the velocity \(v\) of the atoms is very small, we only include terms of order \(v/c\) and ignore all the higher-order terms. Moreover, we have \(m\gg\hbar\omega_{\mathrm A}/c^2\) so that the velocity of the atom nearly does not change after absorbing the photon. Also, the change in frequency due to the Doppler effect is so small compared to \(\omega_{\mathrm A}-\omega_{\mathrm L}\), that the function for \(s\) in the diagram of Figure 3 may be approximated by the following linear function:

\[ s(\omega)=s_{\mathrm L}+\alpha(\omega-\omega_{\mathrm L}) \]

where \(s\) is the number of absorbed photons per unit of time, \(s_{\mathrm L}\) is the value of \(s\) for \(\omega=\omega_{\mathrm L}\), and \(\alpha\) is the slope of the line tangent to the curve at \(\omega_{\mathrm L}\). The frequency of the re-emitted photon is almost equal to the frequency of the incident photon, but it is emitted with equal probability in the positive or negative \(x\)-direction. In fact, up to the order considered here the two frequencies are identical. Note that we are considering the whole process in the atom’s reference frame.

Figure 3

Figure 3 - The rate of photon absorption as a function of the frequency for a particular trap: the frequency corresponding to the energy difference between the two atomic levels is indicated by \(\omega_{\mathrm A}\) and the slighlty smaller frequency of the laser is indicated by \(\omega_{\mathrm L}\).

B-2 a) Assume that the trapped atom is moving with a velocity, \(v=v_x\) in the lab frame. In the frame of reference of the atom, calculate the collision rate of the photons, incident from each of the two directions, with the atoms (denoted by \(s_+\) and \(s_-\)) and the rate of absorption of momentum in each direction (denoted by \(\pi_+\) and \(\pi_-\)).

b) Determine the effective force on the atom as a function of \(v\), \(k_{\mathrm L}=\omega_{\mathrm L}/c\), \(\hbar\), and \(\alpha\), in the reference frame of the laboratory. Assume \(s_{\mathrm L}\ll\alpha\omega_{\mathrm L}\),		 1.7 pt

We would like to find the lowest temperature that can be achieved using this technique. Assume that the velocity of a particular atom has been reduced to zero exactly, and at this very moment it absorbs a photon (incident from any of the two directions), and re-emits the photon randomly in any of the two directions, with almost the same frequency. Assume that this process happens once every \(\tau\) units of time.

B-3 Considering the momentum of the atom after such a process for the two possible outcomes, calculate the average power absorbed by the atom. 1.0 pt
B-4 Consider the force calculated in Task B-2 and calculate the output power. Then, calculate the average value of \(v^2\) at equilibrium. Using your knowledge of the kinetic theory of gases estimate the temperature of the atoms. 0.8 pt
B-5 Estimate this temperature, for an atom 100 times heavier than a hydrogen atom. Assume that \(\omega_{\mathrm L}=2\times10^{16}\ \mathrm{rad/s}\), \(\tau=5\times10^{-9}\ \mathrm s\), and \(\alpha=4\). 0.4 pt
mass of hydrogen atom: \(m_{\mathrm H}=1.674\times10^{-27}\ \mathrm{kg}\)
charge of an electron: \(e=1.602\times10^{-19}\ \mathrm C\)
permittivity of free space: \(\varepsilon_0=8.854\times10^{-12}\ \mathrm{F/m}\)
Boltzmann constant: \(k_{\mathrm B}=1.381\times10^{-23}\ \mathrm{J/K}\)
Planck constant: \(\hbar=1.055\times10^{-34}\ \mathrm{J\cdot s}\)

Official English text of the problem.

Answers

  • A-1(a): \[ \boxed{\mathbf E\simeq \frac{\lambda}{4\varepsilon_0R^2} \left(-x\,\hat{\mathbf x}-y\,\hat{\mathbf y}+2z\,\hat{\mathbf z}\right).} \]

  • A-1(b): All directions in the ring’s plane are stable, with \[ \boxed{\omega_x=\omega_y=\sqrt{\frac{Q\lambda}{4\varepsilon_0mR^2}}.} \] The \(z\)-direction is unstable.

  • A-2: \[ \boxed{a=\frac{Qu}{2\varepsilon_0mR^2\Omega^2}, \qquad k=\sqrt{\frac{Q\lambda_0}{2\varepsilon_0mR^2}}.} \]

  • A-3(a): \(\boxed{\ddot q\simeq a\Omega^2p\cos\Omega t}\).

  • A-3(b): \(\boxed{q(t)\simeq-ap(t)\cos\Omega t}\). For \(p\) treated as constant during a fast cycle, the appropriate initial conditions are \(q(0)=-ap\), \(\dot q(0)=0\).

  • A-4(a): \[ \boxed{\ddot p+\left(\frac{a^2\Omega^2}{2}-k^2\right)p=0.} \]

  • A-4(b): Stable trapping requires, within the stated approximation, \[ \boxed{a^2\Omega^2>2k^2.} \] The slow axial angular frequency is \(\sqrt{a^2\Omega^2/2-k^2}\).

  • A-5: \[ \boxed{k\simeq2.1\times10^5\ \mathrm{s^{-1}}.} \] The estimated threshold is \[ \boxed{\Omega_{\mathrm{th}}\simeq7.4\times10^6\ \mathrm{rad/s}, \qquad f_{\mathrm{th}}\simeq1.2\ \mathrm{MHz}.} \] Choose a frequency above this threshold.

  • B-1: The linewidth scale is \(\Gamma\sim1/\tau\); using the conventional lifetime half-width, \[ \boxed{\Gamma\simeq\frac{1}{2\tau}.} \]

  • B-2(a): Let \(+\) denote photons propagating along \(+x\), and let \(\pi_\pm\) be signed \(x\)-momentum transfer rates. Then \[ \boxed{s_\pm=s_{\mathrm L}\mp\alpha\omega_{\mathrm L}\frac vc,} \] \[ \boxed{\pi_\pm= \pm\hbar k_{\mathrm L} \left[s_{\mathrm L}\mp (s_{\mathrm L}+\alpha\omega_{\mathrm L})\frac vc\right].} \]

  • B-2(b): \[ \boxed{F_x=-2\alpha\hbar k_{\mathrm L}^{\,2}v.} \]

  • B-3: \[ \boxed{P_{\mathrm{in}}=\frac{\hbar^2k_{\mathrm L}^{\,2}}{m\tau}.} \]

  • B-4: The positive rate of energy removal is \[ \boxed{P_{\mathrm{out}}=2\alpha\hbar k_{\mathrm L}^{\,2}v^2.} \] At equilibrium, \[ \boxed{\langle v^2\rangle=\frac{\hbar}{2\alpha m\tau}, \qquad T=\frac{\hbar}{2\alpha k_{\mathrm B}\tau}.} \]

  • B-5: \[ \boxed{T\simeq1.9\times10^{-4}\ \mathrm K =0.19\ \mathrm{mK}.} \]

Solution

A-1(a)

The ring lies in the \(xy\)-plane and carries total charge \(2\pi R\lambda\). By Coulomb’s law, its electric field on the symmetry axis is

\[ E_z(0,0,z) =\frac{1}{4\pi\varepsilon_0} \frac{2\pi R\lambda\,z}{(R^2+z^2)^{3/2}}. \]

For \(|z|\ll R\),

\[ E_z(0,0,z)\simeq\frac{\lambda}{2\varepsilon_0R^2}z. \]

Reflection and rotational symmetry require the general linear field near the center to have the form

\[ E_x=Cx,\qquad E_y=Cy,\qquad E_z=Dz. \]

The axial result gives \(D=\lambda/(2\varepsilon_0R^2)\). There is no charge in the neighborhood of the center, so Gauss’s law gives

\[ \nabla\cdot\mathbf E=2C+D=0. \]

Consequently,

\[ \boxed{\mathbf E\simeq \frac{\lambda}{4\varepsilon_0R^2} \left(-x\,\hat{\mathbf x}-y\,\hat{\mathbf y}+2z\,\hat{\mathbf z}\right).} \]

A-1(b)

Newton’s second law in the plane gives

\[ m\ddot x=-\frac{Q\lambda}{4\varepsilon_0R^2}x, \qquad m\ddot y=-\frac{Q\lambda}{4\varepsilon_0R^2}y. \]

These are harmonic-oscillator equations. Thus every direction in the plane has the same angular frequency,

\[ \boxed{\omega_x=\omega_y= \sqrt{\frac{Q\lambda}{4\varepsilon_0mR^2}}.} \]

Along the axis,

\[ m\ddot z=\frac{Q\lambda}{2\varepsilon_0R^2}z. \]

For \(Q,\lambda>0\), this force points away from the center, so the axial equilibrium is unstable.

A-2

Substituting the time-dependent charge density into the axial equation gives

\[ \ddot z= \frac{Q}{2\varepsilon_0mR^2} \left(\lambda_0+u\cos\Omega t\right)z. \]

Comparison with Equation (1) yields

\[ \boxed{k^2=\frac{Q\lambda_0}{2\varepsilon_0mR^2}, \qquad a=\frac{Qu}{2\varepsilon_0mR^2\Omega^2}.} \]

Here \(a\) is dimensionless and \(k\) has units of inverse time.

A-3(a)

Insert \(z=p+q\) into the equation of motion:

\[ \ddot p+\ddot q =k^2p+k^2q+a\Omega^2p\cos\Omega t +a\Omega^2q\cos\Omega t. \]

Over a fast oscillation, \(p\) is approximately constant. Since \(q\) has small amplitude, \(a\ll1\), and \(a\Omega^2\gg k^2\), the leading rapidly varying equation is

\[ \boxed{\ddot q\simeq a\Omega^2p\cos\Omega t.} \]

A-3(b)

Holding \(p\) constant during the integration gives

\[ q(t)=-ap\cos\Omega t+C_1t+C_2. \]

A rapidly oscillating component cannot contain a secular drift, so \(C_1=0\). Its mean must vanish, which requires \(C_2=0\). Hence

\[ \boxed{q(t)\simeq-ap(t)\cos\Omega t.} \]

For frozen \(p\), the corresponding initial conditions are

\[ q(0)=-ap,\qquad \dot q(0)=0. \]

The amplitude of \(q\) is \(a\) times that of the local slow displacement, consistent with \(a\ll1\). Corrections involving the slow variation of \(p\) are neglected at this order.

A-4(a)

Average the full equation over a drive period, treating \(p\) as constant during that period. Since

\[ \langle q\rangle=0,\qquad \langle\cos\Omega t\rangle=0,\qquad \langle\cos^2\Omega t\rangle=\frac12, \]

we obtain

\[ \ddot p=k^2p+a\Omega^2\langle q\cos\Omega t\rangle. \]

Using \(q=-ap\cos\Omega t\),

\[ \langle q\cos\Omega t\rangle=-\frac{ap}{2}. \]

Therefore,

\[ \boxed{\ddot p= \left(k^2-\frac{a^2\Omega^2}{2}\right)p.} \]

Although \(k^2\) is small compared with \(a\Omega^2\), it must be retained here: the alternating field contributes to the mean motion only at order \(a^2\Omega^2\).

A-4(b)

Define

\[ \omega_{\mathrm s}^2=\frac{a^2\Omega^2}{2}-k^2. \]

If \(\omega_{\mathrm s}^2>0\), the slow displacement oscillates:

\[ p(t)=A\cos\omega_{\mathrm s}t+B\sin\omega_{\mathrm s}t, \]

and the complete approximate motion,

\[ z(t)\simeq p(t)\left(1-a\cos\Omega t\right), \]

remains bounded. Thus the axial stability condition is

\[ \boxed{a^2\Omega^2>2k^2.} \]

For \(a>0\), this is

\[ \boxed{\Omega>\frac{\sqrt2\,k}{a}.} \]

Below this threshold, the slow equation has an exponentially growing solution. At equality, \(\ddot p=0\), so a nonzero slow velocity produces unbounded drift; equality does not provide confinement.

For completeness, the transverse equation is

\[ \ddot x= -\left(\frac{k^2}{2} +\frac{a\Omega^2}{2}\cos\Omega t\right)x. \]

The same averaging procedure gives, for its slow component \(X\),

\[ \ddot X+ \left(\frac{k^2}{2}+\frac{a^2\Omega^2}{8}\right)X=0, \]

and identically for \(y\). These directions are stable within the approximation, so the axial condition supplies the requirement for trapping in all three directions.

A-5

The ion has

\[ Q=e,\qquad m=100m_{\mathrm H}=1.674\times10^{-25}\ \mathrm{kg}, \qquad R=0.10\ \mathrm m. \]

Thus

\[ k= \sqrt{ \frac{(1.602\times10^{-19})(8\times10^{-9})} {2(8.854\times10^{-12})(1.674\times10^{-25})(0.10)^2} } =2.079\times10^5\ \mathrm{s^{-1}}. \]

Therefore,

\[ \boxed{k\simeq2.1\times10^5\ \mathrm{s^{-1}}.} \]

For the specified \(a=0.04\), the estimated threshold angular frequency is

\[ \Omega_{\mathrm{th}} =\frac{\sqrt2\,k}{a} =7.351\times10^6\ \mathrm{rad/s}. \]

The corresponding ordinary frequency is

\[ f_{\mathrm{th}} =\frac{\Omega_{\mathrm{th}}}{2\pi} =1.170\times10^6\ \mathrm{Hz}. \]

Hence,

\[ \boxed{\Omega_{\mathrm{th}}\simeq7.4\times10^6\ \mathrm{rad/s}, \qquad f_{\mathrm{th}}\simeq1.2\ \mathrm{MHz}.} \]

A frequency above this estimated threshold is required. At the threshold,

\[ \frac{\Omega_{\mathrm{th}}}{k}\simeq35, \qquad \frac{a\Omega_{\mathrm{th}}^2}{k^2}=50, \]

consistent with the assumed separation of scales.

B-1

The excited state has a characteristic lifetime \(\tau\). The energy–time uncertainty estimate is

\[ \Delta E\,\tau\gtrsim\frac{\hbar}{2}. \]

Identifying the characteristic energy half-width as \(\Delta E\simeq\hbar\Gamma\) gives

\[ \Gamma\tau\gtrsim\frac12. \]

For a width set by the finite lifetime, the usual estimate is therefore

\[ \boxed{\Gamma\simeq\frac{1}{2\tau}.} \]

The robust conclusion from the uncertainty principle is the scale \(\Gamma\sim1/\tau\); the numerical factor depends on how the linewidth is defined.

B-2(a)

Let \(+\) label photons propagating along \(+x\), and \(-\) label photons propagating along \(-x\). For an atom moving at velocity \(v\), the first-order Doppler shifts give

\[ \omega'_+=\omega_{\mathrm L}\left(1-\frac vc\right), \qquad \omega'_-=\omega_{\mathrm L}\left(1+\frac vc\right). \]

A photon traveling in the same direction as the atom is redshifted; one traveling against it is blueshifted.

The supplied function \(s(\omega)\) already specifies the absorption rate in the atom’s frame. Evaluating it at these frequencies gives

\[ s_+=s_{\mathrm L}+\alpha(\omega'_+-\omega_{\mathrm L}) =s_{\mathrm L}-\alpha\omega_{\mathrm L}\frac vc, \]

\[ s_-=s_{\mathrm L}+\alpha(\omega'_--\omega_{\mathrm L}) =s_{\mathrm L}+\alpha\omega_{\mathrm L}\frac vc. \]

Thus,

\[ \boxed{s_\pm=s_{\mathrm L}\mp\alpha\omega_{\mathrm L}\frac vc.} \]

Each absorbed photon has signed momentum \(\pm\hbar\omega'_\pm/c\). Defining \(\pi_\pm\) as signed \(x\)-momentum transfer rates,

\[ \pi_+=\frac{\hbar\omega'_+}{c}s_+, \qquad \pi_-=-\frac{\hbar\omega'_-}{c}s_-. \]

Expanding and keeping only first-order terms in \(v/c\),

\[ \boxed{\pi_+=\hbar k_{\mathrm L} \left[s_{\mathrm L}-(s_{\mathrm L}+\alpha\omega_{\mathrm L})\frac vc\right],} \]

\[ \boxed{\pi_-=-\hbar k_{\mathrm L} \left[s_{\mathrm L}+(s_{\mathrm L}+\alpha\omega_{\mathrm L})\frac vc\right].} \]

B-2(b)

In the atom’s frame, re-emission has zero mean momentum because the two emission directions are equally probable. Therefore the mean force is

\[ F'_x=\pi_++\pi_- =-\frac{2\hbar\omega_{\mathrm L}}{c^2} (s_{\mathrm L}+\alpha\omega_{\mathrm L})v. \]

Within the approximation used here, the absorbed and emitted energies are equal in the atom’s frame. The completed cycle therefore has zero net energy transfer in that frame when recoil is neglected in calculating the mean force. Transforming to the laboratory leaves the longitudinal force unchanged to the retained order.

Using \(s_{\mathrm L}\ll\alpha\omega_{\mathrm L}\), we obtain

\[ \boxed{F_x\simeq -\frac{2\alpha\hbar\omega_{\mathrm L}^2}{c^2}v =-2\alpha\hbar k_{\mathrm L}^{\,2}v.} \]

The laser frequency lies on the rising side of the absorption curve, so \(\alpha>0\). The force therefore opposes the velocity and removes kinetic energy.

B-3

Let the incident photon propagate along \(+x\) and have momentum magnitude

\[ p_\gamma=\hbar k_{\mathrm L}. \]

The atom initially has zero momentum. Momentum conservation gives the two equally probable outcomes:

Emission direction Final atomic momentum Final kinetic energy Probability
Along \(+x\), the incident direction \(0\) \(0\) \(1/2\)
Along \(-x\), opposite the incident direction \(2p_\gamma\) \(2p_\gamma^2/m\) \(1/2\)

Thus the mean kinetic-energy gain per absorption–emission cycle is

\[ \langle\Delta K\rangle =\frac12(0)+\frac12\frac{(2p_\gamma)^2}{2m} =\frac{p_\gamma^2}{m}. \]

An incident photon traveling along \(-x\) gives the same mean energy gain. Since there is one complete cycle every \(\tau\),

\[ \boxed{P_{\mathrm{in}} =\frac{\langle\Delta K\rangle}{\tau} =\frac{\hbar^2k_{\mathrm L}^{\,2}}{m\tau}.} \]

B-4

The rate at which the damping force changes the atom’s mechanical energy is

\[ F_xv=-2\alpha\hbar k_{\mathrm L}^{\,2}v^2. \]

Hence the positive output power, meaning the rate of energy removal, is

\[ \boxed{P_{\mathrm{out}} =2\alpha\hbar k_{\mathrm L}^{\,2}v^2.} \]

At thermal equilibrium, the mean cooling power balances recoil heating:

\[ 2\alpha\hbar k_{\mathrm L}^{\,2}\langle v^2\rangle =\frac{\hbar^2k_{\mathrm L}^{\,2}}{m\tau}. \]

Cancelling common factors gives

\[ \boxed{\langle v^2\rangle=\frac{\hbar}{2\alpha m\tau}.} \]

The problem restricts motion to one dimension, so equipartition gives

\[ \frac12m\langle v^2\rangle=\frac12k_{\mathrm B}T. \]

Consequently,

\[ \boxed{T=\frac{\hbar}{2\alpha k_{\mathrm B}\tau}.} \]

The atomic mass and laser wave number cancel from the equilibrium temperature in this model.

B-5

Using the given values,

\[ T= \frac{1.055\times10^{-34}} {2(4)(1.381\times10^{-23})(5\times10^{-9})} =1.9098\times10^{-4}\ \mathrm K. \]

Therefore,

\[ \boxed{T\simeq1.9\times10^{-4}\ \mathrm K =0.19\ \mathrm{mK}\simeq190\ \mu\mathrm K.} \]

For the specified mass \(m=1.674\times10^{-25}\ \mathrm{kg}\), the corresponding velocity variance is

\[ \langle v^2\rangle=1.58\times10^{-2}\ \mathrm{m^2/s^2}, \qquad v_{\mathrm{rms}}\simeq0.126\ \mathrm{m/s}. \]

The solution was checked against the official solution.