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IPhO 2023, Теория — Задача 1. Characterization of Soil Colloids

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Theory — Q1-1, English (Official): Characterization of Soil Colloids (10 points) — International Physics Olympiad 2023 Tokyo Japan · 10 т.

Условие

Colloidal science is useful to characterize soil particles because many of them can be regarded as colloidal particles of micrometer size. For example, Brownian motion (random motion of colloidal particles) can be used to measure particle sizes.

Part A. Motions of colloidal particles (1.6 points)

We analyze the one-dimensional Brownian motion of a colloidal particle with mass $M$ . The equation of motion for its velocity $v(t)$ reads:

M\dot{v} = -\gamma v(t) + F(t) + F_{\mathrm{ext}}(t), \qquad (1)

where $\gamma$ is the friction coefficient, $F(t)$ is a force due to random collisions with water molecules, and $F_{\mathrm{ext}}(t)$ is an external force. In Part A, we assume $F_{\mathrm{ext}}(t) = 0$ .

In the following, you may use $\tau$ in your answers.

Part B. Effective equation of motion (1.8 points)

Results so far imply that particle velocities $v(t)$ and $v(t')$ may be regarded as uncorrelated random quantities if $|t - t'| \gg \tau$ . On this basis, we introduce a theoretical model to approximately describe the one-dimensional Brownian motion, where the velocity changes randomly at each time interval $\delta\ (\gg \tau)$ , i.e.,

v(t) = v_n \quad (t_{n-1} < t \leq t_n), \qquad (2)

with $t_n = n\delta\ (n = 0, 1, 2, \cdots)$ and a random quantity $v_n$ . It satisfies

\langle v_n \rangle = 0, \quad \langle v_n v_m \rangle = \begin{cases} C & (n = m), \\ 0 & (n \neq m), \end{cases} \qquad (3)

with a parameter $C$ depending on $\delta$ . Here $\langle X \rangle$ indicates the expectation value of $X$ . That is, if you draw random numbers $X$ infinite times, the mean will be $\langle X \rangle$ .

Now we consider the particle displacement $\Delta x(t) = x(t) - x(0)$ for $t = N\delta$ with an integer $N$ .

Part C. Electrophoresis (2.7 points)

Here we discuss electrophoresis, i.e., transport of charged particles by an electric field. Suspension of colloidal particles with mass $M$ and charge $Q\ (> 0)$ is put in a narrow channel with a cross-section $A$ (Fig.1(a)). We ignore the interaction between particles, effects of the wall, the fluid, the ions therein, and gravity.

Three panels: (a) a 3D rectangular channel along the x-axis containing spherical colloidal particles, with cross-section A marked at the right end; (b) a flat channel with an arrow E pointing in the +x direction and a concentration gradient shaded from dark (left) to light (right); (c) a channel with circles representing particles, arrows showing exchanges across a shaded vertical strip of width vδ around x₀, with an arrow vδ above the strip. — Fig.1: Setting for Part C.

By applying a uniform electric field $E$ in the $x$ -direction, particles are transported and their concentration $n(x)$ (particle number per unit volume) becomes non-uniform (Fig.1(b)). When $E$ is removed, this non-uniformity gradually disappears. This is due to Brownian motion of particles. If $n(x)$ is not uniform, the numbers of right-going and left-going particles may differ (Fig.1(c)). This generates a particle flux $J_D(x)$ , the mean number of particles flowing at $x$ along the $x$ -axis per unit cross-sectional area and unit time. This flux is known to satisfy

J_D(x) = -D\,\frac{dn}{dx}(x), \qquad (4)

where $D$ is called the diffusion coefficient.

Now let's assume, for simplicity, that half of the particles have velocity $+v$ and the other half have velocity $-v$ . Let $N_+(x_0)$ be the number of particles with velocity $+v$ that cross $x_0$ from left to right per unit cross-sectional area and unit time. For particles with velocity $+v$ to cross $x_0$ in the time interval $\delta$ , they should be in the shaded region of Fig.1(c). Since $\delta$ is small, we have $n(x) \simeq n(x_0) + (x - x_0)\frac{dn}{dx}(x_0)$ in this region.

We define $N_-(x_0)$ as the counterpart of $N_+(x_0)$ for the velocity $-v$ . With this, we have $J_D(x_0) = \langle N_+(x_0) - N_-(x_0) \rangle$ . According to Eq.(3), we have $\langle v^2 \rangle = C$ .

Now we discuss the effect of osmotic pressure $\Pi$ . It is given by $\Pi = \frac{n}{N_A}RT = nkT$ with the Avogadro constant $N_A$ , the gas constant $R$ , temperature $T$ , and the Boltzmann constant $k = \frac{R}{N_A}$ . Let us consider the non-uniform concentration formed under the electric field $E$ (Fig.1(b)). Since $n(x)$ depends on $x$ , so does $\Pi(x)$ . Then the forces due to $\Pi(x)$ and $\Pi(x+\Delta x)$ must be balanced with the total force from the field $E$ acting on the particles (Fig.2). Here we consider small $\Delta x$ , so that $n(x)$ can be regarded as constant over this range, while $n(x+\Delta x) - n(x) \simeq \Delta x\,\frac{dn}{dx}(x)$ .

A rectangle representing a segment of the channel between positions x and x+Δx (dashed lines). An arrow E points to the right above the box. Arrow labeled Π(x)A points right into the segment at x, arrow labeled Π(x+Δx)A points left into the segment at x+Δx, and a right-pointing arrow between them represents the electric force. — Fig.2: Force balance.

Let us discuss the balance of the flux now. Besides the flux $J_D(x)$ due to the Brownian motion, there is also a flux due to the electric field, $J_Q(x)$ . It is given by

J_Q(x) = n(x)u, \qquad (5)

where $u$ is the terminal velocity of particles driven by the field.

Part D. Mean square displacement (2.4 points)

Suppose we observed the Brownian motion of an isolated, spherical colloidal particle with radius $a = 5.0\ \mu\mathrm{m}$ in water. Figure 3 shows the histogram of displacements $\Delta x$ measured in the $x$ -direction at every interval $\Delta t = 60\ \mathrm{s}$ . The friction coefficient is given by $\gamma = 6\pi a\eta$ with water viscosity $\eta = 8.9 \times 10^{-4}\ \mathrm{Pa \cdot s}$ and the temperature was $T = 25\ {}^{\circ}\mathrm{C}$ .

Histogram of displacement Δx (μm) from −10 to 10 in 1 μm bins; vertical axis N_count from 0 to 70. The distribution is bell-shaped, peaked at Δx = 0 with about 68 counts. — Fig.3: Histogram of displacements.

$\Delta x\ (\mu\mathrm{m})$	$-10$	$-9$	$-8$	$-7$	$-6$	$-5$	$-4$
$N_{\mathrm{count}}$	0	0	0	2	4	11	15

$\Delta x\ (\mu\mathrm{m})$	$-3$	$-2$	$-1$	$0$	$1$	$2$	$3$
$N_{\mathrm{count}}$	35	44	58	68	59	50	26

$\Delta x\ (\mu\mathrm{m})$	$4$	$5$	$6$	$7$	$8$	$9$	$10$
$N_{\mathrm{count}}$	15	5	4	3	0	1	0

Now we extend the model in Part B to describe the motion of a particle with charge $Q$ under an electric field $E$ . The particle velocity $v(t)$ considered in Eq.(2) should be replaced by $v(t) = u + v_n\ (t_{n-1} < t \leq t_n)$ with $v_n$ satisfying Eq. (3) and $u$ being the terminal velocity considered in Eq.(5).

Next, we consider swimming microbes (Fig.4(a)), in one dimension for simplicity (Fig.4(b)). These are spherical particles with radius $a$ . They swim at velocity either $+u_0$ or $-u_0$ , the sign chosen randomly at every time interval $\delta_0$ without correlation. The observed motion is a combination of displacements due to swimming and those due to the Brownian motion of a spherical particle.

(a) Five green circular microbes with flagella connected by blue line segments showing successive swimming runs in different directions, ending in a blue arrow. (b) A one-dimensional x-axis with a microbe and horizontal double-headed arrows of various lengths representing alternating +u₀ and −u₀ swims. — Fig.4: (a) Motion of microbes. (b) Its one-dimensional version.

Log-log plot of ⟨Δx(t)²⟩ (μm²) versus t (sec) from 10⁻³ to 10² s. An orange curve rises from about 10⁻⁴ μm² to about 10³ μm², with three dashed guide lines of different slopes: shallow at small t, steep (slope 2) at intermediate t, and intermediate slope at large t. — Fig.5: Mean square displacement of the microbes.

Part E. Water purification (1.5 points)

Here we discuss the purification of water including colloid-like soil particles, by adding electrolytes to coagulate them. Particles interact through van der Waals force and electrostatic force, the latter including effects of both surface charges and the surrounding layer of oppositely charged ions (such ions and their layer are called counter-ions and the electric double layer, respectively; see Fig.6(a)). As a result, the interaction potential for particle distance $d$ (Fig.6(b)) is given by

(a) Two grey spheres each carrying plus signs on their surfaces, surrounded by circled minus signs (counter-ions) forming an electric double layer. (b) Two grey spheres separated by a distance d indicated with a double arrow. — Fig.6: (a) Surface charges of colloidal particles and counter-ions. (b) Definition of the distance d.

U(d) = -\frac{A}{d} + \frac{B\epsilon(kT)^2}{q^2}e^{-d/\lambda}, \qquad (6)

where $A$ and $B$ are positive constants, $\epsilon$ is the dielectric constant of water, and $\lambda$ is the thickness of the electric double layer. Assuming that charges of ions are $\pm q$ , we have

\lambda = \sqrt{\frac{\epsilon kT}{2N_A q^2 c}}, \qquad (7)

where $c$ is the molar concentration of ion.

A.1 Consider that a water molecule collides with the particle at $t = t_0$ , giving impulse $I_0$ , and $F(t) = 0$ afterward. If $v(t) = 0$ before the collision, $v(t) = v_0 e^{-(t-t_0)/\tau}$ for $t > t_0$ . Determine $v_0$ and $\tau$ , using $I_0$ and necessary parameters in Eq.(1). [0,8 т.]

A.2 Actually, water molecules collide with the particle one after another. Suppose the $i$ th collision gives the impulse $I_i$ at time $t_i$ and determine $v(t)$ on condition that $t > 0$ and $v(0) = 0$ . Also give the inequality specifying the range of $t_i$ that needs to be considered for a given $t$ . In the answer sheet, it is not necessary to specify this range in the expression for $v(t)$ . [0,8 т.]

B.1 Determine $\langle \Delta x(t) \rangle$ and $\langle \Delta x(t)^2 \rangle$ using $C$ , $\delta$ , and $t$ . [1 т.]

B.2 The quantity $\langle \Delta x(t)^2 \rangle$ is called the mean square displacement (MSD). It is a characteristic observable of the Brownian motion, which corresponds to the limiting case $\delta \to 0$ . From this, we can show $C \propto \delta^{\alpha}$ and $\langle \Delta x(t)^2 \rangle \propto t^{\beta}$ . Determine the values of $\alpha$ and $\beta$ . [0,8 т.]

C.1 Express $N_+(x_0)$ using necessary quantities from $v$ , $\delta$ , $n(x_0)$ , and $\frac{dn}{dx}(x_0)$ . [0,5 т.]

C.2 Determine $J_D(x_0)$ using necessary quantities from $C$ , $\delta$ , $n(x_0)$ , and $\frac{dn}{dx}(x_0)$ . Using this and Eq.(4), express $D$ in terms of $C$ and $\delta$ , and $\langle \Delta x(t)^2 \rangle$ in terms of $D$ and $t$ . [0,7 т.]

C.3 Express $\frac{dn}{dx}(x)$ using $n(x)$ , $T$ , $Q$ , $E$ , and $k$ . [0,5 т.]

C.4 To determine $u$ , we use Eq.(1) with $F_{\mathrm{ext}}(t) = QE$ . Since $v(t)$ is fluctuating, we consider $\langle v(t) \rangle$ . Assuming $\langle v(0) \rangle = 0$ and using $\langle F(t) \rangle = 0$ , evaluate $\langle v(t) \rangle$ and obtain $u = \lim_{t \to \infty} \langle v(t) \rangle$ . [0,5 т.]

C.5 The flux balance reads $J_D(x) + J_Q(x) = 0$ . Express the diffusion coefficient $D$ in terms of $k$ , $\gamma$ , and $T$ . [0,5 т.]

D.1 Estimate the value of $N_A$ without using the fact that it is the Avogadro constant, up to two significant digits from the data in Fig.3. The gas constant is $R = 8.31\ \mathrm{J/K \cdot mol}$ . Do not use the value of the Boltzmann constant $k$ given in General Instructions. As for the Avogadro constant, you might obtain a value different from that in General Instructions. [1 т.]

D.2 Express the MSD $\langle \Delta x(t)^2 \rangle$ in terms of $u$ , $D$ , and $t$ . Obtain approximate power laws for small $t$ and large $t$ , as well as the characteristic time $t_*$ where this change occurs. Draw a rough graph of MSD in a log-log plot, indicating the approximate location of $t_*.$ [0,8 т.]

D.3 Figure 5 displays the MSD $\langle \Delta x(t)^2 \rangle$ of those microbes, showing different power laws for small, large, and intermediate $t$ , as indicated by dashed lines. Obtain the power law for each time range and express it using necessary quantities from $D$ , $u_0$ , $\delta_0$ , and $t$ . [0,6 т.]

E.1 Addition of sodium chloride (NaCl) to the suspension causes colloidal particles to coagulate. Determine the lowest concentration $c$ of NaCl necessary for coagulation. It is sufficient to consider two particles without thermal fluctuations, i.e., $F(t) = 0$ in Eq.(1), and assume that the terminal velocity for the given potential force is reached instantaneously. [1,5 т.]

Решение

Part A. Analysis of motions of colloidal particles (1.6 points)

A.1 The relation between the impulse and the momentum change is given by $Mv_0 = I_0$ . Therefore,

v_0 = \frac{I_0}{M}. \qquad (S1.1)

For the situation considered here, the equation of motion reads

M\dot{v} = -\gamma v(t). \qquad (S1.2)

Substituting the form of the solution given in the question sheet, $v(t) = v_0 e^{-(t-t_0)/\tau}$ , we obtain

\tau = \frac{M}{\gamma}. \qquad (S1.3)

A.1 (0.8 pt) $v_0 = \dfrac{I_0}{M}$ , $\tau = \dfrac{M}{\gamma}$

A.2 Thanks to the linearity of Eq. (S1.2), we can use the superposition principle, which tells us that $v(t)$ is given by the sum of solutions for single collision events that occur before time $t$ . This immediately gives the solution as

v(t) = \sum_i \frac{I_i}{M} e^{-(t-t_i)/\tau}, \qquad (S1.4)

where the sum is taken in the range of $i$ that satisfies $0 < t_i < t$ .
It is also not difficult to figure out this superposition principle, by considering the effect of a single collision as well as the velocity change between two consecutive collisions. From A.1, it is straightforward to show that the velocity right after the $i$ th collision is given by

v(t_i) = v_0(t_i) + \frac{I_i}{M}, \qquad (S1.5)

where $v_0(t_i)$ is the velocity right before the collision. Also, since there is no collision during $t_i < t < t_{i+1}$ , we have

v(t) = \left(v_0(t_i) + \frac{I_i}{M}\right) e^{-(t-t_i)/\tau}. \qquad (S1.6)

In particular,

v_0(t_{i+1}) = \left(v_0(t_i) + \frac{I_i}{M}\right) e^{-(t_{i+1}-t_i)/\tau}. \qquad (S1.7)

Therefore, with $v_0(t_1) = 0$ , we obtain

v_0(t_i) = \sum_{j=1}^{i-1} \frac{I_j}{M} e^{-(t_i - t_j)/\tau} \qquad (S1.8)

and, for $t_i < t < t_{i+1}$ ,

v(t) = \sum_{j=1}^{i} \frac{I_j}{M} e^{-(t-t_j)/\tau}. \qquad (S1.9)

This is equivalent to Eq. (S1.4).

A.2 (0.8 pt) $v(t) = \sum_i \frac{I_i}{M} e^{-(t-t_i)/\tau}$ ; the inequality specifying the range of $t_i$ that needs to be considered: $0 < t_i < t$

Part B. Effective equation of motion (1.8 points)

B.1 From the definition of the model, we have

\Delta x(t) = \sum_{n=1}^{N} v_n \delta. \qquad (S1.10)

Taking the average and using $\langle v_n \rangle = 0$ , we obtain

\langle \Delta x(t) \rangle = 0. \qquad (S1.11)

For the mean square displacement, computing the square of Eq. (S1.10) and taking the average, we obtain

\langle \Delta x(t)^2 \rangle = \sum_{m=1}^{N} \sum_{n=1}^{N} \langle v_m v_n \rangle \delta^2. \qquad (S1.12)

Using $\langle v_m v_n \rangle = C$ for $n = m$ and $0$ otherwise, we find

\langle \Delta x(t)^2 \rangle = \sum_{n=1}^{N} C\delta^2 = NC\delta^2. \qquad (S1.13)

Since $N\delta = t$ , we obtain

\langle \Delta x(t)^2 \rangle = C\delta t. \qquad (S1.14)

B.1 (1.0 pt) $\langle \Delta x(t) \rangle = 0$ ; $\langle \Delta x(t)^2 \rangle = C\delta t$

B.2 As described in the question sheet, the mean square displacement $\langle \Delta x(t)^2 \rangle$ is a characteristic observable of the Brownian motion, which of course takes a finite value for a given $t$ . For the model considered here, we have Eq. (S1.14), but we need to consider the limit $\delta \to 0$ to describe the Brownian motion in this model. This requires that $C\delta$ remains finite, so that $C \propto \delta^{-1}$ . It also follows that $\langle \Delta x(t)^2 \rangle \propto t$ .

Note: The continuous time limit $\delta \to 0$ of the present model corresponds to what is called the overdamped Langevin equation. This reads, in the absence of external force as considered here,
$\gamma \frac{dx}{dt} = \xi(t) \qquad (S1.15)$
with a Gaussian noise $\xi(t)$ that satisfies
$\langle \xi(t) \rangle = 0, \qquad \langle \xi(t)\xi(t') \rangle = 2D\delta(t - t') \qquad (S1.16)$
with the diffusion coefficient $D$ . Here, $\delta(t)$ (not to confuse with $\delta$ in the problem) is called the delta function, which satisfies $\delta(t) = 0$ for $t \neq 0$ and $\delta(0) = \infty$ but $\int_a^b \delta(t)\,dt = 1$ for any $a < 0$ and $b > 0$ .

B.2 (0.8 pt) $\alpha = -1$ ; $\beta = 1$

Part C. Electrophoresis (2.7 points)

C.1 For particles with velocity $v$ ( $> 0$ ), only those in the range $x_0 - v\delta \leq x \leq x_0$ pass the position $x_0$ during a time interval $\delta$ . Therefore, the number of such particles per unit cross-sectional area and per unit time is given by

N_+(x_0) = \frac{1}{\delta} \int_{x_0 - v\delta}^{x_0} \frac{1}{2} n(x)\,dx \qquad (S1.17)

Using the Taylor expansion $n(x) \simeq n(x_0) + (x - x_0)\frac{dn}{dx}(x_0)$ and integrating, we obtain

N_+(x_0) = \frac{1}{2} n(x_0) v - \frac{1}{4} \frac{dn}{dx}(x_0) v^2 \delta. \qquad (S1.18)

C.1 (0.5 pt) $N_+(x_0) = \frac{1}{2} n(x_0) v - \frac{1}{4} \frac{dn}{dx}(x_0) v^2 \delta$

C.2 Let $N_-(x_0)$ be the counterpart of $N_+(x_0)$ for particles with velocity $-v$ , then

N_-(x_0) = \frac{1}{2} n(x_0) v + \frac{1}{4} \frac{dn}{dx}(x_0) v^2 \delta. \qquad (S1.19)

With this equation, Eq. (S1.18), and $J_D(x_0) = \langle N_+(x_0) - N_-(x_0) \rangle$ , we obtain

J_D(x_0) = -\frac{1}{2} \frac{dn}{dx}(x_0) \langle v^2 \rangle \delta = -\frac{1}{2} \frac{dn}{dx}(x_0) C\delta. \qquad (S1.20)

Comparing this with Eq. (4) in the question sheet for $x = x_0$ , $J_D(x_0) = -D\frac{dn}{dx}(x_0)$ , we obtain

D = \frac{1}{2} C\delta. \qquad (S1.21)

Plugging this into the result of B.1, we obtain

\langle \Delta x(t)^2 \rangle = 2Dt. \qquad (S1.22)

C.2 (0.7 pt) $J_D(x_0) = -\frac{1}{2} \frac{dn}{dx}(x_0) C\delta$ ; $D = \frac{1}{2} C\delta$ ; $\langle \Delta x(t)^2 \rangle = 2Dt$

C.3 The force balance sketched in Fig. 2 is expressed by the following equation:

\Pi(x)A + n(x)A\Delta x QE = \Pi(x + \Delta x)A. \qquad (S1.23)

Using the van 't Hoff equation for the osmotic pressure, $\Pi(x) = n(x)kT$ , and carrying out the Taylor expansion of $n(x + \Delta x)$ , we obtain

\frac{dn}{dx} = \frac{n(x)}{kT} QE. \qquad (S1.24)

C.3 (0.5 pt) $\frac{dn}{dx} = \frac{n(x)}{kT} QE$

C.4 The equation of motion for $\langle v(t) \rangle$ is

M\frac{d\langle v(t) \rangle}{dt} = -\gamma \langle v(t) \rangle + QE. \qquad (S1.25)

By solving this with the initial condition $\langle v(0) \rangle = 0$ , we obtain

\langle v(t) \rangle = \frac{QE}{\gamma} \left(1 - e^{-t/\tau}\right). \qquad (S1.26)

Therefore,

u = \lim_{t \to \infty} \langle v(t) \rangle = \frac{QE}{\gamma}. \qquad (S1.27)

Note: The student is expected to surmise that the solution to Eq. (S1.25) has a functional form analogous to that to Eq. (S1.2), whose solution is given in the question sheet.

C.4 (0.5 pt) $\langle v(t) \rangle = \frac{QE}{\gamma}(1 - e^{-t/\tau})$ ; $u = \frac{QE}{\gamma}$

C.5 From the result of C.3 and Eq. (4) in the question sheet, we have

J_D(x) = -\frac{DQE}{kT} n(x). \qquad (S1.28)

From the result of C.4 and Eq. (5) in the question sheet, we have

J_Q(x) = \frac{QE}{\gamma} n(x). \qquad (S1.29)

Plugging these into the flux balance condition, $J_D(x) + J_Q(x) = 0$ , we obtain

D = \frac{kT}{\gamma}. \qquad (S1.30)

C.5 (0.5 pt) $D = \frac{kT}{\gamma}$

Part D. Mean square displacement (2.4 points)

D.1 Combining the results of C.2 and C.5, $k = R/N_A$ , $\gamma = 6\pi a\eta$ , we obtain the following equation that links the mean square displacement to $N_A$ :

\langle \Delta x^2 \rangle = \frac{RT\Delta t}{3\pi a\eta N_A}. \qquad (S1.31)

From the data given in the question sheet, the mean square displacement is estimated at $\langle \Delta x^2 \rangle = 6.34\ \mu\mathrm{m}^2$ . Plugging this and the values of the parameters given in the question sheet, we obtain

N_A = 5.6 \times 10^{23}\ \mathrm{mol}^{-1}. \qquad (S1.32)

Note: In 1908, Jean Baptiste Perrin (1870-1942) carried out such an observation and obtained an estimate of $N_A$ , which turned out to be consistent with the values known at that time by other approaches. This convinced the community of the fact that molecules and hence atoms do exist as constituents of matter. Perrin was awarded the Nobel Prize in Physics in 1926 for his work on the discontinuous structure of matter, and especially for his discovery of sedimentation equilibrium''. For more details, see, e.g., S. G. Brush, A History of Random Processes: I. Brownian Movement from Brown to Perrin'', Archive for History of Exact Sciences, volume 5, pages 1–36 (1968).

Note: On May 20, 2019, the definition of physical constants including the Avogadro constant $N_A$ was changed. As a result, $N_A$ is now defined by a fixed value, not to be determined through measurements.

D.1 (1.0 pt) $N_A = 5.6 \times 10^{23}\ \mathrm{mol}^{-1}$

D.2 Using $\Delta x(t) = \sum_{n=1}^{N} (u + v_n)\delta$ and Eq. (3) in the question sheet, we obtain

\langle \Delta x^2 \rangle = (ut)^2 + 2Dt \qquad (S1.33)

for general $t$ . This can be rewritten as

\langle \Delta x^2 \rangle = u^2 t\left(t + \frac{2D}{u^2}\right) = u^2 t(t + t_*), \qquad (S1.34)

with $t_* = 2D/u^2$ . Therefore,

\langle \Delta x^2 \rangle \propto \begin{cases} t & \text{for } t \ll t_* \\ t^2 & \text{for } t \gg t_* \end{cases} \qquad (S1.35)

D.2 (0.8 pt) $\langle \Delta x^2 \rangle = (ut)^2 + 2Dt$ for general $t$ ; $\langle \Delta x^2 \rangle \propto t$ for small $t$ , $\propto t^2$ for large $t$ ; $t_* = \frac{2D}{u^2}$

An example of the graph to answer:

Log-log plot: log⟨Δx(t)²⟩ versus log t — a curve that first rises linearly (slope 1), then bends upward and rises with slope 2; the dotted line at t_* (log t_*) marks the transition. — An example of the graph to answer:

D.1 Since the microbe does not change the swimming direction for $t \ll \delta_0$ , we can use the result of D.2 just by replacing $u$ by $u_0$ . By contrast, for $t \gg \delta_0$ , the motion of the microbe can be described by the model considered in PART B, though its parameter $\delta$ is not an artificial parameter anymore but is now a quantity that characterizes the microbe's motion, $\delta_0$ . The parameter $C$ is given by $C = u_0^2$ . Plugging this into Eq. (S1.14) and collecting all these results, we obtain

\langle \Delta x^2 \rangle = \begin{cases} 2Dt & \text{for } t \ll 2D/u_0^2 \\ u_0^2 t^2 & \text{for } 2D/u_0^2 \ll t \ll \delta_0 \\ u_0^2 \delta_0 t & \text{for } \delta_0 \ll t \end{cases} \qquad (S1.36)

Note: More precisely, one can show $\langle \Delta x^2 \rangle = (u_0^2\delta_0 + 2D)t$ for $t \gg \delta_0$ . However, in order for the intermediate regime to exist, we have $2D/u_0^2 \ll \delta_0$ , from which it follows that $u_0^2\delta_0 \gg 2D$ and the expression in Eq. (S1.36) is a good approximation.

Note: The motion of the microbe described here is called the run-and-tumble motion, except that it is usually assumed that the microbe changes the swimming direction ("tumbling") at random time intervals. Some bacteria including Escherichia coli is known to swim in this manner.

D.3 (0.6 pt) $\langle \Delta x^2 \rangle = 2Dt$ for small $t$ ; $u_0^2 t^2$ for intermediate $t$ ; $u_0^2 \delta_0 t$ for large $t$

Part E. Water purification (1.5 points)

E.1 The interaction energy $U(d)$ has a barrier if $c$ is small enough, but the barrier disappears if $c$ exceeds a threshold. This threshold is the critical concentration to derive in this question. The condition for the barrier to disappear is given by

\min U'(d) = 0. \qquad (S1.37)

This can be expressed by the following two equations:

U'(d) = \frac{A}{d^2} - \frac{B\epsilon (kT)^2}{q^2\lambda} e^{-d/\lambda} = 0, \qquad (S1.38)

U''(d) = -\frac{2A}{d^3} + \frac{B\epsilon (kT)^2}{q^2\lambda^2} e^{-d/\lambda} = 0. \qquad (S1.39)

Solving these, we obtain

d = 2\lambda = \sqrt{\frac{Aq^2\lambda}{B\epsilon (kT)^2}} \qquad (S1.40)

and therefore

\lambda = \frac{e^2 A q^2}{4B\epsilon (kT)^2}. \qquad (S1.41)

Plugging this into $c = \frac{\epsilon kT}{2N_A q^2}\lambda^{-2}$ , we obtain

c = \frac{8B^2\epsilon^3 (kT)^5}{e^4 N_A A^2 q^6}. \qquad (S1.42)

Note: In the literature, it is also common to consider that the critical concentration is reached when the energy barrier becomes as low as the energy for $d \to \infty$ , i.e., $\max U(d) = 0$ , although this does not meet the requirements given in the question sheet. If this condition is used instead, we find $c = \frac{B^2\epsilon^3 (kT)^5}{2e^2 N_A A^2 q^6}$ . This differs from Eq. (S1.42) only by a factor $e^2/8 \approx 0.92$ .

E.1 (1.5 pt) $c = \dfrac{8B^2\epsilon^3 (kT)^5}{e^4 N_A A^2 q^6}$

Оригинал в Архива: IPhO_2023_Q1.pdf · официални решения: IPhO_2023_S1.pdf