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Условие Отговори Решение

IPhO 2026, Теория — Задача 1.2. Electron-positron pair

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QUESTION T1 [10 pts]. Three problems · 3,5 т.

QUESTION T1 [10 pts]. Three problems

This question consists of three independent problems.

Условие

A positron e⁺ (top) and an electron e⁻ (bottom) separated vertically by a distance 100a₀; the positron's velocity arrow points to the right and the electron's to the left. — Fig. 1b.

At a certain instant of time, a positron $e^+$ is located at a distance of $100a_0$ from an electron $e^-$ . The particles are moving in such a way that their velocities are antiparallel at that instant, perpendicular to their separation, as shown in figure 1b. It is known that each particle has an angular momentum of magnitude $\mu\hbar$ with respect to the system's center of mass, where $\mu$ is a dimensionless numerical factor. Assume that the only interaction between $e^+$ and $e^-$ is electrostatic (both particles have the same mass $m$ and equal magnitude of the charge but of opposite sign). Furthermore, assume that this system remains isolated and that its dynamics are classical and non-relativistic.

The Bohr radius $a_0$ is given by $a_0 = 4\pi\epsilon_0\hbar^2/(me^2)$ .

T1-B1 If $\mu = 4$ , the system is bound, meaning that the particles move in a closed orbit around the system's center of mass. Find the maximum separation distance between $e^+$ and $e^-$ in terms of $a_0$ . [1 т.]

T1-B2 If $\mu = \frac{15}{2}$ , the system is unbound, meaning that the particles are no longer in closed orbit around the system's center of mass. Let $\vec{u}_\infty$ be the velocity of $e^+$ relative to $e^-$ as the separation distance between them tends to infinity. Calculate the angle between $\vec{u}_\infty$ and the initial line of motion of $e^+$ . Give your answer in degrees. [2,5 т.]

Hint 1: The eccentricity $\varepsilon$ of a conic trajectory for the present case is given by

\varepsilon = \sqrt{1 + \frac{4L^2E}{k^2e^4m}}

where E and L are the total energy and the magnitude of the total angular momentum, respectively. k is Coulombs constant $k = 1/(4\pi\epsilon_0)$ .

Hint 2: The equation of a conic in polar coordinates is

r = \frac{a}{1 - \varepsilon\cos(\theta)}

Отговори

Решение

T1-B [2.5 pts]. Electron-positron pair
From the definition of angular momentum,

mv_0(50a_0) = \mu\hbar

v_0 = \frac{\mu\hbar}{50ma_0}

The total energy of the system is

2\left(\frac{1}{2}mv_0^2\right) - \frac{ke^2}{100a_0} = \frac{ke^2}{a_0}\left(\frac{\mu^2}{50^2} - \frac{1}{100}\right)

T1-B1 [1.0 pts].

For $\mu = 4$ ,

E = -\frac{9ke^2}{2500a_0}

SOLUTION 1:
At the critical separation distances, $\vec{v} \perp \vec{r}$ and $L = mvr$ . The energy at $r_{max}$ and $r_{min}$ can be written as

E = 2\frac{L^2}{2mr^2} - \frac{ke^2}{r}

Using $L = 2\mu\hbar = 8\hbar$ and the conservation of energy,

-\frac{9ke^2}{2500a_0} = \frac{64\hbar^2}{mr^2} - \frac{ke^2}{r}

\frac{ke^2}{a_0}\left(\frac{9}{2500}r^2 - a_0r + 64a_0^2\right) = 0

r = \frac{a_0 \pm \sqrt{a_0^2 - \frac{576}{625}a_0^2}}{9/1250}

r = \frac{1250}{9}a_0\left(1 \pm \frac{7}{25}\right)

The maximum separation corresponds to the solution with the plus sign,

\boxed{r_{max} = \frac{1600}{9}a_0}

SOLUTION 2:
Since the energy is negative and the electrostatic force is central and proportional to $r^{-2}$ , the trajectory of the particles will be elliptical with respect to the CM. The energy can be written in terms of the semi-major axis $x$ as

E = -\frac{ke^2}{2x}

x = \frac{1250}{9}a_0

Since $\vec{v}_0$ is perpendicular to the vector that connects the two particles, the initial position must correspond to either $r_{max}$ or $r_{min}$ . The other critical distance, $r_2$ is given by the definition of $x$ :

2x = r_0 + r_2

\boxed{r_2 = \left(\frac{2500}{9} - 100\right)a_0 = \frac{1600}{9}a_0}

Which is greater than $r_0$ and corresponds to $r_{max}$

SOLUTION 3:
The general equation for the trajectory in polar coordinates is

r(\theta) = \frac{\ell}{1 + \varepsilon\cos(\theta - \theta_0)}

with $\ell = \frac{2L^2}{mke^2}$ the semi-latus rectum and $\varepsilon$ the eccentricity, as given by the hint in question T1-B2. The maximum separation distance is reached when $\cos(\theta - \theta_0) = -1$ :

r_{max} = \frac{\ell}{1 - \varepsilon}

Calculating the values of $\ell$ and $\varepsilon$ ,

\ell = \frac{2(8\hbar)^2}{mke^2} = 128a_0

\varepsilon = \sqrt{1 - \frac{36ke^2(8\hbar)^2}{2500a_0m(ke^2)^2}}

\varepsilon = \sqrt{1 - \frac{2304}{2500}} = \frac{7}{25}

Replacing these values in the expression for $r_{max}$ ,

\boxed{r_{max} = \frac{128}{1 - (7/25)}a_0 = \frac{1600}{9}a_0}

T1-B2 [2.5 pts].
For $\mu = 15/2$ ,

E = +\frac{ke^2}{80a_0}

The distance between particles follows

r(\theta) = \frac{\ell}{1 + \varepsilon\cos(\theta - \theta_0)}

Since the initial position corresponds to the closest approach, it is convenient to measure the angle $\theta$ with respect to the y axis so that $\cos\theta_0 = 1$ , $\theta_0 = 0$ . For $r \to \infty$ ,

1 + \varepsilon\cos\theta' \to 0

\theta' = \arccos(-1/\varepsilon)

Calculating the eccentricity,

\varepsilon = \sqrt{1 + \frac{4ke^2(15\hbar^2)}{80a_0m(ke^2)^2}}

\varepsilon = \sqrt{1 + \frac{45}{4}} = \frac{7}{2}

Replacing this value in the expression for $\theta$ ,

\theta' = \arccos(-2/7) = 106.60^{\circ}

Since this is the angle with respect to the $y$ -axis but the initial $e^+$ velocity is directed towards the positive $x$ -axis, heading downward

\boxed{\theta = 90^{\circ} - \theta' = -16.60^{\circ}}

Оригинал в Архива: T1.pdf · официални решения: theory_sol.pdf