IPhO 2026 · Theory problem 1

Three problems (A: Hydrostatic gate; B: Electron-positron pair; C: Photodissociation of ozone)

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QUESTION T1 [10 pts]. Three problems

This question consists of three independent problems.

T1-A [3.0 pts]. Hydrostatic gate

Figure 1

Fig. 1a.

Two water reservoirs are separated by a vertical wall MN, as shown in the figure 1a. The left reservoir can receive water from a source. It must be ensured that the maximum possible difference between the water levels is \(\Delta h = 1.41\ \mathrm{m}\). To achieve this, a slot of vertical size \(a\sqrt{2}/2\) is cut in MN and sealed with a solid cubic block of side length \(a\). The block has density \(3\rho_0\), with \(\rho_0\) the density of water. The block is fully submerged, fixed to the wall at \(O\) and can rotate without friction about the axis perpendicular to the plane of the figure that passes through point \(O\).

T1-A1 [3.0 pts]. Calculate the value of \(a\) that ensures the maximum permissible difference in water levels is not exceeded.

T1-B [3.5 pts]. Electron-positron pair

Figure 2

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 [1.0 pts]. 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 \(a_0\).

T1-B2 [2.5 pts]. 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.

Hint 1: The eccentricty \(\varepsilon\) 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

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

T1-C [3.5 pts]. Photodissociation of ozone

Figure 3

Fig. 1c.

A photon with angular frequency \(\omega\) strikes an ozone molecule \(\mathrm{O}_3\) at rest, dissociating it (breaking it apart) into an oxygen molecule \(\mathrm{O}_2\) and an oxygen atom \(\mathrm{O}\), as shown in figure 1c. The photon is absorbed in this process. Let \(U_i\) and \(U_f\) be the ground state energies of the ozone and oxygen molecules, respectively. Assume that this system remains isolated and that its dynamics is classical (potential energies do not contribute to mass and the motion of the Oxygen atoms are non-relativistic). Use the relation \(p = E/c\) for the linear momentum of the photon, where \(E\) is its energy.

T1-C1 [2.5 pts]. If the angle that the momentum of the outgoing \(\mathrm{O}_2\) with respect to the incident photon is \(\theta\), determine the minimum angular frequency \(\omega_{\min}\) required for this dissociation to occur at this angle. Write your answer in terms of \(\hbar\), \(c\), \(\theta\), \(\Delta U = U_f-U_i\), and the mass \(m\) of an oxygen atom.

T1-C2 [1.0 pts]. If \(\theta = \pi/6\), calculate the value of \(\hbar\omega_{\min}-\Delta U\). Give your answer in electronvolts (eV). Take \(\Delta U = 1.10\ \mathrm{eV}\) and \(m = 16.0\ \mathrm{amu}\).

Hint:

\[ (1+x)^r =1+rx+\frac{r(r-1)}{2!}x^2 +\frac{r(r-1)(r-2)}{3!}x^3+\cdots \]

Physical Constants

Reference values for selected physical constants.

Constant Notation Value Units
Speed of light in vacuum \(c\) \(2.99792458\times10^8\) \(\mathrm{m\cdot s^{-1}}\)
Planck constant \(h\) \(4.135667696\ldots\times10^{-15}\) \(\mathrm{eV\cdot s}\)
Reduced Planck constant \(\hbar=h/2\pi\) \(6.582119569\ldots\times10^{-16}\) \(\mathrm{eV\cdot s}\)
Avogadro number \(N_A\) \(6.02214076\times10^{23}\) \(\mathrm{mol^{-1}}\)
Boltzmann constant \(k_B\) \(8.617333262\times10^{-5}\) \(\mathrm{eV\cdot K^{-1}}\)
Ideal gas constant \(R\) \(8.31446261815324\) \(\mathrm{kg\cdot m^2\cdot s^{-2}\cdot mol^{-1}\cdot K^{-1}}\)
Elementary charge \(e\) \(1.602176634\times10^{-19}\) \(\mathrm{A\cdot s}\)
Universal gravitational constant \(G\) \(6.67430(15)\times10^{-11}\) \(\mathrm{m^3\cdot kg^{-1}\cdot s^{-2}}\)
Stefan Boltzmann constant \(\sigma\) \(5.670374419\times10^{-8}\) \(\mathrm{kg\cdot s^{-3}\cdot K^{-4}}\)
Vacuum permeability \(\mu_0\) \(1.25663706127(20)\times10^{-6}\) \(\mathrm{kg\cdot m\cdot A^{-2}\cdot s^{-2}}\)
Vacuum permittivity \(\epsilon_0\) \(8.8541878188(14)\times10^{-12}\) \(\mathrm{A^2\cdot s^4\cdot kg^{-1}\cdot m^{-3}}\)
Mass of the electron \(m_e\) \(5.1099895069(16)\times10^5\) \(\mathrm{eV}/c^2\)
Mass of the proton \(m_p\) \(9.3827208943(29)\times10^8\) \(\mathrm{eV}/c^2\)
Atomic mass units amu \(9.3149410372(29)\times10^8\) \(\mathrm{eV}/c^2\)

Official English text of the problem.

Answers

  • T1-A1: The largest permissible side length, giving opening at the specified level difference, is \[ \boxed{a=\frac{\Delta h}{2\sqrt{2}}=0.499\ \mathrm{m}.} \]

  • T1-B1: \[ \boxed{r_{\max}=\frac{1600}{9}a_0\approx178a_0.} \]

  • T1-B2: \[ \boxed{\alpha=\sin^{-1}\!\left(\frac{2}{7}\right)=16.6^\circ,} \] directed toward the electron’s initial side of the line of motion.

  • T1-C1: For \(0\leq\theta<\pi/2\), as in the figure, let \(F=1+2\sin^2\theta\). Then \[ \boxed{ \omega_{\min} =\frac{3mc^2}{\hbar F} \left(1-\sqrt{1-\frac{2F\Delta U}{3mc^2}}\right) =\frac{2\Delta U} {\hbar\left(1+\sqrt{1-\frac{2F\Delta U}{3mc^2}}\right)}. } \] For \(\pi/2\leq\theta\leq\pi\), replace \(F\) by \(3\); the result is an infimum approached as the outgoing \(\mathrm{O}_2\) momentum tends to zero.

  • T1-C2: \[ \boxed{\hbar\omega_{\min}-\Delta U \approx\frac{(\Delta U)^2}{4mc^2} =2.03\times10^{-11}\ \mathrm{eV}.} \]

Solution

T1-A1

We calculate the torque about \(O\) at the onset of opening. The hinge force has no moment about \(O\), and the contact force at the lower edge of the slot vanishes at this threshold.

Write the slot height as \[ H=\frac{a}{\sqrt{2}}. \] The marked distance \(a/2\) places \(O\) at the midpoint of the upper-left side of the square cross section. Consequently, the cube’s center is horizontally to the right of \(O\) by \[ d=\frac{a}{2\sqrt{2}}. \]

Decompose the water pressure into two contributions. First, imagine that both reservoirs have the right-hand water level. This hydrostatic pressure gives an upward buoyant force \[ B=\rho_0ga^3 \] through the center of the cube. Its weight is \[ W=3\rho_0ga^3. \] Their combined clockwise torque is therefore \[ \tau_{\mathrm{closing}} =(W-B)d =2\rho_0ga^3\frac{a}{2\sqrt{2}} =\frac{\rho_0ga^4}{\sqrt{2}}. \]

Raising the left water level by \(\Delta h\) adds the uniform pressure \[ \Delta p=\rho_0g\Delta h \] to the cube’s surface on the left of the wall.

The resultant force and torque of this additional pressure are equivalent to those on a vertical rectangle spanning the slot. To see this, close the protruding surface with that rectangle: uniform pressure over the resulting closed surface has zero net force and torque.

Thus the additional pressure produces a horizontal force \[ F_{\Delta p}=\Delta p\,aH \] acting halfway down the slot. Its counterclockwise, opening torque is \[ \tau_{\mathrm{opening}} =\Delta p\,aH\frac{H}{2} =\frac{\rho_0g\Delta h\,a^3}{4}. \]

At the opening threshold, \[ \frac{\rho_0g\Delta h\,a^3}{4} =\frac{\rho_0ga^4}{\sqrt{2}}, \] so \[ \Delta h=2\sqrt{2}\,a. \]

Setting this threshold equal to the permitted level difference gives \[ \boxed{ a=\frac{1.41}{2\sqrt{2}}\ \mathrm{m} =0.498510\ldots\ \mathrm{m} \approx0.499\ \mathrm{m}. } \] Smaller values of \(a\) would make the gate open at a smaller level difference.

T1-B1

Let \(r\) be the separation of the particles and define \[ \kappa=ke^2=\frac{\hbar^2}{ma_0}. \] We work in the center-of-mass frame. The reduced mass is \(m/2\), and the two angular momenta have the same direction, so their total magnitude is \[ L=2\mu\hbar. \]

Conservation of mechanical energy gives \[ E=\frac{m}{4}\dot r^{\,2}+\frac{L^2}{mr^2}-\frac{\kappa}{r}. \] Initially the relative velocity is perpendicular to the separation, hence \(\dot r=0\) at \(r_0=100a_0\). Therefore, \[ E =\frac{4\mu^2\hbar^2}{m(100a_0)^2} -\frac{\kappa}{100a_0} =\frac{\kappa}{a_0} \left(\frac{\mu^2}{2500}-\frac{1}{100}\right). \]

For \(\mu=4\), \[ E=-\frac{9}{2500}\frac{\kappa}{a_0}<0. \]

At either extreme of the separation, \(\dot r=0\). Writing \(x=r/a_0\), the turning-point equation is \[ -\frac{9}{2500}=\frac{64}{x^2}-\frac1x. \] Equivalently, \[ 9x^2-2500x+160000 =(x-100)(9x-1600)=0. \]

The two turning points are therefore \[ r_{\min}=100a_0, \qquad r_{\max}=\frac{1600}{9}a_0. \] Thus the given initial configuration is the point of minimum separation, and \[ \boxed{r_{\max}=\frac{1600}{9}a_0\approx178a_0.} \]

T1-B2

For \(\mu=15/2\), the same expressions give \[ L=15\hbar, \qquad E=\frac{\kappa}{80a_0}>0. \] Using the supplied eccentricity formula, \[ \varepsilon =\sqrt{1+\frac{4L^2E}{m\kappa^2}} =\sqrt{1+\frac{4(15\hbar)^2}{m\kappa^2} \frac{\kappa}{80a_0}}. \] Since \(ma_0\kappa=\hbar^2\), \[ \boxed{\varepsilon=\sqrt{1+\frac{45}{4}}=\frac72.} \]

Choose a polar angle \(\varphi\) measured from the periapsis direction, increasing in the direction of motion. With this choice, the conic equation is \[ r=\frac{p}{1+\varepsilon\cos\varphi}, \qquad p=\frac{L^2}{(m/2)\kappa}=450a_0. \] The periapsis separation is \[ r_{\mathrm p}=\frac{p}{1+\varepsilon} =\frac{450a_0}{1+7/2}=100a_0, \] so the initial position is indeed at periapsis.

On the outgoing branch, \(r\to\infty\) when \[ 1+\varepsilon\cos\varphi_\infty=0, \] giving \[ \cos\varphi_\infty=-\frac1\varepsilon=-\frac27. \]

The transverse relative speed is \(L/[(m/2)r]\), which tends to zero at infinity. Hence the final relative velocity points along the limiting radial direction. The initial velocity is perpendicular to the periapsis direction, so the requested angle is \[ \alpha=\varphi_\infty-\frac{\pi}{2}. \] It follows that \[ \sin\alpha=\frac1\varepsilon=\frac27, \] and therefore \[ \boxed{\alpha=\sin^{-1}\!\left(\frac27\right)=16.6^\circ.} \] In the figure, the outgoing relative velocity points below the positron’s initial horizontal direction.

T1-C1

Let the photon energy and momentum magnitude be \[ E_\gamma=\hbar\omega, \qquad P=\frac{E_\gamma}{c}. \] Take its direction as the \(x\)-axis. Let the outgoing oxygen molecule have momentum \(q\hat{\mathbf n}\), where \[ q\geq0, \qquad \hat{\mathbf n}\cdot\hat{\mathbf x}=\cos\theta. \]

Under the stated mass convention, the oxygen molecule has mass \(2m\), and the atom has mass \(m\). Momentum conservation gives \[ \mathbf p_{\mathrm O}=P\hat{\mathbf x}-q\hat{\mathbf n}. \] At the minimum photon energy, no additional internal excitation is needed. Energy conservation is then \[ E_\gamma-\Delta U =\frac{q^2}{4m} +\frac{P^2+q^2-2Pq\cos\theta}{2m}. \]

Completing the square, \[ E_\gamma-\Delta U = \frac{3}{4m} \left(q-\frac{2P\cos\theta}{3}\right)^2 +\frac{P^2}{6m}\left(3-2\cos^2\theta\right). \]

For the forward-emission geometry, \(0\leq\theta<\pi/2\), the kinetic energy is minimized at the physically allowed momentum \[ q_*=\frac{2P\cos\theta}{3}>0. \] Define \[ F=3-2\cos^2\theta=1+2\sin^2\theta. \] The minimum final kinetic energy for a given photon energy is therefore \[ K_{\min}=\frac{FP^2}{6m} =\frac{F E_\gamma^2}{6mc^2}. \]

At the dissociation threshold, \[ E_{\gamma,\min}-\Delta U =\frac{F E_{\gamma,\min}^2}{6mc^2}. \] Solving this quadratic and taking the lower root gives \[ \boxed{ \omega_{\min} =\frac{3mc^2}{\hbar(1+2\sin^2\theta)} \left[ 1-\sqrt{ 1-\frac{2(1+2\sin^2\theta)\Delta U}{3mc^2} } \right]. } \] The other root has energy of order \(mc^2\) and lies outside the assumed non-relativistic regime.

An equivalent form, useful when \(\Delta U\ll mc^2\), is \[ \boxed{ \omega_{\min} =\frac{2\Delta U} {\hbar\left[ 1+\sqrt{1-\frac{2(1+2\sin^2\theta)\Delta U}{3mc^2}} \right]}. } \] Its expansion is \[ \hbar\omega_{\min} =\Delta U +\frac{(1+2\sin^2\theta)(\Delta U)^2}{6mc^2} +O\!\left(\frac{(\Delta U)^3}{m^2c^4}\right). \]

For completeness, the statement does not explicitly restrict the angular range. If \(\pi/2\leq\theta\leq\pi\), the unconstrained minimizer \(q_*\) is nonpositive. Restricting to \(q\geq0\) instead gives \[ \inf K=\frac{P^2}{2m}. \] Thus the threshold formula uses \(F=3\) for these angles. It is only an infimum for a specified outgoing direction: at \(q=0\) that direction is undefined, while arbitrarily small positive \(q\) permits energies arbitrarily close to the threshold.

T1-C2

For \(\theta=\pi/6\), \[ F=1+2\sin^2\frac{\pi}{6}=\frac32. \] Consequently, \[ \hbar\omega_{\min} =2mc^2\left(1-\sqrt{1-\frac{\Delta U}{mc^2}}\right). \]

Using the supplied atomic mass unit, \[ mc^2 =16.0\left(9.3149410372\times10^8\right)\ \mathrm{eV} =1.49039\times10^{10}\ \mathrm{eV}. \] The expansion parameter is \[ x=\frac{\Delta U}{mc^2} \approx7.38\times10^{-11}. \] Using \[ \sqrt{1-x}=1-\frac{x}{2}-\frac{x^2}{8}+O(x^3), \] we obtain \[ \hbar\omega_{\min} =\Delta U+\frac{(\Delta U)^2}{4mc^2} +O\!\left(\frac{(\Delta U)^3}{m^2c^4}\right). \]

Thus \[ \hbar\omega_{\min}-\Delta U \approx \frac{(1.10\ \mathrm{eV})^2} {4(1.49039\times10^{10}\ \mathrm{eV})}, \] which gives \[ \boxed{\hbar\omega_{\min}-\Delta U =2.03\times10^{-11}\ \mathrm{eV}} \] to three significant figures.

The solution was checked against the official solution. Where this solution differs from the official solution, the official solution contains an error or a misprint.