1... canoeing mystery
3 points
In sunny summer weather, we observe interesting wind behavior on the river during the day. It is cold in the morning at sunrise, and sometimes there is even morning fog. The fog then quickly dissipates, and the air temperature rises. A light wind then blows up the river. In the evening, the situation calms down, and the wind direction turns downstream as the sun lowers toward the horizon. What causes this phenomenon? Explain the ongoing processes in these two cases.
2... rotten apple
3 points
Jarda found an apple in his backpack after the FYKOS camp, which was no longer in good condition. He threw it into a low kitchen trash can $1.0\,\mathrm{m}$ away, and of course, he scored a hit. He threw the apple horizontally from a height $0.5\,\mathrm{m}$, and it landed on the spot where the wall and the base of the trash can meets, where it smashed. The basket with a mass $910\,\mathrm{g}$, was displaced by a distance of $5\,\mathrm{cm}$ after the apple hit. What is the coefficient of friction between the floor and the basket? The apple has a mass of $230\,\mathrm{g}$.
3... repellent resistive bipyramids
5 points
We have a model of a regular $N$-gonal bipyramid made of conducting wires. The connections in the plane of symmetry each have resistance $R_2$, whereas the connections going from one of the vertices to a point in the base have resistance $R_1$. Determine the resistance between
- main vertices (above and below the base plane),
- adjacent vertices in the base plane,
- opposite vertices in the base plane (the most distant ones) for even values of $N$.
4... light faster than light
7 points
There is a laser in the distance $L$ from a large screen. Initially, the laser shines on the screen so that the distance from a laser spot on the screen to the laser is $R > L$. Then at the time $t=0\,\mathrm{s}$, we begin to rotate the laser at a uniform angular speed $\omega$. Consequently, the distance of the spot on the screen from the laser decreases to $L$ and then increases back to $R$. What is the speed of this laser spot relative to the screen? Is it possible that the spot moves at a speed greater than the speed of light in a vacuum? Is there a limit, can it be infinite? How (qualitatively) does this speed depend on the spot's position on the screen? The whole apparatus is in a vacuum.
5... gadolinium sphere
9 points
What is the smallest amount of gadolinium $148$ needed to put together to cause local melting from the heat generated by its nuclear decay? Assume that only $\alpha$ decays take place and the material is at room temperature in the air.
P... Earth at full throttle
10 points
Estimate the upper limit of work that can be done on Earth over the long term. The planet must remain habitable and, if possible, with the same climate for future generations.
E... ripples
13 points
Build an apparatus that can measure the smallest possible ripples on the surface of the liquid. You can choose the container yourself – it can be a cup, a bottle, or something else. Thoroughly describe and take a picture of the whole apparatus. Determine the minimum amplitude you are able to measure.
Instructions for Experimental TasksS... exciting quanta
10 points
The lowest-lying excited singlet state of beta-carotene has an energy $1.8\,\mathrm{eV}$, which is higher than the ground state energy. However, the transition between this state and the ground state is prohibited, so the molecule does not absorb photons at this energy. On the other hand, the transition to the second lowest-lying singlet state with energy $2.4\,\mathrm{eV}$ is allowed and responsible for the bright orange color of the molecule. The lowest-lying triplet level is at $0.9\,\mathrm{eV}$ energy. Draw a Jablonski diagram and use it to explain why beta-carotene does not fluoresce even though it significantly absorbs visible light. $\left(3\,\mathrm{pts}\right)$
Bonus: Why is it so important for life on earth that oxygen is a triplet in the ground state? $\left(+1\,\mathrm{pts}\right)$
Try to calculate the approximate limit on the number of orbitals in the active space with the CASSCF method. Consider that you have as many electrons as orbitals in the active space (which corresponds to the fact that half of them in $\ce{HF}$ will be occupied) and that the most of today's supercomputers have at most $1\,\mathrm{TB}$ of RAM for computing, in which you need to fit a Hamiltonian. $\left(3\,\mathrm{pts}\right)$
For lithographic manufactured modern semiconductor chips, so-called excimer lasers are used to glow with the spectrum far into UV region. They are based on so-called excimers, which are molecules that are stable only in the excited state, while in the ground state, they decay. As a result, the molecule decays after the photon is emitted, ensuring that a larger fraction of the molecules are in the higher state than in the lower state. That is the necessary condition for the laser to work. Try using Psi4 for the helium dimer $\left(\ce{He_2^*}\right)$ to calculate and plot the dissociation curves of the ground and lowest-laying excited states. ($\ce{He_2^*}$ is not yet used for lasers, but for example $\ce{Ar_2^*}$ or $\ce{Kr_2^*}$ are.) At what wavelength would the laser work? Compare it with the experimental wavelength $66\,\mathrm{nm}$. $\left(4\,\mathrm{pts}\right)$
Note: In the problem statement on the website, you will find a prepared input file for one geometry. Do not be surprised that it has a total of three states set up. It needs to have those because we have two excited states close to each other. If we were to include only one of them in the calculations for some internuclear distances, this would lead to problems with convergence.