Find the Rydberg constant's value and determine which hydrogen spectral lines belong to the visible spectrum. These lines are the only ones Rydberg could use to discover his formula, as neither UV nor IR spectra could yet be measured. What color are they, and which transitions in the Bohr model do they correspond to? (3pts)
Calculate de Broglie wavelength of your body. How does this value compare to the size of an atom or atomic nucleus? (3pts)
Assume you have a cuvette with $10 \mathrm{ml}$ of fluorescein water solution. Then, you point an argon laser at the cuvette. The laser is characterized by a wavelength of $488 \mathrm{nm}$ and a power of $10 \mathrm{W}$. At the same time, the fluorescein molecule fluoresces at a wavelength of $521 \mathrm{nm}$ with a quantum yield (proportion of absorbed photons that are emitted back) of $95 \mathrm{\%}$. If the initial temperature of the cuvette is $20 \mathrm{\C }$, how long will it take for its contents to start boiling? Assume that the cuvette is perfectly thermally insulated, the laser beam is fully absorbed in it, and the amount of fluorescein is negligible in terms of heat capacity. (4pts)
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Finally, a bonus for those who would be disappointed if they didn't integrate – the 1s orbital of the hydrogen atom has a spherically symmetric wave function with radial progression $\psi (r) = \frac {e^{-r/a_0}}{\sqrt {\pi }a_0^{3/2}}$, where $a_0=\frac {4\pi \epsilon _0\hbar ^2}{me^2}$ is the Bohr radius. Since the orbitals as functions of three spatial variables would be hard to plot, we prefer to show the region where the electron is most likely to occur. What is the radius of the sphere centered on the nucleus in which the electron will occur with a probability of $95 \mathrm{\%}$? (+2b)
An early Mikuláš's package.
Mikuláš gives presents again, this time at the right time of the year, almost.
English version of the serial will be released soon.
Mikuláš keeps on giving even after Christmas.
The binding energy of a fluorine molecule is approximately $37 \mathrm{kcal/mol}$. Assuming the range of binding interactions to be approximately $3 \mathrm{\AA }$ from the optimum distance, what (average) force do we have to exert to break the molecule? Calculate the „stiffness“ of the fluorine molecule if such an average force was applied in the middle of this range. What would be the vibrational frequency of this molecule? Compare this with the experimental value of $916{,}6 \mathrm{cm^{-1}}$. ($4 \mathrm{pts}$)
Using Psi4, calculate the dissociation curve $\mathrm {F_2}$ and fit a parabola around the minimum. What value will you get for the energy of the vibrational transitions this time? ($3 \mathrm{pts}$)
You are given two bottles of alcohol that you found suspicious, to say the least. After taking them to the lab, you obtain the following Raman spectra from them. Using the Psi4 program, calculate the frequencies at which the vibrational transitions of both the methanol and ethanol molecules occur. Use this to determine which bottle contains methanol and which one contains ethanol. You can use the approximate geometries of ethanol and methanol, which are included in the problem statement on the web. ($3 \mathrm{pts}$)
Alcohol from Mikuláš?!
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. $\(3 \mathrm{pts}\)$
Bonus:: Why is it so important for life on earth that oxygen is a triplet in the ground state? $\(+1 \mathrm{pts}\)$
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. $\(3 \mathrm{pts}\)$
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 ($\ce{He}_{2}^*$) 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 $66 \mathrm{nm}$. $\(4 \mathrm{pts}\)$
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.
Do not worry; the next gift from Mikuláš will not arrive before 5th December.