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Author: lixpas

hex_lattice.png

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+\documentclass{beamer}
+
+\mode<presentation>
+{
+  \usetheme{default}      % or try Darmstadt, Madrid, Warsaw, ...
+  \usecolortheme{default} % or try albatross, beaver, crane, ...
+  \usefonttheme{default}  % or try serif, structurebold, ...
+  \setbeamertemplate{navigation symbols}{}
+  \setbeamertemplate{caption}[numbered]
+} 
+
+\usepackage[english]{babel}
+\usepackage[utf8]{inputenc}
+\usepackage[T1]{fontenc}
+\usepackage{modiagram}
+\usepackage{amsmath}
+\usepackage{graphicx}
+
+\title[Your Short Title]{IChO 2020}
+\author{L.F. Pa\v{s}teka}
+
+
+\begin{document}
+
+\begin{frame}
+  \titlepage
+\end{frame}
+
+
+\begin{frame}{Schr\"{o}dingerova rovnica - voľná častica}
+\begin{align*}
+\hat{H} \Psi(x) = E \Psi(x) \\
+-\frac{\hbar^2}{2m} \frac{d^2\Psi}{dx^2} = E \Psi \\
+-\frac{\hbar^2}{2m} \nabla^2\Psi = E \Psi \\
+\nabla^2\Psi = -\frac{2mE}{\hbar^2} \Psi \\
+\Psi = A \cos(kx) + B \sin(kx)\\
+k=\frac{\sqrt{2mE}}{\hbar}
+\end{align*}
+\end{frame}
+
+
+\begin{frame}{Schr\"{o}dingerova rovnica - potenciálová jama}
+\begin{align*}
+\hat{H} \Psi = E \Psi \\
+-\frac{\hbar^2}{2m} \nabla^2\Psi +\textcolor{red}{V\Psi} = E \Psi \\
+V(x)=0 \quad pre \quad 0<x<L, \quad inak \quad V(x)=\infty
+\end{align*}
+
+V jame: $\Psi = A \cos(kx) + B \sin(kx)$
+
+Mimo jamy: $\Psi = 0$
+\end{frame}
+
+
+\begin{frame}{Schr\"{o}dingerova rovnica - okrajové podmienky}
+\begin{align*}
+\Psi(0)=\Psi(L)=0 \\
+\rightarrow A = 0 \\
+\rightarrow k L=n\pi \\
+k=\frac{n\pi}{L}=\frac{\sqrt{2mE}}{\hbar}\\
+E=\frac{\hbar^2n^2\pi^2}{2mL^2}=\boxed{\frac{h^2n^2}{8mL}}
+\end{align*}
+\end{frame}
+
+
+\begin{frame}{Úlohy - potenciálová jama}
+\textbf{18.1}
+\begin{align*}
+E_{kv}=\frac{h^2}{8mL^2}=\frac{6.626^2}{8 \times 9.1094 \times 8^2 \times 1.4^2} 10^{-34-34+10+10+31}\\
+E_{kv}=4.80 \times 10^{-20} J \\
+E_n=E_{kv} n^2
+\end{align*}
+
+\textbf{18.2}
+\begin{align*}
+E_{tot} = 2(1+4+9+16) E_{kv} = 2.88 \times 10^{-18} J
+\end{align*} 
+
+\textbf{18.3}
+\begin{align*}
+E_{gap} = (25-16) E_{kv} = 4.32 \times 10^{-19} J \\
+c=\lambda\nu \quad E=h\nu \quad \rightarrow \quad \lambda=\frac{h c}{E_{gap}} = 460 \: nm
+\end{align*} 
+\end{frame}
+
+
+\begin{frame}{Úlohy - potenciálová jama}
+\textbf{18.4}
+\begin{align*}
+A_{lattice} = (11 \text{\AA})^2 = 1.21 \times 10^{-18} \: m^2\\
+A_{hex} = \frac{3\sqrt{3}}{2}L^2 = 5.09 \times 10^{-20} \: m^2\\
+N_{hex} = \frac{A_{lattice}}{A_{hex}} \approx 24 \quad \rightarrow \quad 48 e^- \quad \rightarrow \quad 24 \: MO
+\end{align*} 
+
+\includegraphics[width=3cm]{hex_lattice.png} 2 uhlíky na 1 šesťuholník
+\end{frame}
+
+\begin{frame}{Úlohy - potenciálová jama}
+\textbf{18.5-7}
+
+pozri excelovský hárok!
+
+\begin{align*}
+E=E_{kv}(n_1^2+n_2^2) \quad kde \quad E_{kv}=\frac{h^2}{8mL_1^2}=4.98 \times 10^{-20} J \\
+E_{HOMO}=E_{kv}(1^2+6^2)=37 E_{kv} = 1.84 \times 10^{-18} J \\
+E_{LUMO}=E_{kv}(2^2+6^2)=40 E_{kv} = 1.99 \times 10^{-18} J \\
+E_{gap}=(40-37)E_{kv} = 1.5 \times 10^{-19} J
+\end{align*} 
+\end{frame}
+
+
+\begin{frame}{Úlohy - potenciálová jama}
+\textbf{18.8-9}
+
+podobne
+\begin{align*}
+E_{n_1n_2n_3}=E_{kv}(n_1^2+n_2^2+n_3^3) \quad kde \quad E_{kv}=\frac{h^2}{8mL^2} \\
+E_{111}=3E_{kv} \quad 1\times deg.\\
+E_{112}=E_{121}=E_{211}=6E_{kv} \quad 3\times deg.\\
+E_{122}=E_{212}=E_{221}=9E_{kv} \quad 3\times deg.
+\end{align*} 
+atď.
+\end{frame}
+
+
+\begin{frame}{Úlohy - harmonický oscilátor}
+\textbf{19.1-3}
+\begin{align*}
+\mu = \frac{12 \times 16 }{12+16} = 6.86\: amu = 1.14 \times 10^{-26} \: kg\\
+\nu = \frac{1}{2\pi}\sqrt{\frac{k}{\mu}} = 65 \: THz \\
+\tilde{\nu}=\frac{1}{\lambda}=\frac{\nu}{c}=2170 \: cm^{-1} \\
+E_{ZPV}=\frac{1}{2}h\nu=2.16\times 10^{-20} J = 3.1 kcal/mol
+\end{align*}
+\end{frame}
+
+
+\begin{frame}{Úlohy - harmonický oscilátor}
+\textbf{19.4}
+\begin{align*}
+\frac{\nu_2}{\nu_1} = \frac{\frac{1}{2\pi}\sqrt{\frac{k}{\mu_2}}}{\frac{1}{2\pi}\sqrt{\frac{k}{\mu_1}}} = \sqrt{\frac{\mu_1}{\mu_2}} \quad \rightarrow \quad \nu_2=\sqrt{\frac{\mu_1}{\mu_2}} \nu_1\\
+\frac{\mu_1}{\mu_2}=\frac{\frac{12\times16}{12+16}}{\frac{13\times16}{13+16}} = \frac{12\times 29}{13\times 28} = 0.956 \quad \rightarrow \quad \sqrt{\frac{\mu_1}{\mu_2}} = 0.978\\
+\tilde{\nu}_2=0.978\:\tilde{\nu}_1=0.978\times2170=2122 \: cm^{-1}\\
+\end{align*}
+
+\textbf{19.5}
+\begin{align*}
+\frac{\mu_1}{\mu_2}= \frac{16\times 29}{17\times 28} = 0.975 \quad \rightarrow \quad \sqrt{\frac{\mu_1}{\mu_2}} = 0.987\\
+\tilde{\nu}_2=0.987\:\tilde{\nu}_1=0.987\times2170=2142 \: cm^{-1}\\
+\end{align*}
+\end{frame}
+
+
+\begin{frame}{Úlohy - harmonický oscilátor}
+\textbf{19.6}
+\begin{align*}
+\tilde{\nu}_{ZPV}=\frac{1}{2}(1649+3832+3942)=4712\: cm^{-1}\\
+E_{ZPV}=hc_{ZPV}=9.36\times 10^{-20} \:J
+\end{align*}
+
+\textbf{19.7}
+
+\begin{tabular}{ccccccccccc}
+&$\nu_1$&$\nu_2$&$\nu_3$\\
+$\times$&0&0&0&=&0   &+&$\tilde{\nu}_{ZPV}$&=&4712 cm$^{-1}$ \\
+$\times$&1&0&0&=&1649&+&$\tilde{\nu}_{ZPV}$&=&6361 cm$^{-1}$  \\
+$\times$&2&0&0&=&3298&+&$\tilde{\nu}_{ZPV}$&=&8010 cm$^{-1}$  \\
+$\times$&0&1&0&=&3832&+&$\tilde{\nu}_{ZPV}$&=&8544 cm$^{-1}$  \\
+$\times$&0&0&1&=&3943&+&$\tilde{\nu}_{ZPV}$&=&8655 cm$^{-1}$
+\end{tabular}
+\end{frame}
+
+\begin{frame}{Úlohy - tuhý rotor}
+\textbf{19.8}
+
+prepokladáme, že $R$ nezávisí od izotopov
+\begin{align*}
+\Delta E=E_1-E_0=\frac{h^2}{8\pi^2I}\left[1(1+1)-0(0+1)\right]=\frac{h^2}{4\pi^2I}\\
+\Delta E=h\nu=\frac{h^2}{4\pi^2\mu R^2} \quad \rightarrow \quad R=\sqrt{\frac{h}{4\pi^2\mu\nu}}=\frac{1}{2\pi}\sqrt{\frac{h}{\mu\nu}}=1.13\text{\AA}
+\end{align*}
+($\mu$ už máme z úlohy 19.1)
+\\
+\textbf{19.9}
+\begin{align*}
+\nu_{l+1}-\nu_{l}=\frac{h}{8\pi^2I}\left[(l+1)(l+2)-l(l+1)\right]=\frac{h}{4\pi^2I}(l+1)\\
+\rightarrow \nu_{2-1}=2\times115=230\: GHz\\
+\rightarrow \nu_{3-2}=3\times115=345\: GHz\\
+\end{align*}
+\end{frame}
+
+
+\begin{frame}{Úlohy - tuhý rotor}
+\textbf{19.10}
+\begin{align*}
+\frac{\nu_2}{\nu_1}=\frac{\frac{h}{8\pi^2\mu_2R^2}}{\frac{h}{8\pi^2\mu_1R^2}}=\frac{\mu_1}{\mu_2} \quad \rightarrow \quad \nu_2=\frac{\mu_1}{\mu_2} \nu_1\\
+\frac{\mu_1}{\mu_2}=0.975 \quad (\text{poznáme z úlohy 19.5})\\
+\nu_2=0.975\times115=112\: GHz
+\end{align*}
+\end{frame}
+ 
+ \end{document}

main.tex

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+\documentclass{article}
+\usepackage[utf8]{inputenc}
+
+\title{microtest}
+\author{Lukáš F. Pašteka}
+\date{June 2020}
+
+\begin{document}
+
+\maketitle
+
+\section{Preliminaries}
+
+%comment
+\textit{Tri pokusy:}
+
+Matematika: $\frac{\alpha^{123}}{\beta_{456}}$
+
+Diakritika UTF8: Šášov
+
+Dakritika LaTeX: \v{S}\'a\v{s}ov
+
+\end{document}