From b471b033fd7acf8240832d7e60fd31dc277a7ad2 Mon Sep 17 00:00:00 2001 From: Taylor Date: Mon, 20 May 2024 10:04:34 +0200 Subject: [PATCH] Merged textual changes from mol vibration branch. Also fixed some formatting issues still present in mol vibration notebook. --- .../lattice-vibration/Molecule_Vibration.ipynb | 16 ++++++++-------- .../theory/theory_molecular_vibration.ipynb | 2 +- 2 files changed, 9 insertions(+), 9 deletions(-) diff --git a/notebook/lattice-vibration/Molecule_Vibration.ipynb b/notebook/lattice-vibration/Molecule_Vibration.ipynb index e91b7c4..b68ab7c 100644 --- a/notebook/lattice-vibration/Molecule_Vibration.ipynb +++ b/notebook/lattice-vibration/Molecule_Vibration.ipynb @@ -23,9 +23,9 @@ "source": [ "## **Goals**\n", "\n", - "* Understand normal modes of vibrations.\n", - "* Learn about normal modes and their frequency.\n", - "* Explore the vast variety of vibrational modes." + "* Learn about the nature of vibrational modes within molecules\n", + "* Investigate the different frequencies and oscillation patterns of these modes.\n", + "* Explore the variety of molecular vibrations and how they arise from the molecular topology." ] }, { @@ -52,7 +52,7 @@ "2. Compare O$_2$ and OH, what do you observe regarding oscillation amplitudes?\n", "
\n", " Solution\n", - " In a diatomic molecule, the relative amplitudes are given by $A_1=-\\frac{M_2}{M_1}A_2$ (c.f. theory). Therefore, hydrogen being a much lighter element compared to oxygen, its amplitude is much greater. This can be understood intuitively as the hydrogen atom as a much lower inertia than oxygene, and thus when given the same energy, the hydrogen atom will oscillate more easily.\n", + " In a diatomic molecule, the relative amplitudes are given by $A_1=-\\frac{M_2}{M_1}A_2$ (c.f. theory). Therefore, since hydrogen is a much lighter element than oxygen, its amplitude is much greater. This can be understood intuitively: as the hydrogen atom has a much smaller moment of inertia than oxygen, when given the same energy, the hydrogen atom will oscillate more easily.\n", "
\n", "
\n", "3. Compare H$_2$O and CO$_2$, how many vibrational modes does each one have? Are all CO$_2$ vibrations distinct? Can you explain the difference in energy between vibrational modes? \n", @@ -66,10 +66,10 @@ "4. Compute the conversion factor between energy in eV and frequency in cm$^{-1}$.\n", "
\n", " Solution\n", - " The relation between frequency and energy is given by $E=h\\nu$ with $\\nu$ the frequency. With $[h]=[m^2\\cdot kg\\cdot s^{-1}]$ and $\\lambda=\\frac{c}{\\nu}$, with $c$ the speed of light and $\\lambda$ the wavelength, we have :
\n", + " The relation between frequency and energy is given by $E=h\\nu$, with $\\nu$ the frequency. Here, $[h]=[m^2\\cdot kg\\cdot s^{-1}]$ and $\\lambda=\\frac{c}{\\nu}$, with $c$ the speed of light and $\\lambda$ the wavelength, we have :
\n", " $$\\begin{align}\n", - " [eV]&=6.63\\cdot10^{-34}[m^2\\cdot kg\\cdot s^{-1}][Hz]\\\\\n", - " 1.6\\cdot10^{-19}[J]&=6.63\\cdot10^{-34}[m^2\\cdot kg\\cdot s^{-1}]3\\cdot10^8[m\\cdot s^{-1}][m^{-1}]\\\\\n", + " [eV]&=6.63\\cdot10^{-34}[m^2\\cdot kg\\cdot s^{-1}][Hz]\\\n", + " 1.6\\cdot10^{-19}[J]&=6.63\\cdot10^{-34}[m^2\\cdot kg\\cdot s^{-1}]3\\cdot10^8[m\\cdot s^{-1}][m^{-1}]\\\n", " 1.6\\cdot10^{-19}[m^2\\cdot kg\\cdot s^{-2}]&=6.63\\cdot10^{-34}\\cdot3\\cdot10^8[m^2\\cdot kg\\cdot s^{-1}][m\\cdot s^{-1}]10^2[cm^{-1}]\n", " \\end{align}$$\n", "
\n", @@ -263,7 +263,7 @@ "source": [ "
\n", "\n", - "# Using the interactive visualization\n", + "# How to use the interactive visualization\n", "\n", "\n", "### Molecule viewer\n", diff --git a/notebook/lattice-vibration/theory/theory_molecular_vibration.ipynb b/notebook/lattice-vibration/theory/theory_molecular_vibration.ipynb index 75c011c..b03aecb 100644 --- a/notebook/lattice-vibration/theory/theory_molecular_vibration.ipynb +++ b/notebook/lattice-vibration/theory/theory_molecular_vibration.ipynb @@ -65,7 +65,7 @@ "\\begin{equation}\n", " K=\\frac{1}{2} \\sum_{i=1}^{3 N} \\dot{q}_{i}^{2} \\qquad (2)\n", "\\end{equation}\n", - "The potential energy ca be expended as:\n", + "The potential energy can be expended as:\n", "\\begin{equation}\n", " V=V_{0}+\\sum_{i=1}^{3 N}\\left(\\frac{\\partial V}{\\partial q_{i}}\\right)_{0} q_{i}+\\frac{1}{2} \\sum_{i=1}^{3 N}\\left(\\frac{\\partial^{2} V}{\\partial q_{i} \\partial q_{j}}\\right)_{0} q_{i} q_{j}+\\cdots \\qquad (3)\n", "\\end{equation}\n",