Skip to content
This commit does not belong to any branch on this repository, and may belong to a fork outside of the repository.

Commit b471b03

Browse files
committedMay 20, 2024·
Merged textual changes from mol vibration branch. Also fixed some formatting issues still present in mol vibration notebook.
1 parent 09bfbb3 commit b471b03

File tree

2 files changed

+9
-9
lines changed

2 files changed

+9
-9
lines changed
 

‎notebook/lattice-vibration/Molecule_Vibration.ipynb

+8-8
Original file line numberDiff line numberDiff line change
@@ -23,9 +23,9 @@
2323
"source": [
2424
"## **Goals**\n",
2525
"\n",
26-
"* Understand normal modes of vibrations.\n",
27-
"* Learn about normal modes and their frequency.\n",
28-
"* Explore the vast variety of vibrational modes."
26+
"* Learn about the nature of vibrational modes within molecules\n",
27+
"* Investigate the different frequencies and oscillation patterns of these modes.\n",
28+
"* Explore the variety of molecular vibrations and how they arise from the molecular topology."
2929
]
3030
},
3131
{
@@ -52,7 +52,7 @@
5252
"2. Compare O$_2$ and OH, what do you observe regarding oscillation amplitudes?\n",
5353
" <details>\n",
5454
" <summary style=\"color: red\">Solution</summary>\n",
55-
" 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",
55+
" 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",
5656
" </details>\n",
5757
"<br>\n",
5858
"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 @@
6666
"4. Compute the conversion factor between energy in eV and frequency in cm$^{-1}$.\n",
6767
" <details>\n",
6868
" <summary style=\"color: red\">Solution</summary>\n",
69-
" 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 :<br> \n",
69+
" 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 :<br> \n",
7070
" $$\\begin{align}\n",
71-
" [eV]&=6.63\\cdot10^{-34}[m^2\\cdot kg\\cdot s^{-1}][Hz]\\\\\n",
72-
" 1.6\\cdot10^{-19}[J]&=6.63\\cdot10^{-34}[m^2\\cdot kg\\cdot s^{-1}]3\\cdot10^8[m\\cdot s^{-1}][m^{-1}]\\\\\n",
71+
" [eV]&=6.63\\cdot10^{-34}[m^2\\cdot kg\\cdot s^{-1}][Hz]\\\n",
72+
" 1.6\\cdot10^{-19}[J]&=6.63\\cdot10^{-34}[m^2\\cdot kg\\cdot s^{-1}]3\\cdot10^8[m\\cdot s^{-1}][m^{-1}]\\\n",
7373
" 1.6\\cdot10^{-19}[m^2\\cdot kg\\cdot s^{-2}]&=6.63\\cdot10^{-34}\\cdot3\\cdot10^8[m^2\\cdot kg\\cdot s^{-1}][m\\cdot s^{-1}]10^2[cm^{-1}]\n",
7474
" \\end{align}$$\n",
7575
" <br>\n",
@@ -263,7 +263,7 @@
263263
"source": [
264264
"<hr style=\"height:1px;border:none;color:#cccccc;background-color:#cccccc;\" />\n",
265265
"\n",
266-
"# Using the interactive visualization\n",
266+
"# How to use the interactive visualization\n",
267267
"\n",
268268
"\n",
269269
"### Molecule viewer\n",

‎notebook/lattice-vibration/theory/theory_molecular_vibration.ipynb

+1-1
Original file line numberDiff line numberDiff line change
@@ -65,7 +65,7 @@
6565
"\\begin{equation}\n",
6666
" K=\\frac{1}{2} \\sum_{i=1}^{3 N} \\dot{q}_{i}^{2} \\qquad (2)\n",
6767
"\\end{equation}\n",
68-
"The potential energy ca be expended as:\n",
68+
"The potential energy can be expended as:\n",
6969
"\\begin{equation}\n",
7070
" 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",
7171
"\\end{equation}\n",

0 commit comments

Comments
 (0)
Please sign in to comment.