|
23 | 23 | "source": [
|
24 | 24 | "## **Goals**\n",
|
25 | 25 | "\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." |
29 | 29 | ]
|
30 | 30 | },
|
31 | 31 | {
|
|
52 | 52 | "2. Compare O$_2$ and OH, what do you observe regarding oscillation amplitudes?\n",
|
53 | 53 | " <details>\n",
|
54 | 54 | " <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", |
56 | 56 | " </details>\n",
|
57 | 57 | "<br>\n",
|
58 | 58 | "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 | 66 | "4. Compute the conversion factor between energy in eV and frequency in cm$^{-1}$.\n",
|
67 | 67 | " <details>\n",
|
68 | 68 | " <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", |
70 | 70 | " $$\\begin{align}\n",
|
71 |
| - " [eV]&=6.63\\cdot10^{-34}[m^2\\cdot kg\\cdot s^{-1}][Hz]\\\\\n", |
72 |
| - " 1.6\\cdot10^{-19}[J]&=6.63\\cdot10^{-34}[m^2\\cdot kg\\cdot s^{-1}]3\\cdot10^8[m\\cdot s^{-1}][m^{-1}]\\\\\n", |
| 71 | + " [eV]&=6.63\\cdot10^{-34}[m^2\\cdot kg\\cdot s^{-1}][Hz]\\\n", |
| 72 | + " 1.6\\cdot10^{-19}[J]&=6.63\\cdot10^{-34}[m^2\\cdot kg\\cdot s^{-1}]3\\cdot10^8[m\\cdot s^{-1}][m^{-1}]\\\n", |
73 | 73 | " 1.6\\cdot10^{-19}[m^2\\cdot kg\\cdot s^{-2}]&=6.63\\cdot10^{-34}\\cdot3\\cdot10^8[m^2\\cdot kg\\cdot s^{-1}][m\\cdot s^{-1}]10^2[cm^{-1}]\n",
|
74 | 74 | " \\end{align}$$\n",
|
75 | 75 | " <br>\n",
|
|
263 | 263 | "source": [
|
264 | 264 | "<hr style=\"height:1px;border:none;color:#cccccc;background-color:#cccccc;\" />\n",
|
265 | 265 | "\n",
|
266 |
| - "# Using the interactive visualization\n", |
| 266 | + "# How to use the interactive visualization\n", |
267 | 267 | "\n",
|
268 | 268 | "\n",
|
269 | 269 | "### Molecule viewer\n",
|
|
0 commit comments