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",
     "    <details>\n",
     "    <summary style=\"color: red\">Solution</summary>\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",
     "    </details>\n",
     "<br>\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",
     "    <details>\n",
     "    <summary style=\"color: red\">Solution</summary>\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 :<br> \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 :<br> \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",
     "    <br>\n",
@@ -263,7 +263,7 @@
    "source": [
     "<hr style=\"height:1px;border:none;color:#cccccc;background-color:#cccccc;\" />\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",