NUCE2100 HW2 Update

NUCE_2100/HW2.ipynb

@@ -48,10 +48,10 @@
     "    h1_mass = 1007825.03190 #micro AMU\n",
     "    neutron_mass = 1008664.9159 #mirco AMU\n",
     "\n",
-    "    mass_defect = (z*h1_mass + n*neutron_mass - mass)/1e6 #micro amu to u\n",
+    "    mass_defect = (z*h1_mass + n*neutron_mass - mass)/1e6 #calculate mass defect, micro amu to u\n",
-    "    energy = (mass_defect*1.6606e-27)*c**2 #amu to kg, answer in J\n",
+    "    energy = (mass_defect*1.6606e-27)*c**2 #convert amu to kg, answer in J\n",
     "\n",
-    "    return energy*6.242e12 #Convert J to MeV\n",
+    "    return energy*6.242e12 #Convert J to MeV, and return value\n",
     "    \n",
     "twoH_mass = 2014101.77784 #micro AMU\n",
     "print(\"2H Binding Energy:\", binding_energy(twoH_mass,1,1), \"MeV\")\n",
@@ -89,6 +89,7 @@
     "neutron_mass = 1008664.9159 #mirco AMU\n",
     "c = 299792458 # m/s\n",
     "\n",
+    "#Calculate difference in mass of products - reactants, mulitply by c^2\n",
     "Q = (fourHe_mass + neutron_mass - twoH_mass - threeH_mass)/1e6*1.6606e-27 * c**2\n",
     "print(\"Q is \", Q*6.242e12, \"MeV\")"
    ]
@@ -125,6 +126,7 @@
    "id": "e8108f30-f807-4f83-90b8-ce1092404284",
    "metadata": {},
    "source": [
+    "---\n",
     "2. Using atomic mass data, compute the average binding energy per nucleon of the following\n",
     "nuclei:\n",
     "\n",
@@ -153,29 +155,31 @@
      "name": "stdout",
      "output_type": "stream",
      "text": [
-      "6Li Binding Energy: 31.997677545325832 MeV\n",
+      "A: Average 6Li Binding Energy: 5.332946257554306 MeV per nucleon\n",
-      "12C Binding Energy: 92.17236586153292 MeV\n",
+      "B: Average 12C Binding Energy: 7.681030488461077 MeV per nucleon\n",
-      "51V Binding Energy: 445.8977789774969 MeV\n",
+      "C: Average 51V Binding Energy: 8.743093705441115 MeV per nucleon\n",
-      "138Ba Binding Energy: 1158.4258775508786 MeV\n",
+      "D: Average 138Ba Binding Energy: 8.394390417035352 MeV per nucleon\n",
-      "235U Binding Energy: 1784.0708281992524 MeV\n"
+      "E: Average 235U Binding Energy: 7.5917907582946915 MeV per nucleon\n"
      ]
     }
    ],
    "source": [
+    "#For each part, find the binding energy of the nuclide, then divde by the number of nucleons.\n",
+    "\n",
     "sixLi_mass = 6015122.8874 #micro AMU\n",
-    "print(\"6Li Binding Energy:\", binding_energy(sixLi_mass,3,3), \"MeV\")\n",
+    "print(\"A: Average 6Li Binding Energy:\", binding_energy(sixLi_mass,3,3)/6, \"MeV per nucleon\")\n",
     "\n",
     "twelveC_mass = 12000000.0 #micro AMU\n",
-    "print(\"12C Binding Energy:\", binding_energy(twelveC_mass,6,6), \"MeV\")\n",
+    "print(\"B: Average 12C Binding Energy:\", binding_energy(twelveC_mass,6,6)/12, \"MeV per nucleon\")\n",
     "\n",
     "fiftyoneV_mass = 50943957.66 #micro AMU\n",
-    "print(\"51V Binding Energy:\", binding_energy(fiftyoneV_mass,23,28), \"MeV\")\n",
+    "print(\"C: Average 51V Binding Energy:\", binding_energy(fiftyoneV_mass,23,28)/51, \"MeV per nucleon\")\n",
     "\n",
     "onethirtyeightBa_mass = 137905247.06 #micro AMU\n",
-    "print(\"138Ba Binding Energy:\", binding_energy(onethirtyeightBa_mass,56,82), \"MeV\")\n",
+    "print(\"D: Average 138Ba Binding Energy:\", binding_energy(onethirtyeightBa_mass,56,82)/138, \"MeV per nucleon\")\n",
     "\n",
     "twothirtyfiveU_mass = 235043928.1  #micro AMU\n",
-    "print(\"235U Binding Energy:\", binding_energy(twothirtyfiveU_mass,92,143), \"MeV\")"
+    "print(\"E: Average 235U Binding Energy:\", binding_energy(twothirtyfiveU_mass,92,143)/235, \"MeV per nucleon\")"
    ]
   },
   {
@@ -183,6 +187,7 @@
    "id": "df1603d2-3e2c-463f-93eb-bd9aa1514314",
    "metadata": {},
    "source": [
+    "---\n",
     "3. Compute the atom densities of $^{235}U$ and $^{238}U$ in UO2 of physical density 10.8 g/cm3 if the uranium\n",
     "is enriched to 3.5 w/o in $^{235}U$."
    ]
@@ -219,6 +224,7 @@
    "id": "df698b09-ce8d-4b38-92c7-0d96765e1b25",
    "metadata": {},
    "source": [
+    "---\n",
     "4. Calculate the mean free path of 1-eV neutrons in graphite. The total cross section of carbon at\n",
     "this energy is 4.8 b.\n",
     "\n",
@@ -243,11 +249,21 @@
    "source": [
     "graphite_density = 1.82 #g/cm^3\n",
     "graphite_nuclei_density = graphite_density/12.000*0.6022e24 #atoms/cm^3\n",
+    "#Take the density of graphite, divide by 12.000 g 12C / mole, then multiply by Avogadro's Number to get to atoms\n",
     "print(\"The nuclei density of graphite is \",graphite_nuclei_density,\" atoms/cm^3\")\n",
     "\n",
     "mean_free_path = 1/(4.8*10e-24)/graphite_nuclei_density\n",
+    "#Follow the formula from the slides, where we are converting Barns to cm.\n",
     "print(\"The mean free path is \",mean_free_path,\"cm in graphite.\")"
    ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": null,
+   "id": "2097a2cb-619e-410e-bfad-a5ca91849cbc",
+   "metadata": {},
+   "outputs": [],
+   "source": []
   }
  ],
  "metadata": {