diff options
Diffstat (limited to 'Modern_Physics/chapter13_1.ipynb')
| -rwxr-xr-x | Modern_Physics/chapter13_1.ipynb | 248 |
1 files changed, 248 insertions, 0 deletions
diff --git a/Modern_Physics/chapter13_1.ipynb b/Modern_Physics/chapter13_1.ipynb new file mode 100755 index 00000000..fddec6a2 --- /dev/null +++ b/Modern_Physics/chapter13_1.ipynb @@ -0,0 +1,248 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:bee0f8df3068997ca43cf44df4f129c530dc6e34523ca501c99a2183643ee772"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "13: Nuclear Reactions"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example number 13.1, Page number 259"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#import modules\n",
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#Variable declaration\n",
+ "m1=7.0183; #mass of 3Li7(amu)\n",
+ "m2=4.0040; #mass of 2He4(amu)\n",
+ "m3=1.0082; #mass of 1H1(amu)\n",
+ "N=6.026*10**26; #Avgraodo no.(per kg atom)\n",
+ "#rxn = 3Li7 + 1H1 = 2He4 + 2He4 \n",
+ "\n",
+ "#Calculation\n",
+ "delta_m=m1+m3-(2*m2); #deltam(amu)\n",
+ "E=delta_m*931; #energy per disintegration(MeV)\n",
+ "n=0.1*N/7; #no of atoms in 100 gm of lithium\n",
+ "TE=n*E; #Total energy available(MeV) \n",
+ "\n",
+ "#Result\n",
+ "print \"energy available per disintegration is\",round(E,2),\"MeV\"\n",
+ "print \"Total energy available is\",round(TE/1e+25,2),\"*10**25 MeV\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "energy available per disintegration is 17.22 MeV\n",
+ "Total energy available is 14.83 *10**25 MeV\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example number 13.2, Page number 259"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#import modules\n",
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#Variable declaration\n",
+ "m1=6.015126; #mass of 3Li7(a.m.u)\n",
+ "m2=4.002604; #mass oh 2He4(a.m.u)\n",
+ "m3=1.00865; #mass of 0n1(a.m.u)\n",
+ "m4=3.016049; #mass of 1H3(a.m.u)\n",
+ "#rxn = 3Li7 + 0n1 = 2He4 + 1H3 + Q\n",
+ "\n",
+ "#Calculation\n",
+ "dm=m1+m3-(m2+m4);\n",
+ "Q=dm*931; #energy released(MeV)\n",
+ "\n",
+ "#Result\n",
+ "print \"energy released is\",round(Q,4),\"MeV\"\n",
+ "print \"answer given in the book varies due to rounding off errors\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "energy released is 4.7695 MeV\n",
+ "answer given in the book varies due to rounding off errors\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example number 13.3, Page number 260"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#import modules\n",
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#Variable declaration\n",
+ "m1=14.007515; #mass of 7N14(a.m.u)\n",
+ "m2=4.003837; #mass of 2He4(a.m.u)\n",
+ "m3=17.004533; #mass of 8O17(a.m.u)\n",
+ "m4=1.008142; #mass of 1H1(a.m.u)\n",
+ "#rxn = 7N14 + 2He14 = 8O17 + 1H1\n",
+ "\n",
+ "#Calculation\n",
+ "dm=m3+m4-(m1+m2);\n",
+ "Q=dm*931; #Q value of the reaction(MeV)\n",
+ "\n",
+ "#Result\n",
+ "print \"Q value of the reaction is\",round(Q,3),\"MeV\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Q value of the reaction is 1.232 MeV\n"
+ ]
+ }
+ ],
+ "prompt_number": 11
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example number 13.4, Page number 260"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#import modules\n",
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#Variable declaration\n",
+ "m1=14.007520; #mass of 7N14(a.m.u)\n",
+ "m2=1.008986; #mass oh 0n1(a.m.u)\n",
+ "#m3=mass of 6C14 in a.m.u\n",
+ "m4=1.008145; #mass of 1H1(a.m.u)\n",
+ "#rxn = 7N14 + 0n1 = 6C14 + 1H1 + 0.55 MeV\n",
+ "\n",
+ "#Calculation\n",
+ "Q=0.55; #energy(MeV)\n",
+ "dm=Q/931; \n",
+ "m3=dm+m1+m2-m4; #mass of 6C14(a.m.u)\n",
+ "\n",
+ "#Result\n",
+ "print \"mass of 6C14 is\",round(m3,5),\"a.m.u\"\n",
+ "print \"answer given in the book varies due to rounding off errors\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "mass of 6C14 is 14.00895 a.m.u\n",
+ "answer given in the book varies due to rounding off errors\n"
+ ]
+ }
+ ],
+ "prompt_number": 14
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example number 13.6, Page number 261"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#import modules\n",
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#Variable declaration\n",
+ "m0=11.01280; #mass 5B11(a.m.u)\n",
+ "m1=4.00387; #mass of alpha particle(a.m.u)\n",
+ "m2=14.00752; #mass of 7N14(a.m.u)\n",
+ "#m3=mass of neutron \n",
+ "E1=5.250; #energy of alpha particle(MeV)\n",
+ "E2=2.139; #energy of 7N14(MeV)\n",
+ "E3=3.260; #energy of 0n1(MeV)\n",
+ "\n",
+ "#Calculation\n",
+ "m3=(m0*931)+((m1*931)+E1)-((m2*931)+E2)-E3; #mass of neutron(a.m.u)\n",
+ "\n",
+ "#Result\n",
+ "print \"mass of neutron is\",round(m3/931,3),\"a.m.u\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "mass of neutron is 1.009 a.m.u\n"
+ ]
+ }
+ ],
+ "prompt_number": 16
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file |