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+{
+"cells": [
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "# Chapter 44: Energy from the Nucleus"
+ ]
+ },
+{
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 44.1: Sample_Problem_1.sce"
+ ]
+ },
+ {
+"cell_type": "code",
+ "execution_count": null,
+ "metadata": {
+ "collapsed": true
+ },
+ "outputs": [],
+"source": [
+"//Given that\n",
+"M_U = 235.0439 //in Atomic mass unit\n",
+"M_Ce = 139.9054 //in Atomic mass unit\n",
+"M_Zr = 93.9063 //in Atomic mass unit\n",
+"M_n = 1.00867 //in Atomic mass unit\n",
+"uCsqure = 931.5 //in Mev\n",
+"\n",
+"//Sample Problem 44-1\n",
+"txt = mopen('Example44_1_result.txt','wt')\n",
+"mfprintf(txt, '**Sample Problem 44-1**\n')\n",
+"Q = -((M_Ce + M_Zr + M_n) - M_U) * uCsqure\n",
+"mfprintf(txt, 'The disintegration energy is %dMev', Q)\n",
+"mclose(txt)"
+ ]
+ }
+,
+{
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 44.2: Sample_Problem_2.sce"
+ ]
+ },
+ {
+"cell_type": "code",
+ "execution_count": null,
+ "metadata": {
+ "collapsed": true
+ },
+ "outputs": [],
+"source": [
+"//Given that\n",
+"c = 3*10^8 //in m/s\n",
+"e = 1.6*10^-19\n",
+"conv = 3600*24 //day to sec conversion\n",
+"Pgen = 3400*10^6 //in W\n",
+"Pused = 1100*10^6 //in W\n",
+"fuel = 8.60*10^4 //in kg\n",
+"Q = 200*10^6*e //in J\n",
+"Uinitial = 8.6*10^4 //in kg\n",
+"M_u = 3.90*10^-25 //in kg/atom\n",
+"N = 5.70*10^4\n",
+"p = 3/100\n",
+"\n",
+"txt = mopen('Example44_2_result.txt','wt')\n",
+"//Sample Problem 44-2a\n",
+"mfprintf(txt, '**Sample Problem 44-2a**\n')\n",
+"eff = Pused/Pgen*100\n",
+"mfprintf(txt, 'The efficiency of the power plant is %d\%\n', eff)\n",
+"\n",
+"//Sample Problem 44-2b\n",
+"mfprintf(txt, '\n**Sample Problem 44-2b**\n')\n",
+"R = Pgen/Q\n",
+"mfprintf(txt, 'The fissions in the reactor per second is %e\n', R)\n",
+"\n",
+"//Sample Proble 44-2c\n",
+"mfprintf(txt, '\n**Sample Problem 44-2c**\n')\n",
+"RateDay = (1+0.25)*R*M_u*conv\n",
+"mfprintf(txt, 'The uranium use in a day is equal to %fkg/day\n', RateDay)\n",
+"\n",
+"//Sample Problem 44-2d\n",
+"mfprintf(txt, '\n**Sample Problem 44-2d**\n')\n",
+"T = fuel*p/RateDay\n",
+"mfprintf(txt, 'The U will long for %ddays\n', T)\n",
+"\n",
+"//Sample Problem 44-2e\n",
+"mfprintf(txt, '\n**Sample Problem 44-2e**\n')\n",
+"MassConvRate = Pgen/c^2\n",
+"mfprintf(txt, 'The mass conversion rate is %ekg/s', MassConvRate)\n",
+"mclose(txt)"
+ ]
+ }
+,
+{
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 44.3: Sample_Problem_3.sce"
+ ]
+ },
+ {
+"cell_type": "code",
+ "execution_count": null,
+ "metadata": {
+ "collapsed": true
+ },
+ "outputs": [],
+"source": [
+"//Given that\n",
+"ratio = 0.0072\n",
+"T = 2.0*10^9 //in years\n",
+"Th1 = 7.04*10^8 //in years\n",
+"Th2 = 44.7*10^8 //in years\n",
+"\n",
+"//Sample Problem 44-3\n",
+"txt = mopen('Example44_3_result.txt','wt')\n",
+"mfprintf(txt, '**Sample Problem 44-3**\n')\n",
+"l1 = log(2)/Th1\n",
+"l2 = log(2)/Th2\n",
+"ratioEarlier = ratio*%e^((l1 - l2)*T)\n",
+"mfprintf(txt, 'The earlier ratio is equal to %f', ratioEarlier)\n",
+"mclose(txt)"
+ ]
+ }
+,
+{
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 44.4: Sample_Problem_4.sce"
+ ]
+ },
+ {
+"cell_type": "code",
+ "execution_count": null,
+ "metadata": {
+ "collapsed": true
+ },
+ "outputs": [],
+"source": [
+"//Given that\n",
+"R = 10^-15 //in meter\n",
+"e = 1.6*10^-19 //in coloumb\n",
+"q1 = e\n",
+"k = 9*10^9 //in SI unit\n",
+"B = 1.38*10^-23 //in J/K\n",
+"\n",
+"txt = mopen('Example44_4_result.txt','wt')\n",
+"//Sample Problem 44-4a\n",
+"mfprintf(txt, '**Sample Problem 44-4a**\n')\n",
+"K = k*q1^2/(2*R)/2\n",
+"mfprintf(txt, 'The initial kinetic energy is equal to %dKev\n', K/e/10^3)\n",
+"\n",
+"//Sample Problem 44-4b\n",
+"mfprintf(txt, '\n**Sample Problem 44-4b**\n')\n",
+"T = 2*K/(3*B) //B is Boltzman constant\n",
+"mfprintf(txt, 'The temprature required to achieve that energy equal to %eK', T)\n",
+"mclose(txt)"
+ ]
+ }
+,
+{
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 44.5: Sample_Problem_5.sce"
+ ]
+ },
+ {
+"cell_type": "code",
+ "execution_count": null,
+ "metadata": {
+ "collapsed": true
+ },
+ "outputs": [],
+"source": [
+"//Sample Problem 44-5\n",
+"txt = mopen('Example44_5_result.txt','wt')\n",
+"mfprintf(txt, '**Sample Problem 44-5**\n')\n",
+"MassRate = 4*1.67*10^-27/(4.20*10^-12) //mass of proton required to produce 1 unit of energy \n",
+"Ps = 3.90*10^26 //in W\n",
+"Rate = MassRate*Ps\n",
+"mfprintf(txt, 'The rate at which hydrogen is consumed is %ekg/s', Rate)\n",
+"mclose(txt)"
+ ]
+ }
+,
+{
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 44.6: Sample_Problem_6.sce"
+ ]
+ },
+ {
+"cell_type": "code",
+ "execution_count": null,
+ "metadata": {
+ "collapsed": true
+ },
+ "outputs": [],
+"source": [
+"//Given that\n",
+"d = 200 //in kg/m^3\n",
+"fac = 10^3\n",
+"Na = 6.023*10^23\n",
+"Mt = 3*10^-3 //in kg/mol\n",
+"Md = 2*10^-3 //in kg/mol\n",
+"\n",
+"txt = mopen('Example44_6_result.txt','wt')\n",
+"//Sample Problem 44-6a\n",
+"mfprintf(txt, '**Sample Problem 44-6a**\n')\n",
+"n = 2*fac *d *Na /(Mt + Md)\n",
+"mfprintf(txt, 'The number of particle in unit volume is %em^-3\n', n)\n",
+"\n",
+"//Sample Problem 44-6b\n",
+"mfprintf(txt, '\n**Sample Problem 44-6b**\n')\n",
+"TauMin = 10^20/n\n",
+"mfprintf(txt, 'The duration of time, pallet can maintain is of the order of %esec', TauMin)\n",
+"mclose(txt)"
+ ]
+ }
+],
+"metadata": {
+ "kernelspec": {
+ "display_name": "Scilab",
+ "language": "scilab",
+ "name": "scilab"
+ },
+ "language_info": {
+ "file_extension": ".sce",
+ "help_links": [
+ {
+ "text": "MetaKernel Magics",
+ "url": "https://github.com/calysto/metakernel/blob/master/metakernel/magics/README.md"
+ }
+ ],
+ "mimetype": "text/x-octave",
+ "name": "scilab",
+ "version": "0.7.1"
+ }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 0
+}