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   "source": [
    "# Chapter 5 Imperfection in Solids"
   ]
  },
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   "metadata": {},
   "source": [
    "## Example 5_1 pgno:56"
   ]
  },
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     "text": [
      "Example 5.1\n",
      "\n",
      "\n",
      " Equilibrium number of vacancies/m**3 is  for 1273K 2.18444488963e+25\n"
     ]
    }
   ],
   "source": [
    "# given that\n",
    "Na=6.023*10**23 #Avogadro No.\n",
    "rho=8.4e6 #Density of Copper in g/m**3\n",
    "A=63.5 #Atomic weight of Copper\n",
    "Qv=0.9 #Activation energy in eV\n",
    "k=8.62*10**-5 #Boltzmann Constant in eV/K\n",
    "T=1000+273#Temperature in K\n",
    "from math import exp\n",
    "print\"Example 5.1\\n\"\n",
    "N=Na*rho/A #No. of atomic site per cubic meter\n",
    "Nv=N*exp(-Qv/(k*T))\n",
    "print\"\\n Equilibrium number of vacancies/m**3 is  for 1273K\",Nv\n",
    "\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 5_3 pgno:57"
   ]
  },
  {
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   "execution_count": 2,
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   "outputs": [
    {
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     "text": [
      " Example 5.3\n",
      "\n",
      "\n",
      " Atomic  of Al is 98.7039833218\n",
      "\n",
      " Atomic  of Cu is 1.29601667817\n"
     ]
    }
   ],
   "source": [
    "# given that\n",
    "C_Al=97. #Aluminium wt%\n",
    "C_Cu=3. #Copper wt%\n",
    "A_Al=26.98 #Atomic wt of Aluminium\n",
    "A_Cu=63.55 #Atomic wt of Copper\n",
    "\n",
    "print\" Example 5.3\\n\"\n",
    "CAl=C_Al*A_Cu*100/((C_Al*A_Cu)+(C_Cu*A_Al))\n",
    "CCu=C_Cu*A_Al*100/((C_Cu*A_Al)+(C_Al*A_Cu))\n",
    "print\"\\n Atomic  of Al is\",CAl\n",
    "print\"\\n Atomic  of Cu is\",CCu\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 5_4 pgno:58"
   ]
  },
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   "outputs": [
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     "text": [
      "Example 5.4\n",
      "\n",
      "\n",
      " Number of Schottky defects are  defects/m**3. 5.31422380078e+19\n"
     ]
    }
   ],
   "source": [
    "# given that\n",
    "Na=6.023*10**23 #Avogadro No.\n",
    "rho=1.955 #Density of KCl in g/cm**3\n",
    "A_k= 39.10 #Atomic weight of potassium in g/mol\n",
    "A_cl= 35.45 #Atomic weight of Chlorine in g/mol\n",
    "Qs=2.6 #Activation energy in eV\n",
    "k=8.62*10**-5 #Boltzmann Constant in eV/K\n",
    "T=500+273 #Temperature in K\n",
    "from math import exp\n",
    "\n",
    "print\"Example 5.4\\n\"\n",
    "A = A_k+A_cl # Molar mass of KCl in gram\n",
    "N=Na*rho*1e6/A #No. of atomic site per cubic meter\n",
    "Ns=N*exp(-Qs/(2*k*T))\n",
    "print\"\\n Number of Schottky defects are  defects/m**3.\",Ns\n",
    "\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 5_6 pgno:58"
   ]
  },
  {
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   "execution_count": 4,
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   "outputs": [
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     "text": [
      "Example 5.6\n",
      "\n",
      "\n",
      " Part A\n",
      "\n",
      " Grain size number is \n",
      "6.49185309633\n",
      "\n",
      " Part B\n",
      "\n",
      " At magnification of 85x\n",
      "\n",
      " Number of grains per inch square are\n",
      "62.2837370242\n"
     ]
    }
   ],
   "source": [
    "# given that \n",
    "N=45. #Number of grains per square inch\n",
    "M=85. # magnification\n",
    "from math import log\n",
    "print\"Example 5.6\\n\"\n",
    "print\"\\n Part A\"\n",
    "n=(log(N)/log(2))+1 #calculation for grain size no.  N=2**(n-1)\n",
    "print\"\\n Grain size number is \\n\",n\n",
    "print\"\\n Part B\"\n",
    "Nm=(100/M)**2*2**(n-1)\n",
    "print\"\\n At magnification of 85x\\n\"\n",
    "print\" Number of grains per inch square are\\n\",Nm\n",
    "# answer in book is 62.6. It is because of rounding off at intermediate stages\n"
   ]
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