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{
"metadata": {
"name": ""
},
"nbformat": 3,
"nbformat_minor": 0,
"worksheets": [
{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"Chapter 5:Quantum Mechanics"
]
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 5.4,Page no:180"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Variable declaration \n",
"L= 1.0; #assuming Length L of box to be 1, this would not affect the probability\n",
"x1=0.45; #lower bound\n",
"x2=0.55; #upper bound\n",
"\n",
"from scipy.integrate import quad\n",
"import math\n",
"\n",
"#Calculation \n",
"def f(x):\n",
" y=(math.sin(n*(math.pi)*x))**2\n",
" return(y)\n",
"n=1.0;\n",
"I1=quad(f,x1,x2) #for ground state\n",
"P1=(2/L*I1[0])\n",
"n=2.0;\n",
"I2=quad(f,x1,x2) #for ground state\n",
"P2=(2/L*I2[0])\n",
"\n",
"\n",
"\n",
"#Result\n",
"print\"The probability n ground state is: \",round(P1,3),\"=\",round(P1,3)*100,\"percent\"\n",
"print\"The probability in first excited state is: \",round(P2,4),\"=\",round(P2,4)*100,\"percent\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The probability n ground state is: 0.198 = 19.8 percent\n",
"The probability in first excited state is: 0.0065 = 0.65 percent\n"
]
}
],
"prompt_number": 2
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 5.6,Page no:186"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Variable declaration\n",
"#Part (a)\n",
"E1= 1.0 #energy of first electron, eV\n",
"E2= 2.0 #energy of second electron, eV\n",
"Eb= 10.0 #height of barrier, eV\n",
"Wb= 0.50 #width of barrier, nm\n",
"Wb= Wb* 10**(-9) #converting to m\n",
"hbar= 1.054*10**(-34) #reduced Planck's conctaant, J.s\n",
"Me= 9.1*10**(-31) #mass of electron, kg\n",
"e= 1.6*10**(-19) #charge of an electron, J/eV\n",
"import math\n",
"#Calculation\n",
"\n",
"k2= (math.sqrt(2*(Me)*(Eb-E1)*(e)))/(hbar) #for 1.0 eV e-, m**(-1)\n",
"k1= (math.sqrt(2*Me*(Eb-E2)*e))/hbar #for first electron, m**(-1)\n",
"T1= math.exp((-2)*k2*Wb) #transmission probability for first electron\n",
"T2= math.exp((-2)*k1*Wb) #for second electron\n",
"print\"(A.)\\nTransmission probability for electrons with energy 1.0 eV is:%.2g\"%T1\n",
"print\"Transmission probability for electrons with energy 2.0 eV is: %.2g\"%T2\n",
"\n",
"#Part (b)\n",
"Wb= Wb*2; #Barrier width doubled\n",
"T11= math.exp((-2)*k2*Wb) # changed transmission probability for first electron\n",
"T22= math.exp((-2)*k1*Wb) #for second electron\n",
"\n",
"#Result\n",
"print\"\\n\\n(B.):After the barrier width is doubled:\\n\"\n",
"print\"Transmission probability for electrons with energy 1.0 eV is:%.2g\"%T11\n",
"print\"Transmission probability for electrons with energy 2.0 eV is: %.2g\"%T22\n",
"print\"\\n\\nNOTE:Calculation mistake in book in the calculation of k2,\\nit is wrongly written as 1.6e+10,\\nTHAT's Why a change in final answer\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"(A.)\n",
"Transmission probability for electrons with energy 1.0 eV is:2.1e-07\n",
"Transmission probability for electrons with energy 2.0 eV is: 5.1e-07\n",
"\n",
"\n",
"(B.):After the barrier width is doubled:\n",
"\n",
"Transmission probability for electrons with energy 1.0 eV is:4.6e-14\n",
"Transmission probability for electrons with energy 2.0 eV is: 2.6e-13\n",
"\n",
"\n",
"NOTE:Calculation mistake in book in the calculation of k2,\n",
"it is wrongly written as 1.6e+10,\n",
"THAT's Why a change in final answer\n"
]
}
],
"prompt_number": 3
}
],
"metadata": {}
}
]
}
|
