{
"metadata": {
"name": "",
"signature": "sha256:64e0a844a3ba33dead720cfb202f40f7dab27e1a9fab7ab4975bb30624a36195"
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"nbformat": 3,
"nbformat_minor": 0,
"worksheets": [
{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"Chapter 31 Radioactivity"
]
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 31.1 Page no 855"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Given\n",
"T=138.0 #days\n",
"E=0.693\n",
"N=12.5\n",
"N0=100\n",
"\n",
"#Calculation\n",
"import math\n",
"L=E/T\n",
"N1=N/N0\n",
"t=(2.303*math.log10(8))/L\n",
"\n",
"#Result\n",
"print\"Time is\",round(t,2),\"days\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Time is 414.16 days\n"
]
}
],
"prompt_number": 71
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 31.2 Page no 855"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Given\n",
"T=1.42*10**17 #s\n",
"A=6.02*10**23 #mol**-1\n",
"E=0.693\n",
"L=238.0\n",
"\n",
"#Calculation\n",
"N=A/L\n",
"L1=E/T\n",
"Z=L1*N\n",
"\n",
"#Result\n",
"print\"Disintegrations per second occur in 1g of U**238 is\",round(Z*10**-4,3),\"*10**4 s**-1\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Disintegrations per second occur in 1g of U**238 is 1.234 *10**4 s**-1\n"
]
}
],
"prompt_number": 1
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 31.3 Page no 855"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Given\n",
"K=-40 #Cl\n",
"M=0.075 #kg mole**-1\n",
"m=1.2*10**-6 #kg\n",
"A=6.0*10**23\n",
"D=170 #s**-1\n",
"E=0.693\n",
"\n",
"#Calculation\n",
"N=(A/M)*m\n",
"L=D/N\n",
"T=(E/L)\n",
"O=T/31536000.0\n",
"\n",
"#Result\n",
"print\"Half life of K40 atom is\",round(O*10**-9,3),\"*10**9 years\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Half life of K40 atom is 1.241 *10**9 years\n"
]
}
],
"prompt_number": 81
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 31.4 Page no 856"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Given\n",
"T=6.0 #hours\n",
"E=0.693\n",
"A=6.025*10**23\n",
"W=99.0\n",
"S=10**-12\n",
"t=1 #hours\n",
"\n",
"#Calculation\n",
"import math\n",
"L=E/T\n",
"N0=(A/W)*S\n",
"R0=L*N0\n",
"N=N0*math.exp(-L)\n",
"R=L*N\n",
"R1=L*N\n",
"\n",
"#Result\n",
"print\"Activity at beginning is\", round(R0*10**-8,2),\"*10**8 /h\"\n",
"print\"Activity at the end is\",round(R1*10**-8,3),\"*10**8 /h\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Activity at beginning is 7.03 *10**8 /h\n",
"Activity at the end is 6.262 *10**8 /h\n"
]
}
],
"prompt_number": 15
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 31.5 Page no 856"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Given\n",
"t=30.0 #years\n",
"a=16\n",
"\n",
"#Calculation\n",
"import math\n",
"l=(2.3026*math.log10(a))/t\n",
"T=0.693/l\n",
"\n",
"#Result\n",
"print\"Half life period is\", round(T,1),\"years\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Half life period is 7.5 years\n"
]
}
],
"prompt_number": 22
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 31.6 Page no 856"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Given\n",
"m1=238.05081\n",
"m2=234.04363\n",
"m3=4.00260\n",
"A=931.5 #Mev\n",
"\n",
"#Calculation\n",
"Ea=(m1-m2-m3)*A\n",
"\n",
"#Result\n",
"print\"Kinetic energy is\",round(Ea,2),\"Mev\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Kinetic energy is 4.27 Mev\n"
]
}
],
"prompt_number": 25
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 31.7 Page no 856"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Given\n",
"m1=22.994466\n",
"m2=22.989770\n",
"A=931.5\n",
"\n",
"#Calculation\n",
"Eb=(m1-m2)*A\n",
"\n",
"#Result\n",
"print\"Maximum kinetic energy is\",round(Eb,3),\"Mev\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Maximum kinetic energy is 4.374 Mev\n"
]
}
],
"prompt_number": 28
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 31.8 Page no 856"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Given\n",
"E1=0\n",
"E2=0.412*1.6*10**-13\n",
"E3=1.088*1.6*10**-13\n",
"h=6.62*10**-34\n",
"m1=197.968233\n",
"m2=197.966760\n",
"a=931.5\n",
"A=1.088\n",
"A1=0.412\n",
"\n",
"#Calculation\n",
"v1=(E3-E1)/h\n",
"v2=(E2-E1)/h\n",
"v3=(E3-E2)/h\n",
"Eb1=((m1-m2)*a)-A\n",
"Eb2=((m1-m2)*a)-A1\n",
"\n",
"#Result\n",
"print\"Radiation frequencies are\",round(v1*10**-20,2)*10**20,\",\",round(v2*10**-20,3)*10**20,\"and\",round(v3*10**-20,2)*10**20\n",
"print\"Maximum kinetic energies are\",round(Eb1,3),\"Mev and \",round(Eb2,3),\"Mev\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Radiation frequencies are 2.63e+20 , 9.96e+19 and 1.63e+20\n",
"Maximum kinetic energies are 0.284 Mev and 0.96 Mev\n"
]
}
],
"prompt_number": 42
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 31.9 Page no 857"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Given\n",
"a=15\n",
"b=9.0\n",
"T=5730\n",
"\n",
"#Calculation\n",
"import math\n",
"A=a/b\n",
"l=2.303*math.log10(A)\n",
"t=(l*T)/0.693\n",
"\n",
"#Result\n",
"print\"Approximation age of the indus valley civilization is\",round(t,2),\"years\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Approximation age of the indus valley civilization is 4224.47 years\n"
]
}
],
"prompt_number": 46
}
],
"metadata": {}
}
]
}