1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
106
107
108
109
110
111
112
113
114
115
116
117
118
119
120
121
122
123
124
125
126
127
128
129
130
131
132
133
134
135
136
137
138
139
140
141
142
143
144
145
146
147
148
149
150
151
152
153
154
155
156
157
158
159
160
161
162
163
164
165
166
167
168
169
170
171
172
173
174
175
176
177
178
179
180
181
182
183
184
185
186
187
188
189
190
191
192
193
194
195
196
197
198
199
200
201
202
203
204
205
206
207
208
209
210
211
212
213
214
215
216
217
218
219
220
221
222
223
224
225
226
227
228
229
230
231
232
233
234
235
236
237
238
239
240
241
242
243
244
245
246
247
248
249
250
251
252
253
254
255
256
257
258
259
260
261
262
263
264
265
266
267
268
269
270
271
272
273
274
275
276
277
278
279
280
281
282
283
284
285
286
287
|
{
"metadata": {
"name": "",
"signature": "sha256:0e9faaea32136a2f476b53b6ab2d5d2eb5330fb68ebd80aaad9bbe7ed1328c93"
},
"nbformat": 3,
"nbformat_minor": 0,
"worksheets": [
{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"12: Radioactivity"
]
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example number 12.1, Page number 351"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#importing modules\n",
"import math\n",
"from __future__ import division\n",
"\n",
"#Variable declaration\n",
"M7Li3=7.018232; #mass of 7li3(amu)\n",
"Malpha=4.003874; #mass of alpha particle(amu)\n",
"Mpr=1.008145; #mass of proton(amu)\n",
"Ey=9.15; #K.E energy of product nucleus\n",
"\n",
"#Calculation\n",
"#xMy -> x-mass no., M-element, y-atomic no.\n",
"#reaction:- 7li3 + 1H1-> 4He2 + 4He2\n",
"deltaM=M7Li3+Mpr-2*Malpha; #mass defect(amu)\n",
"Q=deltaM*931; #mass defect(MeV)\n",
"Ex=2*Ey-Q; #K.E of incident particle(MeV)\n",
"\n",
"#Result\n",
"print \"kinetic energy of incident proton is\",round(Ex,4),\"MeV\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"kinetic energy of incident proton is 0.9564 MeV\n"
]
}
],
"prompt_number": 2
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example number 12.2, Page number 351"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#importing modules\n",
"import math\n",
"from __future__ import division\n",
"\n",
"#Variable declaration\n",
"M235U=235; #atomic mass of 235U\n",
"m=10**-3; #mass of fissions(gm)\n",
"N=6.023*10**23; #avagadro number\n",
"Eperfi=200*10**6; #energy per fission(eV)\n",
"T=10**-6; #time(s)\n",
"\n",
"#Calculation\n",
"E=Eperfi*1.6*10**-19; #energy per fission(J)\n",
"A=M235U; \n",
"P=((m*N)/A)*(E/T); #power explosion(Watt)\n",
"\n",
"#Result\n",
"print \"power of explosion is\",P,\"Watt\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"power of explosion is 8.20153191489e+13 Watt\n"
]
}
],
"prompt_number": 3
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example number 12.4, Page number 352"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#importing modules\n",
"import math\n",
"from __future__ import division\n",
"\n",
"#Variable declaration\n",
"n=0.4; #efficiency\n",
"N=6.06*10**26; #avagadro number\n",
"Eperfi=200*10**6; #energy per fission(eV)\n",
"P=100*10**6; #electric power(W)\n",
"A=235;\n",
"\n",
"#Calculation\n",
"E=Eperfi*1.6*10**-19; #energy per fission(J)\n",
"T=24*60*60; #time(sec)\n",
"N235=P*T/(E*n); #number of atoms in 235 kg of U235\n",
"m=(A*N235)/N; #mass of 235U consumed/day(kg)\n",
"\n",
"#Result\n",
"print \"mass of 235U consumed/day is\",int(m*10**3),\"g\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"mass of 235U consumed/day is 261 g\n"
]
}
],
"prompt_number": 13
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example number 12.5, Page number 352"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#importing modules\n",
"import math\n",
"from __future__ import division\n",
"\n",
"#Variable declaration\n",
"M2H1=2.01474; #mass of M2H1(amu)\n",
"M3H1=3.01700; #mass of M3H1(amu)\n",
"M1n0=1.008986; #mass of M1n0(amu)\n",
"M4He2=4.003880; #mass of M4He2(amu)\n",
"\n",
"#Calculation\n",
"#thermonuclear reaction in hydrogen bomb explosion \n",
"#2H1 + 3H1 -> 4He2 + 1n0\n",
"Mreac=M2H1+M3H1; #mass of reactants(amu)\n",
"Mprod=M4He2+M1n0; #mass of products(amu)\n",
"Q=Mreac-Mprod; #amount of energy released per reaction(J)\n",
"Q=Q*931; #amount of energy released per reaction(MeV)\n",
"\n",
"#Result\n",
"print \"amount of energy released per reaction is\",round(Q,3),\"MeV\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"amount of energy released per reaction is 17.572 MeV\n"
]
}
],
"prompt_number": 15
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example number 12.6, Page number 353"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#importing modules\n",
"import math\n",
"from __future__ import division\n",
"\n",
"#Variable declaration\n",
"M7Li3=7.01818; #mass of Li atom(amu)\n",
"M1H1=1.0081; #mass of H atom(amu)\n",
"M1n0=1.009; #mass of neutron(amu)\n",
"\n",
"#Calculation\n",
"BEpernu=(1/7)*((3*M1H1)+(4*M1n0)-M7Li3); #binding energy per nucleon(J)\n",
"BEpernu=BEpernu*931; #binding energy per nucleon(MeV)\n",
"\n",
"#Result\n",
"print \"binding energy per nucleon is\",BEpernu,\"MeV\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"binding energy per nucleon is 5.60196 MeV\n"
]
}
],
"prompt_number": 16
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example number 12.7, Page number 353"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#importing modules\n",
"import math\n",
"from __future__ import division\n",
"\n",
"#Variable declaration\n",
"m=10*10**3; #mass of U235(gm)\n",
"N=6.02*10**23; #avagadro number\n",
"Eperfi=200*10**6; #energy per fission(eV)\n",
"A=235;\n",
"\n",
"#Calculation\n",
"E=Eperfi*1.6*10**-19; #energy(J)\n",
"T=24*60*60; #time(s)\n",
"P=((m*N)/A)*(E/T); #power output(Watt)\n",
"\n",
"#Result\n",
"print \"power output is\",round(P/10**9,3),\"*10**9 Watt\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"power output is 9.488 *10**9 Watt\n"
]
}
],
"prompt_number": 19
}
],
"metadata": {}
}
]
}
|
