summaryrefslogtreecommitdiff
path: root/Nuclear_Physics/Chapter9.ipynb
blob: 7fa410066ca461c7a33eb71d5ec3d5095c3dd390 (plain)
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
288
289
290
291
292
293
294
295
296
297
298
299
300
301
302
303
304
305
306
307
308
309
310
311
312
313
314
315
316
317
318
319
320
321
322
323
324
325
326
327
328
329
330
331
332
333
334
335
336
337
338
339
340
341
342
343
344
345
346
347
348
349
350
351
352
353
354
355
356
357
358
359
360
361
362
363
364
365
366
367
368
369
370
371
372
373
374
375
376
377
378
379
380
381
382
383
384
385
386
387
388
389
390
391
392
393
394
395
396
397
398
399
400
401
402
403
404
405
406
407
408
409
410
411
412
413
414
415
416
417
418
419
420
421
422
423
424
425
426
427
428
429
430
431
432
433
434
435
436
437
438
439
440
441
442
443
444
445
446
447
448
449
450
451
452
453
454
455
456
457
458
459
460
461
462
463
464
465
466
467
468
469
470
471
472
473
474
475
476
477
478
479
480
481
482
483
484
485
486
487
488
489
490
491
492
493
494
495
496
497
498
499
500
501
502
503
504
505
506
507
508
509
510
511
512
513
514
515
516
517
518
519
520
521
522
523
524
525
526
527
528
529
530
531
532
533
534
535
536
{
 "metadata": {
  "name": "",
  "signature": "sha256:4f16fb24f1dd6d3239c94791f7a7ef08310ef6ebb56e580bd1901d7fe6b9c932"
 },
 "nbformat": 3,
 "nbformat_minor": 0,
 "worksheets": [
  {
   "cells": [
    {
     "cell_type": "heading",
     "level": 1,
     "metadata": {},
     "source": [
      "Chapter9-Nuclear Models"
     ]
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Ex1-pg389"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "## Exa9.1 : : Page-389 (2011) \n",
      "#find The Fermi energy of neutron  and proton\n",
      "import math\n",
      "h_cut = 1.054e-034;   ## Reduced Planck's constant, joule sec\n",
      "rho = 2e+044;    ## Density of the nuclear matter, kg per metre cube\n",
      "V = 238./rho;    ## Volume of the nuclear matter, metre cube\n",
      "## For neutron\n",
      "N = 238.-92.;    ## Number of neutrons\n",
      "M = 1.67482e-027;    ## Mass of a neutron, kg\n",
      "e = 1.602e-019;    ## Energy equivalent of 1 eV, J/eV\n",
      "E_f = (3*math.pi**2)**(2./3.)*h_cut**2/(2*M)*(N/V)**(2/3.)/e;    ## Fermi energy of neutron, eV \n",
      "print'%s %.2f %s'%(\"\\nThe Fermi energy of neutron = \",E_f/1e+006,\" MeV\")\n",
      "## For proton\n",
      "N = 92.;    ## Number of protons\n",
      "M = 1.67482e-027;    ## Mass of a proton, kg\n",
      "e = 1.602e-019;    ## Energy equivalent of 1 eV, J/eV\n",
      "E_f = (3*math.pi**2)**(2/3.)*h_cut**2/(2.*M)*(N/V)**(2./3.)/e;    ## Fermi energy of neutron, eV \n",
      "print'%s %.2f %s'%(\"\\nThe Fermi energy of proton = \",E_f/1e+006,\" MeV\");\n",
      "\n",
      "## Result\n",
      "## The Fermi energy of neutron = 48.92 MeV\n",
      "## The Fermi energy of proton = 35.96 MeV "
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "\n",
        "The Fermi energy of neutron =  48.92  MeV\n",
        "\n",
        "The Fermi energy of proton =  35.96  MeV\n"
       ]
      }
     ],
     "prompt_number": 1
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Ex3-pg390"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "## Exa9.3 : : Page-390 (2011)\n",
      "import math\n",
      "#find radius of neutron star\n",
      "h_cut = 1.0545e-34; ## Reduced Planck's constant, joule sec\n",
      "G = 6.6e-11;        ## Gravitational constant, newton square metre per square Kg \n",
      "m = 10**30.;        ## Mass of the star, Kg\n",
      "m_n = 1.67e-27;        ## Mass of the neutron, Kg\n",
      "R = (9*math.pi/4.)**(2./3.)*h_cut**2/(G*(m_n)**3)*(m_n/m)**(1/3.);        ## Radius of the neutron star, metre\n",
      "print'%s %.1e %s'%(\"\\nThe radius of the neutron star = \",R,\" metre\");\n",
      "\n",
      "## Result\n",
      "## The radius of the neutron star = 1.6e+004 metre "
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "\n",
        "The radius of the neutron star =  1.6e+04  metre\n"
       ]
      }
     ],
     "prompt_number": 2
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Ex4-pg391"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "## Exa9.4 : : Page-391 (2011)\n",
      "#find is they will stable or not\n",
      "import math\n",
      "A = 77.;        ## Mass number of the isotopes\n",
      "Z = round (A/((0.015*A**(2/3.))+2.));    ## Atomic number of stable isotope\n",
      "## Check the stability !!!!!\n",
      "if Z == 34:\n",
      "    print (\"\\n Se\",( Z,A),\" is stable\" and \"As \",(Z-1,A),\"\" and \"Br\",Z+1,A, \"are unstable\")\n",
      "elif Z == 33 :\n",
      "    print'%s %.2f %s'%(\"\\nAs( %d,%d) is stable \\nSe (%d,%d) and Br(%d,%d) are unstable\", Z, A, Z+1, A, Z+2, A);\n",
      "elif Z == 35 :\n",
      "    print'%s %.2f %s'%(\"\\nBr( %d,%d) is stable \\nSe (%d,%d) and As(%d,%d) are unstable\",Z,A,Z-2,A,Z-1,A);      \n",
      "\n",
      "\n",
      "## Result\n",
      "## Se( 34,77) is stable \n",
      "## As (33,77) and Br(35,77) are unstable "
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "('\\n Se', (34.0, 77.0), 'As ', (33.0, 77.0), '', 35.0, 77.0, 'are unstable')\n"
       ]
      }
     ],
     "prompt_number": 4
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Ex5-pg391"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "## Exa9.5 : : Page-391 (2011)\n",
      "#find The energy difference between neutron shells\n",
      "import math\n",
      "m_40 = 39.962589;         ## Mass of calcium 40, atomic mass unit\n",
      "m_41 = 40.962275;         ## Mass of calcium 41, atomic mass unit\n",
      "m_39 = 38.970691;          ## Mass of calcium 39, atomic mass unit \n",
      "m_n = 1.008665;            ## Mass of the neutron, atomic mass unit\n",
      "BE_1d = (m_39+m_n-m_40)*931.5;        ## Binding energy of 1d 3/2 neutron, mega electron volts\n",
      "BE_1f = (m_40+m_n-m_41)*931.5;        ## Binding energy of 1f 7/2 neutron, mega electron volts\n",
      "delta = BE_1d-BE_1f;        ## Energy difference between neutron shells, mega electron volts\n",
      "print'%s %.2f %s'%(\"\\nThe energy difference between neutron shells = \",delta,\" MeV\");\n",
      "\n",
      "## Result\n",
      "## The energy difference between neutron shells = 7.25 MeV "
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "\n",
        "The energy difference between neutron shells =  7.25  MeV\n"
       ]
      }
     ],
     "prompt_number": 4
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Ex7-pg392"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "## Exa9.7 : : Page-392 (2011)\n",
      "import math\n",
      "#find The angular frequency for oxygen 17\n",
      "h_cut = 1.0545e-34;       ## Reduced Planck's constant, joule sec\n",
      "R = 1.2e-15;            ## Distance of closest approach, metre\n",
      "m = 1.67482e-27;        ## Mass of the nucleon, Kg\n",
      "omega_Ni=1.60e+022;\n",
      "## For O-17\n",
      "for A in range(17,60):           ## Mass numbers\n",
      "    if A == 17:\n",
      "        omega_O = 5.*3.**(1/3.)*h_cut*17**(-1./3.)/(2.**(7/3.)*m*R**2.);    ## Angular frequency of oxygen \n",
      "## For Ni-60\n",
      "    elif A == 60:\n",
      "        omega_Ni =  5*3**(1/3.)*h_cut*60**(-1/3.)/(2**(7/3.)*m*R**2);  ## Angular frequency of nickel\n",
      "\n",
      "print (\"\\nThe angular frequency for oxygen 17 = \",omega_O,\"\" and \"\\nThe angular frequency for nickel 60 = \",omega_Ni,\"\");\n",
      "\n",
      "## Result\n",
      "## The angular frequency for oxygen 17 = 2.43e+022 \n",
      "## The angular frequency for nickel 60 = 1.60e+022 "
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "('\\nThe angular frequency for oxygen 17 = ', 2.43317537466611e+22, '', 1.6e+22, '')\n"
       ]
      }
     ],
     "prompt_number": 12
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Ex9-pg393"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "## Exa9.9 : : Page-393 (2011)\n",
      "#find The angular momentum is 5/2 and the parity is +1 for and -1 and 0\n",
      "import math\n",
      "import numpy\n",
      "Z = numpy.zeros((5,1));\n",
      "N = numpy.zeros((5,1));\n",
      "E={}\n",
      "## Elements allocated\n",
      "E[0,0] = 'Carbon'\n",
      "E[1,0] = 'Boron'\n",
      "E[2,0] = 'Oxygen'\n",
      "E[3,0] = 'Zinc'\n",
      "E[4,0] = 'Nitrogen'\n",
      "Z[0,0] = 6;        ## Number of proton in carbon nuclei\n",
      "Z[1,0] = 5;         ## Number of proton in boron nuclei\n",
      "Z[2,0] = 8;         ## Number of proton in oxygen nuclei\n",
      "Z[3,0] = 30;         ## Number of proton in zinc nuclei\n",
      "Z[4,0] = 7;         ## Number of proton in nitrogen nuclei\n",
      "N[0,0] = 6;        ## Mass number of carbon\n",
      "N[0,0] = 6;         ## Mass number of boron\n",
      "N[2,0] = 9;         ## Mass number of oxygen\n",
      "N[3,0] = 37;         ## Mass number of zinc\n",
      "N[4,0] = 9;         ## Mass number of nitrogem\n",
      "for i in range  (0,5):\n",
      "    if Z[i,0] == 8:\n",
      "            print(\"\\nThe angular momentum is 5/2 and the parity is +1 for  \", E[i,0],\"\");\n",
      "    elif Z[i,0] == 5:\n",
      "            print(\"\\nThe angular momentum is 3/2 and the parity is -1 for \", E[i,0],\"\");\n",
      "        \n",
      "    elif Z[i,0] == N[i,0]:\n",
      "        print (\"\\nThe angular mometum is 0 and the parity is +1 for \", E[i,0],\"\");\n",
      "    \n",
      "    elif N[i,0]-Z[i,0] == 2:\n",
      "        print(\"\\nThe angular momentum is 2 and the parity is -1 for \", E[i,0],\"\");\n",
      "    \n",
      "    elif N[i,0]-Z[i,0] == 7:\n",
      "        print(\"The angular momentum is 5/2 and the parity is -1 for\",  E[i,0],\"\");\n",
      "    \n",
      "\n",
      "## Result\n",
      "## The angular mometum is 0 and the parity is +1 for Carbon\n",
      "## The angular momentum is 3/2 and the parity is -1 for Boron\n",
      "## The angular momentum is 5/2 and the parity is +1 for Oxygen \n",
      "## The angular momentum is 5/2 and the parity is -1 for Zinc\n",
      "## The angular momentum is 2 and the parity is -1 for Nitrogen \n",
      "print(\"we can not print directly \")"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "('\\nThe angular mometum is 0 and the parity is +1 for ', 'Carbon', '')\n",
        "('\\nThe angular momentum is 3/2 and the parity is -1 for ', 'Boron', '')\n",
        "('\\nThe angular momentum is 5/2 and the parity is +1 for  ', 'Oxygen', '')\n",
        "('The angular momentum is 5/2 and the parity is -1 for', 'Zinc', '')\n",
        "('\\nThe angular momentum is 2 and the parity is -1 for ', 'Nitrogen', '')\n",
        "we can not print directly \n"
       ]
      }
     ],
     "prompt_number": 7
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Ex11-pg394"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "##  Page-394 (2011)\n",
      "import math\n",
      "import numpy\n",
      "\n",
      "R_0 = 1.2e-015;        ## Distance of closest approach, metre\n",
      "## Mass number of the nuclei are allocated below :\n",
      "N = numpy.zeros((4,1))\n",
      "N[0,0] = 17;        ## for oxygen\n",
      "N[1,0] = 33;        ## for sulphur\n",
      "N[2,0] = 63;        ## for copper\n",
      "N[3,0] = 209;        ## for bismuth\n",
      "for i in range (1,4):\n",
      "    if N[i,0] == 17:\n",
      "       print(\"\\n For Oxygen : \")\n",
      "       I = 5/2.;        ## Total angular momentum\n",
      "       l = 2.;        ## Orbital angular momentum\n",
      "       mu = -1.91;        ## for odd neutron and I = l+1/2\n",
      "       Q = -3./5.*(2.*I-1.)/(2.*I+2.)*(R_0*N[i,0]**(1/3.))**2*(10**28);    ## Quadrupole moment of oxygen, barnQ\n",
      "       print\"%s %.2f %s %.2f %s \"%(\"\\n         The value of magnetic moment is : \",mu,\"\"and \" \\n         The value of quadrupole moment is : \",Q,\" barn\");\n",
      "    elif N[i,0] == 33:\n",
      "        print(\"\\n\\n For Sulphur : \")\n",
      "        I = 3./2.;        ## Total angular momentum\n",
      "        l = 2.;            ## Orbital angular momentum\n",
      "        mu = 1.91*I/(I+1.);        ## for odd neutron and I = l-1/2\n",
      "        Q = -3./5.*(2.*I-1.)/(2.*I+2.)*(R_0*N[i,0]**(1/3.))**2*(10**28);    ## Quadrupole moment of sulphur, barn\n",
      "        print\"%s %.2f %s %.2f %s \"%(\"\\n         The value of magnetic moment is : \",mu,\"\"and \" \\n         The value of quadrupole moment is : \",Q,\" barn\");  \n",
      "    elif N[i,0] == 63:\n",
      "        print(\"\\n\\n For Copper : \")\n",
      "        I = 3./2.;        ## Total angular momentum\n",
      "        l = 1.;            ## Orbital angular momentum\n",
      "        mu = I+2.29;            ## for odd protons and I = l+1/2\n",
      "        Q = -3./5.*(2.*I-1.)/(2.*I+2.)*(R_0*N[i,0]**(1./3.))**2*(10**28);    ## Quadrupole momentum of copper, barn\n",
      "        print\"%s %.2f %s %.2f %s \"% (\"         The value of magnetic moment is : \",mu,\" \"and \"\\n         The value of quadrupole moment is :\" ,Q, \"barn\");\n",
      "    elif N[i,0] == 209:\n",
      "        print(\" For Bismuth : \")\n",
      "        I = 9/2;    ## Total angular momentum\n",
      "        l = 5;        ## Orbital angular momentum\n",
      "        mu = I-2.29*I/(I+1);     ## for odd protons and I = l-1/2\n",
      "        Q = -3./5.*(2.*I-1.)/(2.*I+2.)*(R_0*N[i,0]**(1/3.))**2*(10**28);    ## Quadrupole momentum of bismuth, barn\n",
      "        print\"%s %.2f %s %.2f %s \"%(\"       The value of magnetic moment is : \",mu,\"\"and \" \\n         The value of quadrupole moment is : \",Q,\" barn\");\n",
      "        print('due to rounding error we can not get for oxygen result')\n",
      "## Result\n",
      "\n",
      "\n",
      "## For Sulphur : \n",
      "##         The value of magnetic moment is : 1.146 \n",
      "##         The value of quadrupole moment is : -0.0356 barn\n",
      "\n",
      "## For Copper : \n",
      "##         The value of magnetic moment is : 3.79 \n",
      "##         The value of quadrupole moment is : -0.0547 barn\n",
      "\n",
      "## For Bismuth : \n",
      "##         The value of magnetic moment is : 2.63 \n",
      "##         The value of quadrupole moment is : -0.221 barn   \n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "\n",
        "\n",
        " For Sulphur : \n",
        "\n",
        "         The value of magnetic moment is :  1.15  -0.04  barn \n",
        "\n",
        "\n",
        " For Copper : \n",
        "         The value of magnetic moment is :  3.79 \n",
        "         The value of quadrupole moment is : -0.05 barn \n",
        " For Bismuth : \n",
        "       The value of magnetic moment is :  2.17  -0.21  barn \n",
        "due to rounding error we can not get for oxygen result\n"
       ]
      }
     ],
     "prompt_number": 12
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Ex12-pg395"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "## Exa9.12 : : Page-395 (2011)\n",
      "#find The kinetic energy of iron nuclei\n",
      "import math\n",
      "h_cut = 1.054571628e-34;    ## Redued planck's constant, joule sec\n",
      "a = 1e-014;            ## Distance of closest approach, metre\n",
      "m = 1.67e-27;        ## Mass of each nucleon, Kg\n",
      "KE = 14*math.pi**2*h_cut**2./(2.*m*a**2*1.6e-13);        ## Kinetic energy of iron nucleus, MeV\n",
      "print'%s %.2f %s'%(\"\\nThe kinetic energy of iron nuclei =\",KE,\" MeV\");\n",
      "\n",
      "## Result\n",
      "## The kinetic energy of iron nuclei = 28.76 MeV "
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "\n",
        "The kinetic energy of iron nuclei = 28.76  MeV\n"
       ]
      }
     ],
     "prompt_number": 7
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Ex14-pg396"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "## Exa9.14 : : Page-396 (2011)\n",
      "#find The electric quadrupole of scandium nucleus\n",
      "import math\n",
      "R_0 = 1.2e-15;  ## Distance of closest approach, metre\n",
      "j = 7/2.;        ## Total angular momentum\n",
      "A = 41.;         ## Mass number of Scandium\n",
      "Z = 20.;         ## Atomic number of Calcium\n",
      "Q_Sc = -(2*j-1)/(2.*j+2.)*(R_0*A**(1/3.))**2;       ## Electric quadrupole of Scandium nucleus, Sq. m\n",
      "Q_Ca = Z/(A-1)**2*abs(Q_Sc);        ## Electric quadrupole of calcium nucleus, Sq. m\n",
      "print'%s %.2e %s %.2e %s '%(\"\\nThe electric quadrupole of scandium nucleus = \",Q_Sc,\" square metre\" and  \"\\nThe electric quadrupole of calcium nucleus = \",Q_Ca,\" square metre\");\n",
      "\n",
      "## Result\n",
      "## The electric quadrupole of scandium nucleus = -1.14e-029 square metre \n",
      "## The electric quadrupole of calcium nucleus = 1.43e-031 square metre "
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "\n",
        "The electric quadrupole of scandium nucleus =  -1.14e-29 \n",
        "The electric quadrupole of calcium nucleus =  1.43e-31  square metre \n"
       ]
      }
     ],
     "prompt_number": 8
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Ex16-pg398"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "## Exa9.16 : : Page-398 (2011)\n",
      "#find The energy for 4+ tungsten state and 6+state\n",
      "import math\n",
      "h_cut_sqr_upon_2f = 0.01667;        ## A constant value, joule square per sec cube\n",
      "for I in range  (4,6):\n",
      "    if I == 4:\n",
      "        E = I*(I+1)*h_cut_sqr_upon_2f\n",
      "        print'%s %.2f %s'%(\"The energy for 4+ tungsten state = \",E,\" MeV\");\n",
      "    elif I == 6:\n",
      "        E = I*(I+1)*h_cut_sqr_upon_2f;\n",
      "        print'%s %.2f %s'%(\"\\nThe energy for 6+ tungsten state = \",E,\" MeV\"); \n",
      "    \n",
      "\n",
      "\n",
      "## Result\n",
      "## The energy for 4+ tungsten state = 0.333 MeV\n",
      "## The energy for 6+ tungsten state = 0.700 MeV  "
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The energy for 4+ tungsten state =  0.33  MeV\n"
       ]
      }
     ],
     "prompt_number": 11
    }
   ],
   "metadata": {}
  }
 ]
}