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-rwxr-xr-xModern_Physics/Chapter13.ipynb430
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diff --git a/Modern_Physics/Chapter13.ipynb b/Modern_Physics/Chapter13.ipynb
index e35d94ac..5e2693aa 100755
--- a/Modern_Physics/Chapter13.ipynb
+++ b/Modern_Physics/Chapter13.ipynb
@@ -1,7 +1,7 @@
{
"metadata": {
- "name": "",
- "signature": "sha256:8b9d9c8500ecb20deeeb6f065d1408d228b300a0b95f2800ee645e0184760d45"
+ "name": "Chapter13",
+ "signature": "sha256:3612dc1ad0c2617d4c1e1da21ea7791334391562bfe93ddc9d0f98037d4fe15d"
},
"nbformat": 3,
"nbformat_minor": 0,
@@ -13,7 +13,7 @@
"level": 1,
"metadata": {},
"source": [
- "Chapter 13: Nuclear Structure "
+ "Chapter 13:Nuclear Reaction and Application"
]
},
{
@@ -21,111 +21,27 @@
"level": 2,
"metadata": {},
"source": [
- "Example 13.1, page no. 466"
+ "Example 13.1, Page 417"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
+ "#initiation of variable\n",
+ "v=1*1.0*10**-6.0*10**2; p=7.9; m=p*v;Na=6.023*10**23 #given values and various constants in suitable units\n",
+ "M=56.0;N=m*Na/M; #number of atoms\n",
+ "i=3.0*10**-6;\n",
+ "q=1.6*10**-19;\n",
"\n",
+ "#calculation\n",
+ "Io=i/q; #intensity\n",
+ "s=0.6*10**-24;S=1; #given values in suitable units\n",
+ "R=N*s*Io/S; #rate of neutrons\n",
"\n",
- "#Variable Declaration\n",
- "\n",
- "NA = 6.02 * 10 ** 23 # Avogadro's number\n",
- "m = 0.012 # one mole of carbon (kg)\n",
- "mC = 12 # mass of one atom(1u)\n",
- " \n",
- "#Calculation\n",
- "\n",
- "ma = m/NA\n",
- "u = ma / mC\n",
- "\n",
- "#Results\n",
- "\n",
- "print \"The atomic mass unit is \",round(u/10**-27,2),\"X 10^-27 kg.\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The atomic mass unit is 1.66 X 10^-27 kg.\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 13.2, page no. 468"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "\n",
- "import math\n",
- "\n",
- "#Variable Declaration\n",
- "\n",
- "m = 1.67 * 10 ** -27 # mass of neutron(kg)\n",
- "r0 = 1.2 * 10 ** -15 # radius os the nucleus(m)\n",
- "\n",
- "#Calculation \n",
- "\n",
- "pn = 3* m /(4 * math.pi * r0**3)\n",
- "\n",
- "#Results\n",
- "\n",
- "print \"The nuclear density is\",round(pn/10**17,1),\"X 10^17 kg/m^3.\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The nuclear density is 2.3 X 10^17 kg/m^3.\n"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 13.3, page no. 473"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- " \n",
- "\n",
- "#Variable Declaration\n",
- "\n",
- "MH = 1.007825 #mass of hydrogen (u)\n",
- "mn = 1.008665 #mass of neutron (u)\n",
- "M2 = 2.014102 #mass of deuteron (u)\n",
- "\n",
- "#Calculation\n",
- "\n",
- "Eb = (MH + mn - M2) * 931.494\n",
- "\n",
- "#results\n",
- "\n",
- "print \"The binding energy of deuteron is\",round(Eb,3),\"MeV.\""
+ "#result\n",
+ "print\"The rate of neutrons emitted from the target in particles per second is %.1e\" %round(R,3);\n",
+ "print\"Slight difference in answer due to approximation error\""
],
"language": "python",
"metadata": {},
@@ -134,41 +50,38 @@
"output_type": "stream",
"stream": "stdout",
"text": [
- "The binding energy of deuteron is 2.224 MeV.\n"
+ "The rate of neutrons emitted from the target in particles per second is 9.6e+07\n",
+ "slight difference due to approximation error\n"
]
}
],
- "prompt_number": 5
+ "prompt_number": 4
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
- "Example 13.4, page no. 482"
+ "Example 13.2, Page 419"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
- "\n",
- "#Variable declaration\n",
- "\n",
- "Thalf = 5730 #Half-life (yr)\n",
- "N = 1000 #No of carbon nuclei\n",
- "T = 22920 #(yr)\n",
+ "#initiation of variable\n",
+ "A=197.0; m=30*10**-3;phi=3.0*10**12; #given values and various constants taken in suitable units\n",
+ "Ar=99.0*10**-24; Na=6.023*10**23\n",
"\n",
"#calculation\n",
- "n=N\n",
- "t = 0\n",
- "while(t!=T):\n",
- " n = n/2\n",
- " t = t + Thalf\n",
+ "R=(phi*Na*Ar*m/A); #rate or production of gold\n",
+ "t=2.7*24*60 # time of decay\n",
+ "Act=R*(0.693/t); #activity /sec\n",
+ "ActCi=Act/(3.7*10**4); # in terms of curie(Ci)\n",
"\n",
"#result\n",
- "\n",
- "print \"The number of Carbon nuclei left after 22,920 yr is\",n"
+ "print\"The activity is found out to be %.1e\" %round(Act,3),\"/sec i.e \" ,round(ActCi,3),\"muCi\"\n",
+ "print\"Slight difference in answer due to approximation error\""
],
"language": "python",
"metadata": {},
@@ -177,7 +90,8 @@
"output_type": "stream",
"stream": "stdout",
"text": [
- "The number of Carbon nuclei left after 22,920 yr is 62\n"
+ "The activity is found out to be 4.9e+06 /sec i.e 131.229 muCi\n",
+ "slight difference due to approximation error\n"
]
}
],
@@ -188,162 +102,37 @@
"level": 2,
"metadata": {},
"source": [
- "Example 13.5, page no. 483"
+ "Example 13.3, Page 423"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
+ "#initiation of variable\n",
+ "from math import exp\n",
+ "v=1.5*1.5*2.5*(10**-6)*10**2; #volume in cm3\n",
+ "p=8.9; #density in g/cm3\n",
+ "m=p*v;Na=6.023*10**23 #mass and Avagadro's number\n",
+ "M=58.9; #Given values\n",
"\n",
- "\n",
- "import math\n",
- "\n",
- "#Variable declaration\n",
- "\n",
- "Thalf = 1.6 * 10 ** 3 #Half-life (yr)\n",
- "s = 3.16 * 10 ** 7 #number of seconds in a year (s/yr)\n",
- "\n",
- "#Calculation\n",
- "\n",
- "lamda = 0.693 / (Thalf * s)\n",
+ "#calculation\n",
+ "N=m*Na/M;\n",
+ "i=12*10**-6; #thickness of beam\n",
+ "q=1.6*10**-19;\n",
+ "Io=i/(2*q); #intensity\n",
+ "s=0.64*10**-24; #Given values\n",
+ "S=1.5*1.5;\n",
+ "R=N*s*Io/S; #rate of production of 61Cu\n",
"\n",
"#result\n",
+ "print \"The rate of neutrons emitted from the target in particles/second is %.1e\" %round(R,3);\n",
"\n",
- "print \"(a) The decay constant is\",round(lamda/10**-11,1),\"X 10^-11 s^-1.\"\n",
- "\n",
- "\n",
- "#Variable Declaration\n",
- "\n",
- "N0 = 3.0 * 10 ** 16 #number of radioactive nuclei at t=0\n",
- "Ci = 3.7 * 10 **10 \n",
- "\n",
- "#Calculation\n",
- "\n",
- "R0 = lamda * N0\n",
- "\n",
- "#results\n",
- "\n",
- "print \"(b) Its activity is\",round(R0/Ci/10**-6,1),\"X 10^-6 Ci.\"\n",
- "\n",
- "\n",
- "#Variable Declaration\n",
- "\n",
- "T = 2.0 * 10 ** 3 #(yr)\n",
- "\n",
- "#Calculation\n",
- "\n",
- "R = R0 * math.exp(-lamda*(T * s))\n",
- "\n",
- "#results\n",
- "\n",
- "print \"(c) The decay rate after 2.0 X 10^3 yr is\",round(R/10**5,1),\"X 10^5 decays/s\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(a) The decay constant is 1.4 X 10^-11 s^-1.\n",
- "(b) Its activity is 11.1 X 10^-6 Ci.\n",
- "(c) The decay rate after 2.0 X 10^3 yr is 1.7 X 10^5 decays/s\n"
- ]
- }
- ],
- "prompt_number": 10
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 13.6, page no. 483"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "\n",
- "\n",
- "import math\n",
- "\n",
- "#Variable Declaration\n",
- "\n",
- "A = 6.02 * 10 ** 23 #avogadro's number\n",
- "m = 3.50 * 10 ** -6 # mass of carbon(g)\n",
- "ma = 11.0 #atomic mass of carbon (g)\n",
- "\n",
- "#Calculation\n",
- "\n",
- "N = m * A / ma\n",
- "\n",
- "#Result\n",
- "\n",
- "print \"(a) The number of nuclei samples at t=0 is\",round(N/10**17,2),\"X 10^17 nuclei.\"\n",
- "\n",
- "\n",
- "\n",
- "#Variable declaration\n",
- "\n",
- "Thalf = 20.4 * 60 #half-life (s)\n",
- "T = 8.00 #(h)\n",
- "\n",
- "#Calculation\n",
- "\n",
- "lamda = 0.693 / Thalf\n",
- "R0 = lamda * N\n",
- "R = R0 * math.exp(-lamda* T*60*60)\n",
+ "#part b\n",
+ "act=R*(1-(exp((0.693)*(-2/3.41)))); #activity\n",
"\n",
"#result\n",
- "\n",
- "print \"(b) The activity initially is\",round(R0/10**14,2),\"X 10^14 decay/s and after t=8.0h is\",round(R/10**6,2),\"X 10^6 decay/s\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(a) The number of nuclei samples at t=0 is 1.92 X 10^17 nuclei.\n",
- "(b) The activity initially is 1.08 X 10^14 decay/s and after t=8.0h is 8.99 X 10^6 decay/s\n"
- ]
- }
- ],
- "prompt_number": 12
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 13.7, page no. 484"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "\n",
- "\n",
- "import math\n",
- "\n",
- "#Variable Declaration\n",
- "\n",
- "Thalf = 8.04 # half-life (days)\n",
- "R0 = 5.0 #Activity at t=0 (mCi)\n",
- "R = 4.2 #Activity (mCi)\n",
- "\n",
- "#Calculation\n",
- "\n",
- "t = - (Thalf/0.693)* math.log(R/R0)\n",
- "\n",
- "#Result\n",
- "\n",
- "print \"The time elapsed is\",round(t,2),\"days.\""
+ "print\"The activity after 2.0h in /sec is %.1e\" %round(act,3),\"=4.9mCi\";\n"
],
"language": "python",
"metadata": {},
@@ -352,83 +141,41 @@
"output_type": "stream",
"stream": "stdout",
"text": [
- "The time elapsed is 2.02 days.\n"
+ "The rate of neutrons emitted from the target in particles/second is 5.5e+08\n",
+ "The activity after 2.0h in /sec is 1.8e+08 =4.9mCi\n"
]
}
],
- "prompt_number": 16
+ "prompt_number": 6
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
- "Example 13.8, page no. 486"
+ "Example 13.4, Page 425"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
- " \n",
- "\n",
- "#Variable Declaration\n",
- "\n",
- "Mx = 226.025406 #atomic mass of 226Ra(u)\n",
- "My = 222.017574 #atomic mass of 222Rn(u)\n",
- "Ma = 4.002603 #atomic mass of 4He (u)\n",
+ "#initiation of variable\n",
+ "m2H=2.014102; #mass of various particles\n",
+ "mn=1.008665;m63Cu=62.929599;\n",
+ "m64Zn=63.929145;c2=931.5; #c^2=931.5 MeV\n",
+ "Q=(m2H+m63Cu-mn-m64Zn)*c2; #Q of the reaction\n",
"\n",
- "#Calculation\n",
- "\n",
- "Q = (Mx - My - Ma) * 931.494\n",
- "\n",
- "#Result\n",
- "\n",
- "print \"Q value is\",round(Q,2),\"MeV.\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Q value is 4.87 MeV.\n"
- ]
- }
- ],
- "prompt_number": 18
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 13.9, page no. 487"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "\n",
- "import math\n",
- "#Variable declaration\n",
- "\n",
- "r0 = 7.25 * 10**-5 #Bohr radius for alpha particle(A')\n",
- "Z = 86 #daughter nucleus radon's atomic number\n",
- "A = 222 #radon's mass number\n",
- "ke2 = 14.40 #boltzmann constant X (charge of electron)^2 (eV.A')\n",
- "E = 5 #disintegration energy (MeV)\n",
+ "#result\n",
+ "print\"The value of Q is in MeV\",round(Q,3);\n",
"\n",
- "#Calculation\n",
"\n",
- "E0 = ke2 / (2.0*r0) * 10**-6 #Energy unit analogous to Rydberg (MeV)\n",
- "R = (1.2 * 10**-5) *(A)**(1.0/3.0) #radius of Radon nucleus(A')\n",
- "Te = math.exp(round(-4*math.pi*Z*math.sqrt(E0/E)+8*math.sqrt(Z*R/r0)))\n",
+ "#part b\n",
+ "Kx=12.00;Ky=16.85;\n",
+ "Ky=Q+Kx-Ky #kinetic energy of 64Zn\n",
"\n",
"#result\n",
- "print \"The probability of escape of alpha particle is\",round(Te/10**-34,2),\"X 10^-34\""
+ "print\"The value of Ky was found out to be in MeV\",round(Ky,3);"
],
"language": "python",
"metadata": {},
@@ -437,49 +184,40 @@
"output_type": "stream",
"stream": "stdout",
"text": [
- "The probability of escape of alpha particle is 1.33 X 10^-34\n"
+ "The value of Q is in MeV 5.487\n",
+ "The value of Ky was found out to be in MeV 0.637\n"
]
}
],
- "prompt_number": 21
+ "prompt_number": 5
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
- "Example 13.11, page no. 490"
+ "Example 13.5, Page 425"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
+ "#initiation of variable\n",
+ "mp=1.007825;m3H=3.016049; #mass of the particle\n",
+ "m2H=2.014102;c2=931.5; #constant\n",
+ "Q=(mp+m3H-(2*m2H))*c2; #Q of the reaction\n",
"\n",
+ "#result\n",
+ "print\"The value of q was found out to be in MeV\",round(Q,3);\n",
"\n",
- "import math\n",
- "\n",
- "#Variable Declaration\n",
- "\n",
- "Thalf = 5730 # half-life (yr)\n",
- "s = 3.16 * 10 ** 7 # (s/yr)\n",
- "A = 6.02 * 10 ** 23 # Avogadro's number\n",
- "mw = 12.0 # molar weight of carbon(g)\n",
- "m = 25.0 # mass of carbon(g) \n",
- "r = 1.3 * 10 ** -12 # ratio of 14C to 12C\n",
- "R = 250 # activity observed (decays/min)\n",
- "\n",
- "#Calculation\n",
- "\n",
- "lamda = 0.693 /(Thalf * s)\n",
- "N1 = A * m/mw\n",
- "N0 = r * N1\n",
- "R0 = N0 * lamda\n",
- "t = -(1/lamda)*math.log(R/(R0*60))\n",
- " \n",
- "#Result\n",
+ "#partb\n",
+ "Kth1= -Q*(1+(mp/m3H)); #threshold energy of kinetic energy\n",
+ "Kth2=-Q*(1+(m3H/mp)); #threshold kinetic energy in case2\n",
"\n",
- "print \"The tree has been dead for\",round(round(t/10**11)*10**11/s/10**3,1),\"X 10^3 yr.\""
+ "#result\n",
+ "print\"The threshold kinetic energy in case-1 in MeV\",round(Kth1,3);\n",
+ "print\"The threshold kinetic energy in case-2 in MeV\",round(Kth2,3);"
],
"language": "python",
"metadata": {},
@@ -488,11 +226,13 @@
"output_type": "stream",
"stream": "stdout",
"text": [
- "The tree has been dead for 3.2 X 10^3 yr.\n"
+ "The value of q was found out to be in MeV -4.033\n",
+ "The threshold kinetic energy in case-1 in MeV 5.381\n",
+ "The threshold kinetic energy in case-2 in MeV 16.104\n"
]
}
],
- "prompt_number": 25
+ "prompt_number": 6
}
],
"metadata": {}
generated by cgit (git 2.34.1) at 2024-12-20 21:58:58 +0000