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|
{
"metadata": {
"name": "",
"signature": "sha256:2b8beebd2e19262e31711f6c9b2785f35ddc221da552f5b3fe2ed984a495062e"
},
"nbformat": 3,
"nbformat_minor": 0,
"worksheets": [
{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"Chapter8:X-RAY"
]
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Ex8.1:pg-240"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"#to calculate value of planck's constant\n",
"e=1.6*10**-19 #in C\n",
"V=100*10**3 #voltage in KV\n",
"c=3*10**8 #light speed in m/s\n",
"lamdamin=12.35*10**-12 #wavelength in m\n",
"h=e*V*lamdamin/c\n",
"print \"the value of plancks constant is h=\",\"{:.2e}\".format(h),\"J-s\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"the value of plancks constant is h= 6.59e-34 J-s\n"
]
}
],
"prompt_number": 1
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Ex8.2:pg-240"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"#to calculate maximum frequency\n",
"h=6.6*10**-34 #planck's constant in J-s\n",
"c=3.0*10**8 #light speed in m/s\n",
"Ve=50000 #accelerating potential in V\n",
"lamdamin=h*c/Ve #wavelength in m\n",
"numax=c/lamdamin\n",
"print \"maximum frequency present in the radiation from an X-ray tube is numax=\",\"{:.2e}\".format(numax),\"Hz\"\n",
"#answer is given in thec book is incorrect =1.2*10**19 Hz\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"maximum frequency present in the radiation from an X-ray tube is numax= 7.58e+37 Hz\n"
]
}
],
"prompt_number": 2
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Ex8.3:pg-240"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"#to calculate number of electrons \n",
"I=2*10**-3 #current in mA\n",
"e=1.6*10**-19 \n",
"n=I/e\n",
"print \"number of electrons striking the target per second is n=\",n,\"unitless\"\n",
"#to calculate speed\n",
"m=9.1*10**-31 #mass of electron in kg\n",
"V=12.4*10**3 #potential difference in V\n",
"v=math.sqrt(2*V*e/m)\n",
"print \"the speed with which electrons strike the target is v=\",\"{:.1e}\".format(v),\"m/s\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"number of electrons striking the target per second is n= 1.25e+16 unitless\n",
"the speed with which electrons strike the target is v= 6.6e+07 m/s\n"
]
}
],
"prompt_number": 3
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Ex8.4:pg-240"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"#to calculate wavelength\n",
"n=2 #second order for longest wavelength\n",
"d=2.82*10**-10 # spacing in angstrom\n",
"sintheta=1 \n",
"lamdamax=2*d*sintheta/n\n",
"print \"the longest wavelength that can be analysed by a rock salt crystal is lamdamax=\",lamdamax,\"m\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"the longest wavelength that can be analysed by a rock salt crystal is lamdamax= 2.82e-10 m\n"
]
}
],
"prompt_number": 5
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Ex8.5:pg-241"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"#to calculate spacing of the crystal\n",
"h=6.62*10**-34 #planck's constant in J-s\n",
"m=9.1*10**-31 #mass of electron in kg\n",
"V=344 #voltage in V\n",
"e=1.6*10**-19\n",
"lamda=h/math.sqrt(2*m*e*V) #wavelength in m\n",
"#according to Bragg's law\n",
"n=1\n",
"#formula is 2*d*sintheta=n*lamda\n",
"d=n*lamda/(2*math.sin(math.pi/6))\n",
"print \"the spacing of the crystal is d=\",round(d/1e-10,2),\"angstrom\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"the spacing of the crystal is d= 0.66 angstrom\n"
]
}
],
"prompt_number": 5
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Ex8.6:pg-241"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"#to calculate wavelength of Kalpha line for an atom\n",
"R=1.1*10**5\n",
"z=92\n",
"#Ka line is emitted when electron jumps from l shell(n2=2) to k shell(n1=1)\n",
"#formula is 1/alphaa=R*(z-b)*((1/n1**2)-(1/n2)**2)\n",
"alphaa=4/(3*R*(z-1)**2)\n",
"print \"wavelength of Kalpha line for an atom is alphaa=\",\"{:.3e}\".format(alphaa),\"cm\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"wavelength of Kalpha line for an atom is alphaa= 1.464e-09 cm\n"
]
}
],
"prompt_number": 7
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Ex8.7:pg-241"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"#to calculate thickness\n",
"#mass absorption coefficient mum of an absorber is related with linear absorption coefficient mu and density of the material rho is given by\n",
"#mu=rho*mum=2.7*0.6=1.62 cm**-1\n",
"mu=1.62\n",
"#if initial intensity Io of the X-ray beam is reduced to I in traversing a distance x in absorber I=Io*e**-mu*x\n",
"#where I/Io=20\n",
"#put above values in the below equation , we get\n",
"x=(2.3026*(math.log(20)/math.log(10)))/1.62\n",
"print \"thickness is x=\",round(x,2),\"cm\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"thickness is x= 1.85 cm\n"
]
}
],
"prompt_number": 8
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Ex8.8:pg-242"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"#to calculate atomic number of the element\n",
"#equation for balmer series in hydrogen spectrum is 1/lamda=R*((1/2**2)-(1/n**2))\n",
"#for series limit n=infinity ,R=4/lamdainfinity i.e. R=4/364.6nm\n",
"#X-ray wavelength of K series is 1/lamda=R*(z-1)**2*((1/1**2)-(1/n**2))\n",
"lamda=0.1*10**-9\n",
"R=4/(364.6*10**-9)\n",
"#for n=infinity ,minimum wavelength of k series is given by\n",
"z=math.sqrt(1/(lamda*R))+1\n",
"print \"atomic number is z=\",int(z),\"unitless\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"atomic number is z= 31 unitless\n"
]
}
],
"prompt_number": 9
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Ex8.9:pg-242"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"#to calculate wavelength\n",
"d=1.87*10**-10 #spacing in angstrom\n",
"n=2 \n",
"#formula is lamda=2*d*sintheta/n\n",
"lamda=2*d*math.sin(math.pi/6)/n\n",
"print \"the wavelength of X-rays is lamda=\",round(lamda/1e-10,3),\"angstrom\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"the wavelength of X-rays is lamda= 0.935 angstrom\n"
]
}
],
"prompt_number": 11
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Ex8.10:pg-242"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"#to calculate wavelength of second X-ray beam\n",
"#from bragg's law\n",
"#lamda=(d*math.sin(math.pi/3))/n eq(1)\n",
"#it is given that,theta=60,n=3,lamda=1.97 angstrom\n",
"#from eq(1) we get,2*d*sin60degree=3*0.97 eq(2)\n",
"#let lamda' be the second X-ray beam \n",
"#we get 2*d'*sin theta'=n'*lamda' eq(3)\n",
"#from eq(2) and eq(3),we get\n",
"lamda1=math.sin(math.pi/6)*3*0.97/math.sin(math.pi/3) #where lamda1=lamda'\n",
"print \"wavelength of X-ray is lamda1=\",round(lamda1,2),\"angstrom\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"wavelength of X-ray is lamda1= 1.68 angstrom\n"
]
}
],
"prompt_number": 12
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Ex8.11:pg-243"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"#to calculate wavelength\n",
"d=2.82*10**-10 #spacing in m\n",
"n=1 \n",
"lamda=2*d*math.sin(10*math.pi/180)/n\n",
"print \"wavelength of X-ray is lamda=\",round(lamda/1e-10,3),\"angstrom\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"wavelength of X-ray is lamda= 0.979 angstrom\n"
]
}
],
"prompt_number": 13
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Ex8.12:pg-243"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"#deduce possible spacing of the set of planes\n",
"#for first order , 2*d*sintheta1=1*lamda...eq(1)\n",
"#for second order ,2*d*sintheta2=2*lamda..eq(2)\n",
"#for third order, 2*d*sintheta3=3*lamda......eq(3)\n",
"#for fourth order, 2*d*sintheta4=4*lamda..............eq(4)\n",
"#divide eq(2) by eq(1),we get sintheta2=2*sintheta1\n",
"#similarly,sintheta3=3*sintheta1,sintheta4=4*sintheta1\n",
"lamda=1.32*10**-10\n",
"sintheta1=0.1650\n",
"d1=lamda/(2*sintheta1)#for first order n=1,d1=d/n\n",
"d2=lamda/(2*2*sintheta1) #for second order n=2,d2=d/n\n",
"d3=lamda/(2*3*sintheta1) #for third order n=3,d3=d/n\n",
"d4=lamda/(2*4*sintheta1) #for fourth order n=4,d4=d/n\n",
"print \"d1=\",d1,\"m\"\n",
"print \"d2=\",d2,\"m\"\n",
"print \"d3=\",round(d3,2),\"m\"\n",
"print \"d4=\",d4,\"m\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"d1= 4e-10 m\n",
"d2= 2e-10 m\n",
"d3= 0.0 m\n",
"d4= 1e-10 m\n"
]
}
],
"prompt_number": 14
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Ex8.13:pg-248"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"#to calculate compton shift and wavelength\n",
"h=6.63*10**-34 #planck's constant in J-s\n",
"m0=9.11*10**-31 #mass of electron\n",
"c=3*10**8 #light speed in m/s\n",
"dellamda=h*(1-(1/math.sqrt(2)))/(m0*c)\n",
"lamda0=2*10**-10\n",
"lamda=dellamda+lamda0\n",
"print \"compton shift is dellamda=\",round(dellamda/1e-10,4),\"angstrom\"\n",
"print \"wavelength of the scattered X-rays is lamda=\",round(lamda/1e-10,4),\"angstrom\"\n",
"#to calculate fraction of energy lost by the photon in the collision\n",
"#energy lost =E0-E/E0=(hc/lamda0)-(hc/lamda)/(ha/lamda0)\n",
"#we get,\n",
"energylost=dellamda/lamda\n",
"print \"energylost =\",round(energylost,5),\"unitless\" \n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"compton shift is dellamda= 0.0071 angstrom\n",
"wavelength of the scattered X-rays is lamda= 2.0071 angstrom\n",
"energylost = 0.00354 unitless\n"
]
}
],
"prompt_number": 18
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Ex8.14:pg-249"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"#to calculate wavelength and energy\n",
"#formula is lamda'-lamda=h*(1-cos phi)/(m0*c),where phi=90 degree, lamda'=2lamda ---------------eq(1)\n",
"#dellamda=2lamda-lamda=lamda ----------------------------eq(2)\n",
"h=6.62*10**-34 #planck's constant\n",
"c=3*10**8 #light speed in m.s\n",
"m0=9*10**-31 #mass of electron in kg\n",
"#from eq(1) and eq(2) ,we get\n",
"lamda=h/(m0*c)\n",
"print \"wavelength is lamda=\",round(lamda/1e-10,4),\"angstrom\"\n",
"E=h*c/lamda\n",
"print \"energy of the incident photon is E=\",E,\"J\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"wavelength is lamda= 0.0245 angstrom\n",
"energy of the incident photon is E= 8.1e-14 J\n"
]
}
],
"prompt_number": 19
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Ex8.15:pg-249"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"#to calculate wavelength of radiation and direction of emission\n",
"h=6.6*10**-34 #planck's constant in J-s\n",
"c=3*10**8 #speed of light in m/s\n",
"energy=510*10**3 #energy of photon in eV\n",
"lamda=h*c/(energy*1.6*10**-19)\n",
"mo=9.1*10**-31 #mass of electron in Kg\n",
"lamda1=lamda+h*(1-math.cos(math.pi/2))/(mo*c)\n",
"print \"wavelength of radiation is lamda1=\",\"{:.2e}\".format(lamda1),\"m\"\n",
"theta=math.degrees(math.atan(lamda*math.sin(math.pi/2)/(lamda1-lamda*math.cos(math.pi/2))))\n",
"print\"direction of emission of electron is theta=\",round(theta,2),\"degree\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"wavelength of radiation is lamda1= 4.84e-12 m\n",
"direction of emission of electron is theta= 26.61 degree\n"
]
}
],
"prompt_number": 21
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Ex8.16:pg-249"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"#to calculate wavelength of two X-rays\n",
"h=6.6*10**-34 #planck's constant in J-s\n",
"c=3.0*10.0**8 #light speed in m/s\n",
"mo=9.1*10**-31 #mass of electron in kg\n",
"lamda=10.0*10**-12 #wavelength in pm\n",
"lamda1=lamda+((h/(mo*c))*(1-(1/math.sqrt(2))))\n",
"print \"wavelength of two X-rays is lamda1=\",round(lamda1*(1e12),1),\"picometer\"\n",
"#to calculate maximum wavelength\n",
"lamda2=lamda+((2*h)/(mo*c))\n",
"print \"maximum wavelength present in the scattered X-rays is lamda2=\",round(lamda2*(1e12),2),\"picometer\"\n",
"#to calculate maximum kinetic energy \n",
"Kmax=(h*c)*((1/lamda)-(1/lamda2))/(1.6*10**-19)\n",
"print \"maximum kinetic energy of the recoil electrons is Kmax=\",round(Kmax/1000.0,1),\"KeV\"\n",
"\n",
"# the answer is slightly different due to approximation\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"wavelength of two X-rays is lamda1= 10.7 picometer\n",
"maximum wavelength present in the scattered X-rays is lamda2= 14.84 picometer\n",
"maximum kinetic energy of the recoil electrons is Kmax= 40.3 KeV\n"
]
}
],
"prompt_number": 6
}
],
"metadata": {}
}
]
}
|
