5: Principles of Quantum Mechanics

Example number 5.1, Page number 85

In [2]:
#importing modules
import math
from __future__ import division

#Variable declaration
e=1.6*10**-19;    #charge(c)
m=9.1*10**-31;    #mass(kg)
h=6.626*10**-34;   #plank constant
E=2000;            #energy(eV)

#Calculation
lamda=h/math.sqrt(2*m*E*e);    #wavelength(m)

#Result
print "wavelength is",round(lamda*10**9,4),"nm"

wavelength is 0.0275 nm

Example number 5.2, Page number 85

In [3]:
#importing modules
import math
from __future__ import division

#Variable declaration
V=1600;      #potential energy of electron(V)

#Calculation
lamda=12.27/math.sqrt(V);      #wavelength(m)

#Result
print "wavelength is",lamda,"angstrom"
print "answer given in the book is wrong"

wavelength is 0.30675 angstrom
answer given in the book is wrong

Example number 5.3, Page number 85

In [4]:
#importing modules
import math
from __future__ import division

#Variable declaration
me=9.1*10**-31;    #mass(kg)
h=6.62*10**-34;   #plank constant
mn=1.676*10**-27;    #mass(kg)
c=3*10**8;     #velocity of light(m/s)

#Calculation
lamda=h*10**10/math.sqrt(4*mn*me*c**2);     #de broglie wavelength(angstrom)  

#Result
print "de broglie wavelength is",round(lamda*10**4,1),"*10**-4 angstrom"

de broglie wavelength is 2.8 *10**-4 angstrom

Example number 5.4, Page number 85

In [6]:
#importing modules
import math
from __future__ import division

#Variable declaration
a=2*10**-10;    #length(m)
n1=2;
n2=4;
m=9.1*10**-31;    #mass(kg)
e=1.6*10**-19;    #charge(c)
h=6.626*10**-34;   #plank constant

#Calculation
E2=n1**2*h/(8*m*e*a);      #energy of second state(eV)
E4=n2**2*h/(8*m*e*a);      #energy of fourth state(eV)

#Result
print "energy of second state is",round(E2*10**-26,5),"*10**26 eV"
print "energy of second state is",round(E4*10**-26,5),"*10**26 eV"

energy of second state is 0.11377 *10**26 eV
energy of second state is 0.45508 *10**26 eV

Example number 5.5, Page number 86

In [7]:
#importing modules
import math
from __future__ import division

#Variable declaration
V=344;    #accelerated voltage(V)
n=1;
theta=60;   #glancing angle(degrees)

#Calculation
theta=theta*math.pi/180;    #glancing angle(radian)
lamda=12.27/math.sqrt(V);
d=n*lamda/(2*math.sin(theta));    #spacing of crystal(angstrom)

#Result
print "spacing of crystal is",round(d,4),"angstrom"

spacing of crystal is 0.3819 angstrom

Example number 5.6, Page number 86

In [8]:
#importing modules
import math
from __future__ import division

#Variable declaration
lamda=1.66*10**-10;    #wavelength(m)
m=9.1*10**-32;    #mass(kg)
e=1.6*10**-19;    #charge(c)
h=6.626*10**-34;   #plank constant

#Calculation
E=h**2/(4*m*e*lamda**2);   #kinetic energy(eV)
v=h/(m*lamda);      #velocity(m/s)

#Result
print "kinetic energy is",round(E,2),"eV"
print "velocity is",round(v*10**-6,2),"*10**5 m/s"

kinetic energy is 273.57 eV
velocity is 43.86 *10**5 m/s

Example number 5.7, Page number 87

In [10]:
#importing modules
import math
from __future__ import division

#Variable declaration
a=1*10**-10;    #length(m)
n2=2;
n3=3;
m=9.1*10**-31;    #mass(kg)
e=1.6*10**-19;    #charge(c)
h=6.626*10**-34;   #plank constant

#Calculation
E1=h**2/(8*m*e*a**2);
E2=n2**2*E1;      #energy of 1st excited state(eV)
E3=n3**2*E1;      #energy of 2nd excited state(eV)

#Result
print "ground state energy is",round(E1,2),"eV"
print "energy of 1st excited state is",round(E2,2),"eV"
print "energy of 2nd excited state is",round(E3,2),"eV"
print "answer in the book varies due to rounding off errors"

ground state energy is 37.69 eV
energy of 1st excited state is 150.77 eV
energy of 2nd excited state is 339.23 eV
answer in the book varies due to rounding off errors

Example number 5.8, Page number 88

In [11]:
#importing modules
import math
from __future__ import division
from sympy import Symbol

#Variable declaration
n=Symbol('n');
a=4*10**-10;    #width of potential well(m)
m=9.1*10**-31;    #mass(kg)
e=1.6*10**-19;    #charge(c)
h=6.626*10**-34;   #plank constant

#Calculation
E1=n**2*h**2/(8*m*e*a**2);     #maximum energy(eV)

#Result
print "maximum energy is",round(E1/n**2,4),"*n**2 eV"

maximum energy is 2.3558 *n**2 eV

Example number 5.10, Page number 88

In [12]:
#importing modules
import math
from __future__ import division

#Variable declaration
delta_x=10**-8;     #length of box(m)
m=9.1*10**-31;    #mass(kg)
h=6.626*10**-34;   #plank constant

#Calculation
delta_v=h/(m*delta_x);     #uncertainity in velocity(m/s)

#Result
print "uncertainity in velocity is",round(delta_v/10**3,1),"km/s"

uncertainity in velocity is 72.8 km/s

Example number 5.11, Page number 89

In [13]:
#importing modules
import math
from __future__ import division

#Variable declaration
me=9.1*10**-31;    #mass(kg)
mp=1.6*10**-27;    #mass(kg)
h=6.626*10**-34;   #plank constant
c=3*10**10;    #velocity of light(m/s)

#Calculation
lamda=h/math.sqrt(2*mp*me*c**2);    #de broglie wavelength(m)

#Result
print "de broglie wavelength is",round(lamda*10**10*10**5,5),"*10**-5 angstrom"

de broglie wavelength is 0.40929 *10**-5 angstrom

Example number 5.12, Page number 89

In [14]:
#importing modules
import math
from __future__ import division

#Variable declaration
m=1.675*10**-27;    #mass(kg)
h=6.626*10**-34;   #plank constant
E=0.04;     #kinetic energy(eV)
e=1.6*10**-19;    #charge(c)
n=1;
d110=0.314*10**-9;   #spacing(m)

#Calculation
E=E*e;       #energy(J)
lamda=h/math.sqrt(2*m*E);
theta=math.asin(n*lamda/(2*d110));     #glancing angle(radian)
theta=theta*180/math.pi;        #glancing angle(degrees)
theta_m=60*(theta-int(theta));

#Result
print "glancing angle is",int(theta),"degrees",int(theta_m),"minutes"
print "answer given in the book is wrong"

glancing angle is 13 degrees 10 minutes
answer given in the book is wrong