10: Quantum Mechanics

Example number 10.1, Page number 196

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

#Variable declaration
v=10**7;                  #speed of electron(m/s)
h=6.626*10**-34;          #plancks constant
m=9.1*10**-31;            #mass of electron(kg)

#Calculation                        
lamda=h/(m*v);           #de Broglie wavelength(m)

#Result
print "The de Broglie wavelength is",round(lamda*10**11,2),"*10**-11 m"

The de Broglie wavelength is 7.28 *10**-11 m

Example number 10.2, Page number 196

In [5]:
#importing modules
import math
from __future__ import division

#Variable declaration
h=6.626*10**-34;                     #plancks constant
lamda=0.3;                          #de Broglie wavelength(nm)
#For electron
me=9.1*10**-31;                      #mass of electron(kg)
#For proton
mp=1.672*10**-27;                    #mass of proton(kg)

#Calculation                        
p=h/(lamda*10**-9);                 #uncertainity in determining momentum(kg m/s)
K1=p**2/(2*me);                      #kinetic energy of electron(J)
K2=p**2/(2*mp);                      #kinetic energy of proton(J)

#Result
print "The kinetic energy of electron is",round(K1*10**18,1),"*10**-18 J"
print "The kinetic energy of proton is",round(K2*10**21,2),"*10**-21 J"

The kinetic energy of electron is 2.7 *10**-18 J
The kinetic energy of proton is 1.46 *10**-21 J

Example number 10.3, Page number 196

In [9]:
#importing modules
import math
from __future__ import division

#Variable declaration
#K=p^2/(lambda^2*2*m) where K is kinetic energy
h=6.626*10**-34;                  #plancks constant
lamda=10**-14;                   #de Broglie wavelength(m)
m=9.1*10**-31;                    #mass of electron(kg)
e=1.6*10**-19;

#Calculation                        
K=(h**2/((lamda**2)*2*m*e))*10**-9;      

#Result
print "The kinetic energy is",int(K),"GeV"
print "It is not possible to confine the electron to a nucleus."

The kinetic energy is 15 GeV
It is not possible to confine the electron to a nucleus.

Example number 10.4, Page number 197

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

#Variable declaration
m=9.1*10**-31;                #mass of electron(kg)
v=6*10**3;                    #speed of electron(m/s)
h=6.626*10**-34;              #plancks constant
a=0.00005; 

#Calculation                        
p=m*v;                        #uncertainity in momentum(kg m/s)
deltap=a*p;             #uncertainity in p
deltax=(h/(4*math.pi*deltap))*10**3       #uncertainity in position(mm)

#Result
print "The uncertainity in position is",round(deltax,3),"mm"

The uncertainity in position is 0.193 mm

Example number 10.5, Page number 197

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

#Variable declaration
L=3*10**-5;                   #diameter of the sphere(nm)
h=6.626*10**-34;              #plancks constant
m=1.67*10**-27;               #mass of the particle(kg)
n=1;
e=1.6*10**-19;

#Calculation                        
E1=((h**2)*(n**2))/(8*m*(L**2)*e)*10**12        #first energy level(MeV)
E2=E1*2**2;                                     #second energy level(MeV)

#Result
print "The first energy level is",round(E1,3),"MeV"
print "The second energy level is",round(E2,4),"MeV"

The first energy level is 0.228 MeV
The second energy level is 0.9128 MeV

Example number 10.6, Page number 197

In [16]:
#importing modules
import math
from __future__ import division

#Variable declaration
h=6.626*10**-34;               #plancks constant
a=2*10**12;                    #angular frequency(rad/s)
e=1.6*10**-19;

#Calculation                        
E0=(0.5*(h/(2*math.pi*e))*a)*10**3;   #ground state energy(MeV)
E1=(1.5*(h/(2*math.pi*e))*a)*10**3;   #first excited state energy(MeV)

#Result
print "The ground state energy is",round(E0,3),"MeV" 
print "The first excited state energy is",round(E1,3),"MeV"

The ground state energy is 0.659 MeV
The first excited state energy is 1.977 MeV

Example number 10.7, Page number 197

In [21]:
#importing modules
import math
from __future__ import division

#Variable declaration
h=6.626*10**-34;                 #plancks constant
E=85;                            #Energy(keV)
c=3*10**8;                       #speed of light(m/s)
e=1.6*10**-19;

#Calculation                        
lamda=(h*c)/(E*10**3*e);       #de Broglie wavelength(m)
m=9.1*10**-31;                  #mass of electron(kg)
K=((h**2)/((lamda**2)*2*m*e));  #kinetic energy of electron(keV)

#Result
print "The kinetic energy of the electron is",round(K*10**-3,2),"keV"
print "answer in the book varies due to rounding off errors"

The kinetic energy of the electron is 7.06 keV
answer in the book varies due to rounding off errors

Example number 10.8, Page number 198

In [28]:
#importing modules
import math
from __future__ import division

#Variable declaration
lamda=0.08;                          #de Briglie wavelength(nm)
m=9.1*10**-31;                        #mass of electron(kg)
h=6.626*10**-34;                      #plancks constant

#Calculation                        
v=h/(m*lamda*10**-9);                 #velocity of the electron(m/s)

#Result
print "The velocity of the electron is",round(v/10**6,1),"*10**6 m/s"

The velocity of the electron is 9.1 *10**6 m/s

Example number 10.9, Page number 198

In [31]:
#importing modules
import math
from __future__ import division

#Variable declaration
h=6.626*10**-34;                  #plancks constant
lamda=589*10**-9;                #wavelength(m)
m=9.1*10**-31;                    #mass of electron(kg)
e=1.6*10**-19;

#Calculation                        
V=((h**2)/((lamda**2)*2*m*e))*10**6;    #potential diference(micro V)

#Result
print "The potential difference through which an electron should be accelerated is",round(V,2),"micro V"

The potential difference through which an electron should be accelerated is 4.35 micro V

Example number 10.10, Page number 198

In [33]:
#importing modules
import math
from __future__ import division

#Variable declaration
deltax=0.92*10**-9;                  #uncertainity in position(m)
m=9.1*10**-31;                       #mass of electron(kg)
h=6.626*10**-34;                     #plancks constant

#Calculation                        
deltav=h/(4*math.pi*m*deltax);       #uncertainity in velocity(m/s)

#Result
print "The uncertainity in velocity is",round(deltav/10**4,1),"*10**4 m/s"

The uncertainity in velocity is 6.3 *10**4 m/s

Example number 10.11, Page number 198

In [35]:
#importing modules
import math
from __future__ import division

#Variable declaration
h=6.626*10**-34;                     #plancks constant
n=3;                                 #for second excited state
m=1.67*10**-27;                      #mass of proton(kg)
E=0.5;                               #energy(MeV)
e=1.6*10**-19;

#Calculation                        
L=((h*n)/math.sqrt(8*m*E*10**6*e))*10**15;      #length of the box(fm)

#Result
print "The length of the box for proton in its second excited state is",round(L,1),"fm"

The length of the box for proton in its second excited state is 60.8 fm