6: Schroedinger Wave Equation

Example number 1, Page number 211

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

#Variable declaration    
h=6.63*10**-34;        #plancks constant(J sec)
m=9.11*10**-31;        #mass(kg)
a=10**-10;             #width of box(m)
e=1.602*10**-19;       #charge(coulomb)

#Calculations
E1=(h**2)/(8*m*e*a**2);    #least energy(eV)

#Result
print "least energy is",round(E1,2),"eV"
print "answer given in the book is wrong"

least energy is 37.65 eV
answer given in the book is wrong

Example number 2, Page number 211

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

#Variable declaration    
delta_x=5*10**-10;     #interval(m)
a=25*10**-10;          #width(m)

#Calculations
P=2*delta_x/a;         #probability of finding the particle

#Result
print "probability of finding the particle is",P

probability of finding the particle is 0.4

Example number 3, Page number 212

In [11]:
#importing modules
import math
from __future__ import division
from scipy.integrate import quad

#Variable declaration    
a=1;        #assume

#Calculations
def zintg(x):
    return (2/a)*(1/2)*(1-math.cos(2*math.pi*x/a))

P1=quad(zintg,0.45,0.55)[0]

#Result
print "probability of finding the particle is",round(P1*100,2),"%"

probability of finding the particle is 19.84 %

Example number 4, Page number 213

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

#Variable declaration    
h=6.63*10**-34;      #planks constant(Js)
m=9.1*10**-31;       #mass(kg)
a=2.5*10**-10;       #width(m)
e=1.6*10**-19;       #charge(coulomb)
n1=1;                      
n2=2;
n3=3;                #energy states

#Calculations
E1=n1**2*(h**2)/(8*m*e*a**2);    #least energy(eV)
E2=n2**2*E1;                     #energy in 2nd excited state(eV)
E3=n3**2*E1;                     #energy in 3rd excited state(eV)
delta_E=E2-E1;                   #difference of energy between 2nd and 1st excited states(eV)

#Result
print "least energy is",int(E1),"eV"
print "energy in 2nd excited state is",int(E2),"eV"
print "energy in 3rd excited state is",int(E3),"eV"
print "difference of energy between 2nd and 1st excited states is",int(delta_E),"eV"

 least energy is 6 eV
energy in 2nd excited state is 24 eV
energy in 3rd excited state is 54 eV
difference of energy between 2nd and 1st excited states is 18 eV

Example number 5, Page number 213

In [23]:
#importing modules
import math
from __future__ import division

#Variable declaration    
h=6.63*10**-34;      #planks constant(Js)
m=9.1*10**-31;       #mass(kg)
a=10;                #width(angstrom)
e=1.6*10**-19;       #charge(coulomb)
n1=1;                      
n2=2;
n3=3;                #energy states

#Calculations
lamda1=2*a/n1;       #de-Broglie wavelength of first energy state(angstrom)
lamda2=2*a/n2;       #de-Broglie wavelength of second energy state(angstrom)
lamda3=2*a/n3;       #de-Broglie wavelength of third energy state(angstrom)
E1=n1**2*(h**2)/(8*m*e*(a*10**-10)**2);    #energy in 1st excited state(eV)
E2=n2**2*E1;                     #energy in 2nd excited state(eV)
E3=n3**2*E1;                     #energy in 3rd excited state(eV)

#Result
print "de-Broglie wavelength of first three energy states are",int(lamda1),"angstrom",int(lamda2),"angstrom",round(lamda3,2),"angstrom"
print "energies of first three energy states are",round(E1,2),"eV",round(E2,4),"eV",round(E3,3),"eV"
print "answer in the book varies due to rounding off errors"

de-Broglie wavelength of first three energy states are 20 angstrom 10 angstrom 6.67 angstrom
energies of first three energy states are 0.38 eV 1.5095 eV 3.396 eV
answer in the book varies due to rounding off errors

Example number 6, Page number 214

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

#Variable declaration    
h=6.63*10**-34;      #planks constant(Js)
a=0.2*10**-9;        #width(m)
e=1.602*10**-19;     #charge(coulomb)
n5=5;                #energy state
E5=230*e;            #energy 0f 5th state(J)
En=10**3*e;          #energy(eV)

#Calculations
E1=E5/n5**2;         #energy in 1st excited state(eV)
m=h**2/(8*E1*a**2);    #mass(kg)
n=math.sqrt(En/E1);    #quantum state

#Result
print "energy in 1st excited state is",round(E1*10**19,1),"*10**-19 J"
print "mass is",round(m*10**31,1),"*10**-31 kg"
print "quantum state is",round(n,1)
print "as n is not an integer, En is not permitted value of energy"

energy in 1st excited state is 14.7 *10**-19 J
mass is 9.3 *10**-31 kg
quantum state is 10.4
as n is not an integer, En is not permitted value of energy

Example number 7, Page number 225

In [48]:
#importing modules
import math
from __future__ import division

#Variable declaration 
e=1.6*10**-19;     #charge(coulomb)
E1=0.04*e;         #energy(J)
V=0.03*e;          #energy barrier(J) 
E2=0.025*e;        #energy(J)
E3=0.03*e;         #energy(J)
m=1;               #assume 
k1=1;              #assume

#Calculations
x=math.sqrt(E1-V);
y=math.sqrt(E1+V);
R1=((math.sqrt(E1)-x)/(math.sqrt(E1)+y))**2;     #reflection coefficient
T1=1-R;                                          #transmission coefficient
k2=math.sqrt(2*m*(E3-V));                       
R2=((k1-k2)/(k1+k2))**2;                         #reflection coefficient
T2=4*k1*k2/(k1+k2)**2;                           #transmission coefficient 

#Result
print "reflection coefficient in 1st case is",round(R1,2)
print "answer given in the book is wrong"
print "transmission coefficient in 1st case is",round(T1,2)
print "for E=0.025, E<V. so transmission coefficient is 0 and reflection coefficient is 1"
print "reflection coefficient in 3rd case is",int(R2)
print "transmission coefficient in 3rd case is",int(T2)

reflection coefficient in 1st case is 0.05
answer given in the book is wrong
transmission coefficient in 1st case is 0.95
for E=0.025, E<V. so transmission coefficient is 0 and reflection coefficient is 1
reflection coefficient in 3rd case is 1
transmission coefficient in 3rd case is 0

Example number 8, Page number 226

In [54]:
#importing modules
import math
from __future__ import division

#Variable declaration    
T=0.5;            #transmission coefficient 
a=1;
b=6;
c=1;

#Calculations
k1byk2=(b+math.sqrt((b**2)-(4*a*c)))/(2*a);  
x=k1byk2**2;
EbyV=x/(x-1);        #value of E/V

#Result
print "value of E/V is",round(EbyV,2)

value of E/V is 1.03

Example number 9, Page number 226

In [40]:
#importing modules
import math
from __future__ import division

#Variable declaration    
m=9.1*10**-31;       #mass(kg)
a1=5*10**-10;         #width(m)
a2=10*10**-10;         #width(m)
e=1.6*10**-19;       #charge(coulomb)
V0=5;                #energy barrier(eV)
E1=1;                #energy of electron(eV)
E2=2;
chi=1.054*10**-34;   #plancks constant(Js)

#Calculations
beta1=math.sqrt(2*m*(V0-E1)*e/chi**2);      #value of beta(m-1)
x1=int(-2*a1*beta1);
beta2=math.sqrt(2*m*(V0-E2)*e/chi**2);      #value of beta(m-1)
x2=round(-2*a1*beta2,1);
T1=math.exp(x1);               #transmission probability in 1st case
T2=math.exp(x2);               #transmission probability in 1st case
x3=int(-2*a2*beta1);
x4=round(-2*a2*beta1,1);
T1dash=math.exp(x3);               #transmission probability in 2nd case
T2dash=math.exp(x4);               #transmission probability in 1st case

#Result
print "transmission probabilities in 1st case are",round(T1*10**5,1),"*10**-5 and",round(T2*10**4,1),"*10**-4"
print "transmission probabilities in 2nd case are",round(T1dash*10**9,1),"*10**-9 and",round(T2dash*10**9,2),"*10**-9"
print "answer for transmission probability in 2nd case given in the book is wrong"

transmission probabilities in 1st case are 4.5 *10**-5 and 1.4 *10**-4
transmission probabilities in 2nd case are 2.1 *10**-9 and 1.25 *10**-9
answer for transmission probability in 2nd case given in the book is wrong

Example number 10, Page number 227

In [59]:
#importing modules
import math
from __future__ import division

#Variable declaration    
m=9.1*10**-31;       #mass(kg)
a=10**-10;           #width(m)
e=1.6*10**-19;       #charge(coulomb)
V0=5;                #energy barrier(eV)
E=2.5;               #energy of electron(eV)
chi=1.05*10**-34;    #plancks constant(Js)

#Calculations
x=16*E*(V0-E)/V0**2;
y=-2*a*math.sqrt(2*m*(V0-E)*e/chi**2);
#T=x*math.exp(y);     #transmission coefficient

#Result
print "transmission coefficient is",int(x),"math.exp (",round(y,3),")"

transmission coefficient is 4 math.exp ( -1.625 )

Example number 11, Page number 227

In [76]:
#importing modules
import math
from __future__ import division

#Variable declaration    
P=10**21;        #probability(T per sec)
m=4*1.6*10**-27;       #mass(kg)
a=2*10**-14;           #width(m)
e=1.67*10**-19;        #charge(coulomb)
V0=30;                 #energy barrier(eV)
E=4.2;                 #energy of electron(eV)
chi=1.05*10**-34;      #plancks constant(Js)

#Calculations
x=P*16*E*(V0-E)/V0**2;
y=-2*a*math.sqrt(2*m*(V0-E)*10**6*e/chi**2);
T=x*math.exp(y);     #transmission coefficient
tow=1/T;             #average lifetime of nucleus(years) 

#Result
print "average lifetime of nucleus is",round(tow/10**17,1),"*10**17 years"
print "answer given in the book is wrong"

average lifetime of nucleus is 3.7 *10**17 years
answer given in the book is wrong