Chapter 5: The Schrodinger Equation

Example 5.2 Page 150

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#initiation of variable
#parta
from math import pi, sin
import math
from scipy import integrate
h=1.05*10**-34;m=9.11*10**-31;L=10.0**-10;      # all the values are taken in SI units
E1=h**2*pi**2/(2*m*L**2); E2=4*E1;            #Energies are calculated
delE=(E2-E1)/(1.6*10**-19);                  #Difference in energy is converted to eV

#result
print "Energy to be supplied in eV. is ",round(delE,3);

#partb
x1=0.09*10**-10;x2=0.11*10**-10                 #limits of the given region
def integrand(x,L):
    return 2.0/L*(sin(pi*x/L))**2

probGnd=integrate.quad(integrand,x1,x2,args=(L))

#result
print "The percentage probability of finding an electron in the ground state is \n",round(probGnd[0]*100,3);

#partc
k1=0.0;k2=0.25*10**-10;
def integrand(k,L):
    return 2.0/L*(sin(2*pi*k/L))**2

probExc=integrate.quad(integrand,k1,k2,args=(L))

#result
print "The probability of finding an electron in the excited state is",round(probExc[0],3);
 Energy to be supplied in eV. is  111.978
The percentage probablility of finding an electron in the ground state is 
0.383
The probablility of finding an electron in the excited state is 0.25
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