Chapter 8 - Quantum chemistry

Example 1 - pg 460

In [1]:
#calculate the Wavelength
#initialisation of variables
v= 299.8 #V
e= 4.802*10**-10 #ev
h= 6.624*10**-27 #ergs sec
c= 3*10**10 #cm/sec
#CALCULATIONS
E= e/v
l= h*c*10**8/(2*E)
#RESULTS
print '%s %.1f %s' % (' Wavelength =',l,'A')
print 'The answers are a bit different due to rounding off error in textbook'

 Wavelength = 6203.3 A
The answers are a bit different due to rounding off error in textbook

Example 2 - pg 462

In [2]:
#calculate the value of numerical coefficient
#initialisation of variables
u= 109677.583 #cm**-1
#RESULTS
print '%s %.1f %s' % (' value of numerical coefficient =',u,' cm')

 value of numerical coefficient = 109677.6  cm

Example 3 - pg 464

In [3]:
#calculate the wavelength in both cases
#initialisation of variables
import math
h= 6.6234*10**-27 #ergs sec
m= 2.59 #gms
v= 3.35*10**4 #cm sec **-1
e= 4.8*10**-10 #ev
V= 40000. #volts
M= 300. #gms
L= 1836. #A
N= 6*10**23 #molecules
#CALCULATIONS
p= m*v
l= h/p
E= V*e/M
P= math.sqrt(2*E*(1/(L*N)))
L1= h*10**8/P
#RESULTS
print '%s %.2e %s' % (' wavelength =',l,'cm')
print '%s %.4f %s' % (' \n wavelength =',L1,'A')

 wavelength = 7.63e-32 cm
 
 wavelength = 0.0614 A

Example 4 - pg 471

In [4]:
#calculate the lifetime of this excited state
#initialisation of variables
import math
h= 6.624*10**-27 #ergs sec
c= 3*10**10 #cm/sec
u= 5 #cm**-1
#CALCULATIONS  
T= h/(h*2*math.pi*c*u)
#RESULTS
print '%s %.1e %s' % (' lifetime of this excited state =',T,'sec')

 lifetime of this excited state = 1.1e-12 sec

Example 5 - pg 471

In [5]:
#calculate the lifetime
#initialisation of variables
import math
V= 2.5*10**4 #m/sec
m= 30 #gms
s= 10*10**-16 #cm**2
N= 6.023*10**23 #molecules
T= 300 #K
k= 8.3*10**7
#CALCULATIONS
t= math.sqrt((m/(math.pi*k*T)))*(V/(4*s*N))
#RESULTS
print '%s %.1e %s' % (' lifetime =',t,' sec')

 lifetime = 2.0e-10  sec

Example 7 - pg 494

In [6]:
#calculate the internuclear distances
#initialisation of variables
import math
h= 6.6238*10**-27 #ergssec
N= 6.0254*10**23 #molecules
c= 2.9979*10**10
Be= 60.809
mh= 1.00812 #gms
#CALCULATIONS
u= mh/2.
Re= math.sqrt(h*N/(c*8*math.pi**2*Be*u))
#RESULTS
print '%s %.4e %s' % (' internuclear distances =',Re,'cm ')

 internuclear distances = 7.4168e-09 cm

Example 8 - pg 497

In [7]:
#calculate the Resonance energy
#initialisation of variables
H= 19.8 #kcal
H1= -0.8 #kcal
H2= -29.4 #kcal
#CALCULATIONS
H3= -85.8
H4= -49.2
H5= -H3+H4
#RESULTS
print '%s %.1f %s' % (' Resonance energy =',H5,'cal')

 Resonance energy = 36.6 cal

Example 9 - pg 500

In [8]:
#calculate the no of bonds
#initialisation of variables
import math
R= 1.69 #A
l= 1.49 #A
r= 0.706
#CALCULATIONS
n= 10**((R-l)/r)
#RESULTS
print '%s %.2f' % (' no of bonds = ',n)

 no of bonds =  1.92

Example 10 - pg 504

In [9]:
#calculate the lattice energy
#initialisation of variables
N= 6.*10**23 #molecules
R= 2.82 #A
e= 4.8*10**-10 #ev
n= 9.
z= 1.748
#CALCULATIONS
U= (N*z*e**2*(1-(1/n)))*182.2/(R*10**-8*7.63*10**12)
#RESULTS
print '%s %.1f %s' % (' lattice energy =',U,'kcal mole**-1')

 lattice energy = 181.9 kcal mole**-1

Example 11 - pg 507

In [10]:
#calculate the least energy required for transfer
#initialisation of variables
import math
k= 13
e= 4.8*10**-10 #ev
h= 6.624*10**-27 #ergs sec
N= 6.023*10**23 #molecules
l= 1836 #A
#CALCULATIONS
I= e**4*0.080/(l*N*1.28*10**-13*2*k**2*(h/(2*math.pi))**2)
#RESULTS
print '%s %.2f %s' % (' least energy required for transfer=',I,' ev')

 least energy required for transfer= 0.08  ev

Example 12 - pg 509

In [11]:
#calculate the difference between potentials
#initialisation of variables
i= 54.4 #ev
i1= 24.6 #ev
k= 2.5 
#CALCULATIONS
I= i/(4*k**2)
I1= i1/(4*k**2)
d= I-I1
#RESULTS
print '%s %.1f %s' % (' difference between first and second potential=',d,'ev')

 difference between first and second potential= 1.2 ev