Molecular Structure

Example 9.1 Page 270

In [3]:
#initiation of variable
E=-2.7;
K=9.0*(10**9)*((1.6*(10**-19))**2)/(0.106*10**-9);# taking all the values in meters. 1/(4*pi*e0)= 9*10^9 F/m

#calculation
q=((K-E*10**-9)/(4*K))*10**-9;                #balancin by multiplying 10^-9 on numerator. to eV.vm terms

#result
print"Charge on the sphere required is",round(q,4)," times the charge of electron.";

Charge on the sphere required is 0.3105  times the charge of electron.

Example 9.2 Page 273

In [4]:
#initiation of variable
K=1.44; Req=0.236; # K=e^2/(4*pi*e0)=1.44 eV.nm

#calculation
Uc=-K/(Req);        #coulomb energy

#result
print"The coulomb energy at an equilirium separation distance in eV is",round(Uc,3);

E=-4.26; delE=1.53;    #various standars values of NaCl
Ur=E-Uc-delE; 

#result
print"The pauli''s repulsion energy in eV is",round(Ur,3);

#partb
Req=0.1;         #pauli repulsion energy
Uc=-K/(Req);
E=4; delE=1.53;
Ur=E-Uc-delE;

#result
print"The pauli''s repulsion energy in eV is",round(Ur,3);

The coulomb energy at an equilirium separation distance in eV is -6.102
The pauli''s repulsion energy in eV is 0.312
The pauli''s repulsion energy in eV is 16.87

Example 9.3 Page 276

In [6]:
#initiation of variable
from math import pi, sqrt
delE=0.50; delR=0.017*10**-9;      #delE= E-Emin; delR=R-Rmin;
k=2*(delE)/(delR**2);c=3*10**8;     #force constant
m=(1.008)*(931.5*10**6)*0.5;       #mass of molecular hydrogen
v= sqrt(k*c**2/m)/(2*pi);          #vibrational frequency
h=4.14*(10**-15);

#calculation
E=h*v;

#result
print"The value of corresponding photon energy in eV is",round(E,3);

The value of corresponding photon energy in eV is 0.537

Example 9.4 Page 280

In [9]:
#initiation of variable
from math import pi, sqrt
hc=1240.0;        #in eV.nm
m=0.5*1.008*931.5*10**6;              #mass of hydrogen atom
Req=0.074;                          #equivalent radius

#calculation
a=((hc)**2)/(4*(pi**2)*m*(Req**2));  #reduced mass of hydrogen atom
for L in range(1,4):
        delE= L*a; 
        print"The value of energy in eV is",round(delE,4);                  
        w=(hc)/delE;
        print"The respective wavelength in um is",round(w*10**-3,3);  

The value of energy in eV is 0.0151
The respective wavelength in um is 81.849
The value of energy in eV is 0.0303
The respective wavelength in um is 40.925
The value of energy in eV is 0.0454
The respective wavelength in um is 27.283

Example 9.5 Page 283

In [16]:
#initiation of variable
from math import pi
delv=6.2*(10**11);     #change in frequency
h=1.05*(10**-34);        #value of h in J.sec

#calculation
I= h/(2*pi*delv);      #rotational inertia
I1=I/(1.684604*10**-45); #to change units

#result
print"The value of rotational inertia in kg m2 is %.1e" %I;
print"which in terms of amu in u.nm2 is",round(I1,3);

The value of rotational inertia in kg m2 is 2.69536597172e-47
which in terms of amu in u.nm2 is 0.016

Example 9.6 Page 286

In [19]:
#initiation of variable
from math import pi
delE=0.358;hc=4.14*10**-15;          #hc in eV.nm and delE=1.44eV(given values)

#calculation
f=(delE)/hc;                        #frequency 

#result
print"The frequency of the radiation is ",f;


m=0.98;                            #mass in terms of u
k=4*pi**2*m*f**2;                   #value of k in eV/m^2

#result
print"The force constant is",k; 

#partb
hc=1240.0; m=0.98*1.008*931.5*10**6; Req=0.127;      #various constants in terms of  
s=((hc)**2)/(4*(pi**2)*m*(Req**2));                # expeted spacing 

#result
print"The spacing was found out to be",round(s,3),"which is very close to the graphical value of 0.0026 eV."

The frequency of the radiation is  8.64734299517e+13
The force constant is 2.89301831756e+29
The spacing was found out to be 0.003 which is very close to the graphical value of 0.0026 eV.