# Chapter 5:Raman Spectra¶

## Example no:1,Page no:592¶

In [107]:

#Variable declaration
r=1.21*10**-10                                                                   #internuclear distance (meter)
m=2.7*10**-26                                                                    #mass of oxygen atom (kg)
h=6.626*10**-34                                                                  #Plank's constant (joule second)
c=3.0*10**8                                                                        #speed of light (meter/second)

#Calculation
def F(j):
wave=B*j*(j+1)
return(wave)
import math
#(a) moment of inertia
mu=m/2.0                                                                          #reduced mass (kg)
I=mu*r**2                                                                        #moment of inertia (kg m**2)
#(b) rotational constant
B=h/(8*math.pi**2*I*c)                                                               #rotational constant (m-1)

#(c) wave number
waveno=F(1)-F(0)                                                                #wave no. of the line corresponding to the transition J=0 to J=1 (m-1)

#Result
print"\n(a) I = %.3e"%I,"kg m**2\n(b) B =",round(B,1),"m-1\n(c) wave number =",round(waveno),"m**-1"

(a) I = 1.977e-46 kg m**2
(b) B = 141.5 m-1
(c) wave number = 283.0 m**-1


## Example no:2,Page no:592¶

In [108]:
#Variable declaration
m=1.6738*10**-27                                                                 #mass of hydrogen atom (kg)
r=0.74*10**-10                                                                   #intermolecular distance of hydrogen molecule (meter)
h=1.054*10**-34                                                                  #Planck's constant (joule second)
e=1.6*10**-19                                                                    #Charge of electron (coulombs)

#Calculation
def F(j):
energy=a*j*(j+1)
return(energy)
mu=m/2.0                                                                          #reduced mass of hydrogen atom (kg)
I=mu*r**2                                                                        #moment of inertia of molecule (kg meter**2)
a=(h**2)/(2.0*I*e)                                                                   #constant (eV)
E0=F(0)                                                                         #energy of level 0 (eV)
E1=F(1)                                                                         #energy of level 1 (eV)
E2=F(2)                                                                         #energy of level 2 (eV)
E3=F(3)                                                                         #energy of level 3 (eV)

#Result
print"\nE0 =",E0,"\nE1 =%.2e"%E1,"eV\nE2 =%.2e"%E2,"eV\nE3 =%.2e"%E3,"eV"
print"NOTE:Wrong answer for E3 in book"

E0 = 0.0
E1 =1.52e-02 eV
E2 =4.55e-02 eV
E3 =9.09e-02 eV
NOTE:Wrong answer for E3 in book


## Example no:3,Page no:593¶

In [109]:
#Variable declaration
u=1.68*10**-27                                                                   #mass of hydrogen atom (kg)
m1=16.0                                                                           #mass of oxygen atom in terms u
m2=1.0                                                                            #mass of hydrogen atom in terms of u
I=1.48*10**-47                                                                   #moment of inertia of OH-radical (kg m**2)
h_bar=1.054*10**-34                                                              #Planck's constant (joule second)
c=3.0*10**8                                                                        #speed of light (meter/second)
h=6.626*10**-34                                                                  #Plank's constant (joule second)

#Calculation
import math
#(a) internuclear distance
mu=((m1*m2)/(m1+m2))*u                                                          #reduced mass of the molecule (kg)
r=(math.sqrt(I/mu))*10.0**10                                                            #internuclear distance of molecule (Å)
#(b) angular momentum
P=h_bar*math.sqrt(j*(j+1))                                                           #angular momentum of molecule (joule second)
#(c) angular velocity
omega=P/I                                                                       #angular velocity of molecule (radian/second)
#(d) wave number
B=h/(8*math.pi**2*I*c)                                                               #rotational constant (m-1)
no=2*B*(j+1)                                                                    #wave no. of line corresponding to transition j=5 to j=6 (m-1)
#(e) energy absorbed
E=c*h*no                                                                        #energy absorbed in the transition j=6 to j=5 (joule)

#Result
print"\n(a) r =",round(r,2),"Å\n(b) J =%.2e"%P," joule second\n"
print"(c) ω =%.2e"%omega,"rad/s\n(d) wave number =%.2e"%no,"m**-1\n(e) E =%.1e"%E,"J"

(a) r = 0.97 Å
(b) J =5.77e-34  joule second

(d) wave number =2.27e+04 m**-1
(e) E =4.5e-21 J


## Example no:4,Page no:593¶

In [110]:
#Variable declaration
h=6.63*10**-34                                                                   #Plank's constant (joule second)
v=1.153*10**11                                                                   #Frequency of absorption line (Hz)
mu=11.38*10**-27                                                                 #Recuced mass of the molecule (kg)

#Calculation
import math
I=h/(4.0*math.pi**2*v)                                                                 #moment of inertia of CO molecule (kg m**2)
r=math.sqrt(I/mu)*10**10                                                              #Internuclear distance (Å)

#Result
print"\n Internuclear distance =",round(r,2)," Å"

 Internuclear distance = 1.13  Å


## Example no:5,Page no:594¶

In [111]:
#Variable declaration
mu=1.62*10**-27                                                                  #Reduced mass of HCL (kg)
c=3*10**8                                                                        #Velocity of light (m/s)
h=6.62*10**-34                                                                   #Plank's constant (joule second)
v1_P=2906.3                                                                     #Wave no. of P branch (cm-1)
v2_P=2927.5                                                                     #Wave no. of P branch (cm-1)
v3_P=2948.7                                                                     #Wave no. of P branch (cm-1)
v4_P=2969.9                                                                     #Wave no. of P branch (cm-1)
v1_R=3012.2                                                                     #Wave no. of R branch (cm-1)
v2_R=3033.4                                                                     #Wave no. of R branch (cm-1)
v3_R=3054.6                                                                     #Wave no. of R branch (cm-1)
v4_R=3078.8                                                                     #Wave no. of R branch (cm-1)

#Calculation
import math
#(i) Equilibrium internuclear distance
delta_v=v2_P-v1_P                                                               #Separation of successive line of P and R branch (cm-1)
B=delta_v/2.0                                                                     #rotational constant (cm-1)
I=h/(8.0*math.pi**2*B*10**2*c)                                                          #Moment of inertia (kg m**2)
r=math.sqrt(I/mu)*10**10                                                              #Equilibrium internuclear distance (Å)
#(ii) Force constant
v0=(v4_P+v1_R)/2.0                                                                #Equlibrium frequency (cm-1)
k=4*math.pi**2*mu*c**2*v0**2*10**4                                                      #Force constant of HCl (N/m)

#Result
print"\n(i) Equilibrium internuclear distance =",round(r,2),"Å\n(ii) Force constant =",round(k),"N/m"

print"Note: the answer of (ii) part is wrong in the book"

(i) Equilibrium internuclear distance = 1.28 Å
(ii) Force constant = 515.0 N/m
Note: the answer of (ii) part is wrong in the book


## Example no:6,Page no:594¶

In [112]:
#Variable declaration
mu=8.37*10**-28                                                                  #Reducec mass of hydrogen molecule (kg)
h=6.58*10**-16                                                                   #Plank's constant (eV s)
E0=0.273                                                                        #Ground state vibrational energy of hydrogen molecule (eV)

#Calculation
k=mu*((2*E0)/h)**2                                                               #force constant of hydrogen molecule (N/m)

#Result
print"\n Force constant =",round(k),"N/m"

 Force constant = 576.0 N/m


## Example no:7,Page no:595¶

In [113]:
#Variable declaration
m1=1.0                                                                            #molar mass of H atom (amu)
m2=35.0                                                                           #molar mass of Cl atom (amu)
u=1.68*10**-27                                                                   #atomic mass unit (kg)
v=2885.9*100                                                                    #wave no. of line (m-1)
c=3.0*10**8                                                                        #Velocity of light (m/s)

#Calculation
mu=((m1*m2)/(m1+m2))*u                                                          #reduced mass of HCl molecule (kg)
mu=round(mu,29)
k=4*(math.pi*c*v)**2*mu                                                              #force constant of HCl molecule (N/m)

#Result
print"Force constant =",round(k),"N/m"
print"NOTE:Approximate value of pi is used in book,that's why different answer"

Force constant = 482.0 N/m
NOTE:Approximate value of pi is used in book,that's why different answer


## Example no:8,Page no:595¶

In [114]:
#Variable declaration
m1=12.0                                                                           #molar mass of C atom (amu)
m2=16.0                                                                           #molar mass of O atom (amu)
u=1.68*10**-27                                                                   #atomic mass unit (kg)
k=1870.0                                                                          #force constant of CO molecule (N/m)
h=6.6*10**-34                                                                    #Plank's constant (joule second)
e=1.602*10**-19                                                                  #charge of electron (Coulomb)

#Calculation
def F(V):
energy=((V+.5)*h*v)/e
return(energy)

#def G(V):
#    energy=((V+.5)*h*v*8065)/e
#    return(energy)
import math
mu=((m1*m2)/(m1+m2))*u                                                          #reduced mass of CO molecule (kg)
v=(1.0/(2.0*math.pi))*math.sqrt(k/mu)                                                        #frequency of vibration of CO molecule (Hz)
e1=F(0)                                                                         #energy of 1st level of CO molecule (eV)
#E1=G(0)                                                                         #energy of 1st level of CO molecule (cm-1)
E1=round(e1,3)*8065
e2=F(1)                                                                         #energy of 2nd level of CO molecule (eV)
E2=round(e2,3)*8065                                                                         #energy of 2nd level of CO molecule (cm-1)
e3=F(2)                                                                         #energy of 3rd level of CO molecule (eV)
E3=round(e3,3)*8065                                                                       #energy of 3rd level of CO molecule (cm-1)

#Result
print"\nE = ",round(e1,3),"eV,",round(e2,3),"eV,",round(e3,3),"eV.........\n  =",E1,"cm**-1,",E2,"cm**-1,",E3,"cm**-1........."

E =  0.132 eV, 0.396 eV, 0.66 eV.........
= 1064.58 cm**-1, 3193.74 cm**-1, 5322.9 cm**-1.........


## Example no:9,Page no:595¶

In [115]:
#Variable declaration
mu=1.61*10**-27                                                                  #reduced mass of HCl molecule (kg)
c=3*10**8                                                                        #speed of light (m/s)
lembda=3.465*10**-6                                                              #wavelength of infrared radiation (m)

#Calculation
import math
k=4*(math.pi*v)**2*mu                                                                #force constant of HCl molecule (N/m)

#Result
print"Force constant =",round(k),"N/m"

Force constant = 476.0 N/m


## Example no:10,Page no:596¶

In [116]:
#Variable declaration
h=6.6*10**-34                                                                    #Plank's constant (joule second)
mu=1.62*10**-27                                                                  #reduced mass of HCl molecule (kg)
c=3*10**8                                                                        #speed of light (m/s)
v=2.886*10**5                                                                    #wave no. of absorption line in infrared spectrum (m-1)

#Calculation
import math
k=4*(math.pi*c*v)**2*mu                                                              #force constant of HCl molecule (N/m)
amp=math.sqrt((h*c*v)/k)*10**10                                                       #amplitude of vibration in the ground state (Å)

#Result
print"Amplitude of vibration =",round(amp,2),"Å"

Amplitude of vibration = 0.11 Å


## Example no:11,Page no:596¶

In [117]:
#Variable declaration
v1=214330.0                                                                       #fundamental band for CO molecule (m-1)
v2=425970.0                                                                       #first overtone for CO molecule (m-1)

#calculation
import numpy as np
a=np.array([[1,-2],[2,-6]])
b=np.array([v1,v2])
x=np.linalg.solve(a,b)
we=x[0]
wexe=x[1]

#Result

print"We find we=",we,"m**-","xe.we=",wexe,"m**-1"
print"NOTE:Wrong answer for xe.we in book"

We find we= 217020.0 m**- xe.we= 1345.0 m**-1
NOTE:Wrong answer for xe.we in book


## Example no:12,Page no:596¶

In [147]:
#Variable declaration
v1=2886.0                                                                       #intense absorption (m-1)
v2=5668.0                                                                       #intense absorption (m-1)
v3=8347.0                                                                       #intense absorption (m-1)

#Calculation
import math
from numpy.linalg import inv
A=np.array([[1,-2],[2,-6]])                                                                   #coefficient matrix
b=np.array([[v1],[v2]])                                                                       #right hand side matrix
mu=1.61*10**-27                                                                  #reduced mass (kg)
c=3*10**8                                                                        #speed of light (m/s)
x=np.dot(inv(A),b)                                                                      #values of omega and x*omega (m-1)
k=4*(math.pi*c*x[0]*100)**2*mu                                                           #force constant (N/m)

#Result
print"we =",round(x[0]),"cm**-1\nxe*we =",round(x[1]),"cm**-1\nforce constant =",round(k),"N/m"

we = 2990.0 cm**-1
xe*we = 52.0 cm**-1
force constant = 511.0 N/m


## Example no:13,Page no:597¶

In [118]:
#Variable declaration
v1=8.657*10**13                                                                  #frequency of rotation absorption spectrum (Hz)
v2=8.483*10**13                                                                  #frequency of rotation absorption spectrum (Hz)
h=6.6*10**-34                                                                    #Plank's constant (joule second)
mu=1.544*10**-27                                                                 #Recuced mass of CH molecule (kg)

#Calculation
#(i) equilibrium separation
import math
I=h/(2*math.pi**2*(v1-v2))                                                           #Moment of inertia (kg m**2)
r=math.sqrt(I/mu)                                                                    #equilibrium internuclear distance (m)
#(ii) force constant of molecule
v0=(v1+v2)/2.0                                                                    #Central frequency (Hz)
k=4*mu*(math.pi*v0)**2                                                               #Force constant of CH molecule (N/m)

#Result
print"\n (i) equilibrium separation =%.2e"%r,"m\n (ii) force constant =",round(k),"N/m"

 (i) equilibrium separation =1.12e-10 m
(ii) force constant = 448.0 N/m


## Example no:14,Page no:597¶

In [119]:
#Variable declaration
k=448.0                                                                           #force constant of CH molecule (N/m)
mu=4.002*10**-27                                                                 #reduced mass of CH molecule (kg)
r=0.112*10**-9                                                                   #internuclear distance (m)
h=6.6*10**-34                                                                    #Plank's constant (joule second)

#Calculation
import math
v0=(1/(2*math.pi))*math.sqrt(k/mu)                                                       #central frequency (s-1)
I=mu*r**2                                                                        #moment of inertia of molecule (kg m**2)
v1=v0+x                                                                         #peak frequency (Hz)
v2=v0-x                                                                         #peak frequency (Hz)

#Result
print"\n Peak frequencies =%.3e"%v1,"Hz,%.3e"%v2,"Hz"

 Peak frequencies =5.358e+13 Hz,5.292e+13 Hz


## Example no:15,Page no:598¶

In [120]:
#Variable declaration
v1=2174.07                                                                      #peak wave number (cm-1)
v2=2166.35                                                                      #peak wave number (cm-1)
h=6.6*10**-34                                                                    #Plank's constant (joule second)
c=3*10**8                                                                        #Speed of light (m/s)
mu=1.145*10**-26                                                                 #Reduced mass of CO molecule (kg)

#Calculation
import math
#(a) central frequency
B=(v1-v2)/4                                                                     #Rotational constant (cm-1)
v0=(v1+v2)/2                                                                    #Central frequency (cm-1)
#(b) internuclear distance
I=h/(8*math.pi**2*B*100*c)                                                           #moment of inertia of molecule (kg m**2)
r=math.sqrt(I/mu)*10**10                                                              #equilibrium internuclear distance (Å)
#(c) force constant
k=4*mu*(math.pi*c*v0*100)**2                                                             #force constant (N/m)

#Result

print"\n(a) central frequency =",v0,"cm**-1\n(b) internuclear distance =",round(r,2),"Å\n(c) force constant =%d"%k,"N/m"
print"NOTE:Wrong answer for 'k' in book"

(a) central frequency = 2170.21 cm**-1
(b) internuclear distance = 1.12 Å
(c) force constant =1916 N/m
NOTE:Wrong answer for 'k' in book


## Example no:16,Page no:598¶

In [121]:
#Variable declaration
mu=3.142*10**-27                                                                 #reduced mass of the molecule (kg)
r=1.288*10**-10                                                                  #internuclear distance (m)
h=6.6*10**-34                                                                    #Plank's constant (joule second)
c=3*10**8                                                                        #Speed of light (m/s)
v0=201100.0                                                                       #central frequency (m-1)

#Calculation
import math
I=mu*r**2                                                                        #moment of inertia of molecule (kg m**2)
B=h/(8.0*math.pi**2*I*c)                                                               #Rotational constant (m-1)
vR0=v0+(2*B)                                                                    #wave no. of 1st line of R-branch (m-1)
vR1=v0+(4*B)                                                                    #wave no. of 2nd line of R-branch (m-1)
vP1=v0-(2*B)                                                                    #wave no. of 1st line of P-branch (m-1)
vP2=v0-(4*B)                                                                    #wave no. of 2nd line of P-branch (m-1)

#Result
print"\n V_R(0) =",round(vR0),"m**-1\n V_R(1) =",round(vR1),"m**-1\n V_P(1) =",round(vP1),"m**-1\n V_P(2) =",round(vP2),"m**-1"

print"NOTE:Very approximate value of 'B' is calculated in book,that's why difference in answers"

 V_R(0) = 202169.0 m**-1
V_R(1) = 203238.0 m**-1
V_P(1) = 200031.0 m**-1
V_P(2) = 198962.0 m**-1
NOTE:Very approximate value of 'B' is calculated in book,that's why difference in answers


## Example no:17,Page no:599¶

In [122]:
#Variable declaration
mu=1.62*10**-27                                                                  #reduced mass of HCl molecule (kg)
r=1.293*10**-10                                                                  #internuclear distance (m)
h=6.6*10**-34                                                                    #Plank's constant (joule second)
c=3*10**8                                                                        #Speed of light (m/s)

#Calculation
I=mu*r**2                                                                        #moment of inertia of molecule (kg m**2)
I=round(I,48)

B=h/(8*math.pi**2*I*c)                                                               #Rotational constant (m-1)
B=round(B)
sep=4*B                                                                         #separation b/w lines R(0) and P(1) (m-1)

#Result
print"\nΔν =",sep,"m**-1"
print"NOTE:Note:Again I is wrongly approximated due to which 'B' and thus fina answer does not match"

Δν = 4128.0 m**-1
NOTE:Note:Again I is wrongly approximated due to which 'B' and thus fina answer does not match


## Example no:18,Page no:599¶

In [123]:
#Variable declaration
import math
a=214.6*100                                                                     #(m-1)
b=0.6*100                                                                       #(m-1)
h=6.62*10**-34                                                                    #Plank's constant (joule second)
c=3.0*10**8                                                                        #Speed of light (m/s)
no=1.0/(math.e)                                                                       #number of molecules in state with respect to ground state
k=1.38*10**-23                                                                   #Boltzmann constant (J K-1)

#Calculation
deltaE=h*c*(a-2*b)                                                              #difference in the energies of state 0 and state 1 (J)
deltaE=round(deltaE,24)
T1=deltaE/k                                                                     #temperature at which number of molecules in state 1 is 1/e times of state 0 (K)
T2=deltaE/(k*math.log(10.0))                                                           #temperature at which number of molecules in state 1 is 10% of state 0 (K)

#Result
print"n(i) T =",round(T1),"K\n(ii) T =",round(T2),"K"
print"NOTE:Calculation mistake in book"

n(i) T = 307.0 K
(ii) T = 133.0 K
NOTE:Calculation mistake in book


## Example no:19,Page no:599¶

In [124]:
#Variable declaration
vexc=4358.0*10**-10                                                                #wavelength of exciting line (m)
vsto=4458.0*10**-10                                                                #wavelength of Stokes line (m)

#Calculation
vbar_exc=1/vexc                                                                 #wave number of exciting line (m-1)
vbar_sto=1/vsto                                                                 #wave number of Stokes line (m-1)
delta_vbar=vbar_exc-vbar_sto                                                    #Raman shift (m-1)
vbar_antistoke=vbar_exc+delta_vbar                                              #Wave number of Anti-Stokes line (m-1)
lembda_antistoke=(1/vbar_antistoke)*10**10                                       #Wavelength of Anti-Stokes line (Å)

#Result
print"\nwavelength of Anti-stokes line =",round(lembda_antistoke,1),"Å"

wavelength of Anti-stokes line = 4262.4 Å


## Example no:20,Page no:600¶

In [125]:
#Variable declaration
h=6.62*10**-34                                                                   #Plank's constant (joule second)
c=3.0*10**8                                                                        #Speed of light (m/s)
x=62.4*100                                                                      #(m-1)
y=41.6*100                                                                      #(m-1)

#Calculation
import math
B=y/4.0                                                                           #Rotational constant of HCl (m-1)
I=h/(8*math.pi**2*B*c)                                                               #Moment of inertia (kg m**2)

#Result
print"\n I =%.1e"%I,"kg m**2"

 I =2.7e-47 kg m**2


## Example no:21,Page no:¶

In [126]:
def F(v):
energy=(((v+.5)*a)-(((v+.5)**2)*b))*h*c
return(energy)
#Variable declaration
h=6.62*10**-34                                                                   #Plank's constant (joule second)
c=3.0*10**8                                                                        #Speed of light (m/s)
a=1580.36*100                                                                   #value of ωe (m-1)
b=12.07*100                                                                     #value of ωexe (m-1)

#Calculation
E0=F(0)                                                                         #Zero point energy of the molecule (J)
shift=(F(1)-F(0))/(h*c)                                                         #Expected vibrational Raman shift (m-1)

#Result
print"\nZero point energy =%.3e"%E0,"J\nExpected vibrational Raman shift =",shift/100,"cm**-1"

Zero point energy =1.563e-20 J
Expected vibrational Raman shift = 1556.22 cm**-1