#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"
#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"
#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)
j=5.0 #energy level of OH-radical
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"
#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"
#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"
#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"
#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"
#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........."
#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
v=c/lembda #frequency of radiation (s-1)
k=4*(math.pi*v)**2*mu #force constant of HCl molecule (N/m)
#Result
print"Force constant =",round(k),"N/m"
#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"
#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"
#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"
#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"
#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)
x=h/(4.0*math.pi**2*I) #additional frequency (s-1)
v1=v0+x #peak frequency (Hz)
v2=v0-x #peak frequency (Hz)
#Result
print"\n Peak frequencies =%.3e"%v1,"Hz,%.3e"%v2,"Hz"
#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"
#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"
#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"
#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"
#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"
#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"
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"