Rohlf Chapter 10, #5,10,11,14,17,20,21

Physical constants
Boltzmann's constant  k= 1.38066E-23 J/K
k= 0.00008617 eV/K
Atomic mass unit   u= 1.66E-27 kg
umev= 931.5 MeV/c^2
Wien constant   2.90E-03
Planck's constant  h= 4.14E-15 eV s
h= 6.63E-34 J s
h/ 2 pi hbar 1.05E-34 J s
h/ 2 pi hbev 6.59E-16 eV s
Speed of light  c= 3.00E+08 m/s
hc= 1.24E+03 eV nm
hbarc= 1.97E+02 eV nm
Electron charge  e= 1.60E-19 C
electron volt  eV= 1.60E-19 joule
electron mass  mel 9.11E-31 kg
mel 5.11E-01 MeV/c^2
proton mass  mp 1.67E-27 kg
mpev 9.38E+02 MeV/c^2
neutron mass  mn 1.67E-27 kg
mnev 9.40E+02 MeV/c^2
Stefan's constant=  sigma= 5.67E-08 watt/m^2K^4
Nuclear distance unit  fermi 1.00E-15 m (a femtometer)
cross section unit  barn= 1.00E-28 m^2
barn= 100 fm^2
common combo for scattering  ke^2 1.44 MeV fm

First Bohr orbit radius a0  a0 0.0529 nm
Bohr magnetion  mub 5.79E-05 eV/T
muba 9.27E-24 A m^2 A=ampere

#5 Amount of energy to break both bonds in the water molecule
Burning two hydrogen molecules to produce two water molecules releases 5.7 eV
of energy. From table 10-1 we find that the binding energy of H2 is 4.5 eV.
We also find that the H-OH bond energy is 5.2 eV.
Treating the energy invested in bonds as a negative energy and using an energy balance:
-2*4.5 eV (to break hydrogen bonds) - 5.2 eV (to break oxygen bond) = 2x (water molecules) + 5.7 eV net yield.
2x = -9 -5.2 - 5.7 = -19.9 eV for the binding energy of two water molecules.
So it takes 9.95 eV to break both bonds of the water molecule.

#10 Binding energy of LiF
The ionization energy of Li is 5.4 eV and the electron affinity of F is 3.4 eV
The LiF bond length is 0.16 nm.
An estimate of the binding energy is -5.4 +3.4 +1.44 eV nm/0.16nm =     7 eV

#11 Molecular bond length of KI
The ionization energy of K is 4.3 eV and the electron affinity of I is 3.1eV.
The binding energy of the KI molecule is 3.3 eV .
3.3 eV binding energy= -4.3 eV + 3.1 eV + 1.44 eV nm/bond length,
Solving for bond length gives  0.32 nm

#14 Vibrational frequency of HCl molecule.
Force constant for HCl is 480 N/m
The freqency is given by sqrt(k/m)/2*pi, but the mass is the reduced mass.
Mass of hydrogen = 1.00783, masses of chlorine =34.969 and 36.966 amu.
Reduced masses for HCl35 = (1.00783*34.969)/(1.00783+34.969)*1.66E-27 =     1.62613E-27
Reduced masses for HCl37 = (1.00783*36.966)/(1.00783+36.966)*1.66E-27 =     1.6286E-27
Calculated frequency for HCl35 = sqrt(480/1.62613E-27)/(2*pi()) =     8.64695E+13 Hz
Calculated frequency for HCl35 = sqrt(480/1.62863E-27)/(2*pi()) =     8.64031E+13 Hz
So the isotopic difference is very small.

#17 Fractional occupation of first excited vibrational state of CO
Given a vibrational frequency of 6.42E13 Hz, the energy difference for the first excited state is
h*frequency =  0.265788 eV
The excited state fraction is given by exp(-delta E/kT)
Fraction = exp(-.2658 eV/(8.617E-5eV/K*300K)=    3.42598E-05
2.41429E-05 if .025 eV used for kT

#20 CO bond length from rotational spectrum
The absorption corresponding to the J=0->1 transition is at 2.61 mm.
The transition frequency corresponds to 2B where B= hbar^2/(2*I), so B= hc/(2*wavelength) =      3.80528E-23
The moment of inertia is then hbar^2/2*B=   1.46E-46 kgm^2
The reduced mass is (12*15.9949)/(12+15.9949)=    6.856205952 amu
Reduced mass=  1.14E-26 kg
Moment of inertia =reduced mass * bond length squared, so that the bond length
is given by sqrt(2.92e-46 kg m^2/1.14E-26kg) =    1.1321E-10 m

#21 Wavelength for J=1->2 rotational transition of N2.
Bond length for N2 giv 2 given as 0.11 nm
The reduced mass is just half of the mass of N or 14.003/2 amu =    1.16E-26 kg
The moment of inertia is then reduced mass times bond length squared =     1.40632E-46 kg m^2
The energy of the J=1->2 transition is 2*hbar^2/I =   1.58E-22 joules= 9.88E-04 eV
The wavelength is then 1240 eV nm/9.88E-4 eV =   1.25E+06 nm = 1.25E+00 mm