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