1. Given sodium ionization energy of 5.1 eV and chlorine electron affinity

of 3.6 eV, calculate the bond length. Given a measured dissociation
energy of 3.6 eV.

The approach is ionization energy - electron affinity -ke^2/r = -3.6
eV

ke^2 = 1.44 eV nm, so r is given by

r = 1.44/(5.1 -3.6 +3.6) = 0.282352941 nm

2. Given a j=0-1 transition frequency of 1.31E10, calculate bond length
of NaCl.

The masses of Na and Cl are 23 and 34.97 respectively.

The reduced mass is 13.87459031 amu = 2.30318E-26 kg

Using amu = 1.66E-27 kg

The transition energy is hbar^2/moment of inertia = h*frequency =
8.67954E-24 joule

Using h= 6.6256E-34 J s

The momentu of inertia is then = I = 1.28113E-45 kg m^2

Using hbar= 1.0545E-34 J s

The bond length can then be calculated from sqrt(moment of inertia/reduced
mass) =

r = sqrt(1.2811E-45/2.303E-26) m = 2.35855E-10 m = 0.235854706
nm

3. From the vibrational frequency 3E13 Hz, calculate the bond force
constant of NaCl

The vibrational angular frequency is equal to sqrt(k/m) where in this
case

the m is the reduced mass calculated above.

The bond force constant is then equal to (f*2*pi)^2 * m =

For frequency 3.00E+13 Hz

k= (3e13*2*pi())^2*2.303E-26 = 818.2691617 N/m

Hanging a 1 kg mass on such a spring would stretch it 9.8 N/818 N/m
= 0.011976499 m

So a hanging kilogram would stretch this spring only 1.2 cm.