Physics 3402, Chapter 10 Quiz

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.