Question 5: How do molecular force constants k compare with those of ordinary springs? What do you conclude from this comparison?

Consider the case in Example 9.3 on page 289, where k (H2) = 36 ´ 10²⁰ eV/m². Converting this to SI units using 1 eV = 1.6 ´ 10^-19 Joules (noting 1 Joule = 1 Nt-m), this yields

k = (36 ´ 10²⁰ eV/m²)x( 1.6 ´ 10^-19 Joules/eV) = 576 Nt/m

Converting this to US units (1 Nt = 2.2 lb/9.8; 1 m=39.34 in)

576 x 2.2 / 9.8 /39 = 3.3 lb/in

The fact that this value is similar to macroscopic spring constants is not surprising, since solid objects consist of molecules arranged in parallel and series configurations. When two equal springs are connected in parallel, half the force is across each; when two equal springs are connected in series, each contributes half the elongation. Thus one can rewrite the individual molecular spring constant

k = F/x

macroscopically as Young’s modulus Y

Y = kL/A = [F/A]/[x/L]

where the area A takes into account the parallel springs and the length L takes into account the series springs.

Conclusion:

These are very strong springs, and the energies associated with vibration can be expected to be much larger than those arising from rotations, which will scale as

ħ²/MR² = (ħc)²/Mc²R² = (1240 eV-nm) ² / 4π² / 931.5 x10⁶ eV/ (0.1 nm) ² = 0.003 eV