Question 2 of 7: Lennard-Jones Potential and Cohesive Energy of Solid H₂

Question 2: Lennard-Jones Potential and Cohesive Energy of Solid H₂ (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Normalized pair-potential curve $U(R)/\epsilon$ vs. $R/\sigma$ (Fig. P2), minimum just above $R/\sigma=1$, diverging steeply below $R/\sigma\approx0.9$. For H₂: $\epsilon=40\times10^{-16}\ \text{erg}=4\times10^{-22}\ \text{J}$, $\sigma=3\ \text{Å}$, pair separation $R=3.6\ \text{Å}$, hard-sphere fcc packing.

Find. (a) the potential's name; (b) physical meaning of $U<0$; (c)(i) $U(3.6\,\text{Å})$ in J; (c)(ii) the molar cohesive energy in kJ/mol.

Approach. Identify the curve shape (steep short-range repulsion, shallow long-range attractive tail) as the Lennard-Jones form given on the paper as eq. (8); evaluate it directly for (c)(i); use the paper's own lattice-sum result, eq. (9), for the cohesive energy in (c)(ii).

1. Part (a) — name the potential. The functional form printed as eq. (8), $U(R)=4\epsilon\!\left[(\sigma/R)^{12}-(\sigma/R)^6\right]$, with a steep $R^{-12}$ repulsive core and an $R^{-6}$ attractive tail, is the Lennard-Jones (6–12) potential — the standard model for closed-orbital (van der Waals) inert-gas/molecular interactions.
2. Part (b) — meaning of the negative region. For $R/\sigma\gtrsim1$, the $R^{-6}$ (induced-dipole, van der Waals dispersion) attraction dominates over the rapidly-vanishing $R^{-12}$ Pauli-repulsion term, so $U(R)<0$: the pair of atoms is in a lower energy state than when infinitely separated. This is the bound, attractive regime — the net force is attractive ($-dU/dR<0$ is wrong sign convention; correctly, $F=-dU/dR>0$ pulls the atoms together) — and it is precisely this negative well that holds an inert-gas (or, here, H₂) solid together at low temperature.
3. Part (c)(i) — evaluate $U(R)$ at $R=3.6\,\text{Å}$. With $\sigma/R=3/3.6=5/6=0.8333$: $$\frac{U(R)}{\epsilon}=4\left[(0.8333)^{12}-(0.8333)^{6}\right]=4(0.1122-0.3349)=-0.891$$ Converting $\epsilon=40\times10^{-16}\ \text{erg}=4\times10^{-22}\ \text{J}$: $$U(3.6\,\text{Å})=(-0.891)(4\times10^{-22}\ \text{J})$$ $\boxed{U(3.6\,\text{Å})\approx-3.56\times10^{-22}\ \text{J}}$ — negative, confirming the pair sits in the attractive well found in Part (b).
4. Part (c)(ii) — cohesive energy per mole via eq. (9). For $N$ atoms in an fcc lattice of hard-sphere inert-gas-type molecules, the paper's own lattice-sum result at equilibrium spacing is eq. (9): $U_{tot}=-(2.15)(4N\epsilon)=-8.6\,N\epsilon$. Taking $N=N_A=6.02217\times10^{23}\ \text{mol}^{-1}$ (i.e. evaluating cohesive energy per mole of H₂ molecules): $$U_{tot}=-8.6\,(6.02217\times10^{23}\ \text{mol}^{-1})(4\times10^{-22}\ \text{J})=-2071.6\ \text{J/mol}$$ $\boxed{U_{tot}\approx-2.07\ \text{kJ/mol}}$ — small but physically reasonable: real solid H₂ has a cohesive energy of order 1 kJ/mol, so this order-of-magnitude match supports the hard-sphere/fcc model used here.