Lennard-Jones Potential Minimum and Interatomic Force in Krypton: 17-Phys-A6 Solid State Physics, May 2014, Q2

Question 2: Lennard-Jones Potential Minimum and Interatomic Force in Krypton (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. The paper's own Lennard-Jones (6–12) potential, eq. (8): $U(R)=4\epsilon\!\left[(\sigma/R)^{12}-(\sigma/R)^6\right]$, $F(R)=-dU/dR$. For Kr (Table T2): $\epsilon=225\times10^{-16}\ \text{erg}$, $\sigma=3.65\ \text{Å}$. Separation for part (b): $R=5\ \text{Å}$.

Find. (a) the ratio $R/\sigma$ at which $U(R)$ is minimized; (b) the force $F(R)$ between two Kr atoms at $R=5\,\text{Å}$.

Fig. 2 — the normalized Lennard-Jones (6–12) curve $U(R)/\epsilon=4[(\sigma/R)^{12}-(\sigma/R)^6]$, matching Figure P2's shape: steep repulsive wall below $R/\sigma\approx1$, a minimum near $R/\sigma\approx1.12$, and a slowly-vanishing attractive tail.

Approach. (a) set $dU/dR=0$ and solve for $R$ symbolically; (b) differentiate the same functional form to get $F(R)$, then substitute Kr's tabulated $\epsilon,\sigma$ and the given $R$.

Part (a) — locate the minimum. Differentiate eq. (8): $$\frac{dU}{dR}=4\epsilon\left[-\frac{12\sigma^{12}}{R^{13}}+\frac{6\sigma^6}{R^7}\right]$$ Setting $dU/dR=0$ and cancelling common factors, $$\frac{12\sigma^{12}}{R^{13}}=\frac{6\sigma^6}{R^7}\ \Longrightarrow\ R^6=2\sigma^6\ \Longrightarrow\ R_{\min}=2^{1/6}\sigma$$ $$\boxed{R_{\min}/\sigma=2^{1/6}=1.1225\approx1.12}$$ confirming the printed minimum location; substituting back gives $U(R_{\min})/\epsilon=4\!\left[\tfrac14-\tfrac12\right]=-1$, the curve's minimum value shown in Figure P2.
Part (b) — force between two Kr atoms at $R=5\,\text{Å}$. From the same derivative, $F(R)=-dU/dR=\dfrac{24\epsilon}{R}\!\left[2(\sigma/R)^{12}-(\sigma/R)^6\right]$. With $\sigma/R=3.65/5=0.7300$: $$(\sigma/R)^6=0.15133,\quad (\sigma/R)^{12}=0.022902$$ $$F=\frac{24(225\times10^{-16}\ \text{erg})}{5\times10^{-8}\ \text{cm}}\Big[2(0.022902)-0.15133\Big]$$ $$F=(1.080\times10^{-5}\ \text{erg/cm})(-0.10553)$$ $$\boxed{F=-1.140\times10^{-6}\ \text{dyne}=-1.140\times10^{-11}\ \text{N}}$$ The negative sign (in the $F=-dU/dR$ convention) means the force pulls the two Kr atoms together: at $R=5\,\text{Å}>R_{\min}=1.1225(3.65)=4.10\,\text{Å}$, the pair sits on the attractive ($R^{-6}$-dominated) tail of the curve, past the potential well minimum.

Quantity	Result
Minimum location	$R/\sigma=2^{1/6}=1.1225$
$U(R_{\min})/\epsilon$	$-1$
Kr–Kr force at $R=5\,\text{Å}$	$F=-1.140\times10^{-6}\ \text{dyne}=-1.140\times10^{-11}\ \text{N}$ (attractive)

Topic: Lennard-Jones (6–12) pair potential for van der Waals-bonded (inert-gas) solids.

Key relations: $U(R)=4\epsilon[(\sigma/R)^{12}-(\sigma/R)^6]$; $F(R)=-dU/dR$; minimum at $R_{\min}=2^{1/6}\sigma$, $U_{\min}=-\epsilon$.

Why this works: the $R^{-12}$ term is a steep, short-range model of Pauli (electron-overlap) repulsion, while the $R^{-6}$ term is the London dispersion (induced-dipole–induced-dipole) attraction; their competition produces exactly one bound minimum, which is why closed-orbital atoms condense into solids at all despite having no covalent or ionic bonding available.

Common pitfall: dropping the minus sign in $F=-dU/dR$ (reporting a repulsive force where the pair is actually attractive), or mixing up $\sigma$ (the potential's zero-crossing) with $R_{\min}$ (its minimum) — they differ by the factor $2^{1/6}$.

Source: Kittel, Introduction to Solid State Physics, 8th ed., Ch. 3 (Lennard-Jones potential, cohesive energy of inert-gas crystals).

Question 2 of 7: Lennard-Jones Potential Minimum and Interatomic Force in Krypton

Question 2: Lennard-Jones Potential Minimum and Interatomic Force in Krypton (20 marks)