Question

The Lennard-Jones (LJ) potential of interaction between two molecules as a function of distance ($r$), is given by $V_{LJ}(r) = 4\epsilon \left[ \left(\frac{\sigma}{r}\right)^{12} - \left(\frac{\sigma}{r}\right)^6 \right]$, where $\epsilon$ and $\sigma$ are constantsThe expression of $r$ at which $V_{LJ}(r)$ reaches minimum is _ _ _.

• A. $\left(\frac{11}{16}\right)^{1/4} \sigma^{3/2}$
• B. $2^{1/6}\sigma$
• C. $\sigma$
• D. $4^{1/5} e^{1/6}$

Hint:

For Lennard-Jones potential, minimum occurs at $r = 2^{1/6}\sigma$ and zero potential occurs at $r = \sigma$

Answer 1

The Correct Option is B

Step 1: Condition for minimum potential.
The potential is minimum when derivative of $V_{LJ}(r)$ with respect to $r$ is zero
\[ \frac{dV_{LJ}}{dr} = 0 \]

Step 2: Differentiate the given expression.
\[ V_{LJ}(r) = 4\epsilon \left[ \left(\frac{\sigma}{r}\right)^{12} - \left(\frac{\sigma}{r}\right)^6 \right] \]
Differentiating and setting equal to zero:
\[ -12\left(\frac{\sigma}{r}\right)^{12} + 6\left(\frac{\sigma}{r}\right)^6 = 0 \]

Step 3: Simplify equation.
\[ 2\left(\frac{\sigma}{r}\right)^{12} = \left(\frac{\sigma}{r}\right)^6 \]
\[ \Rightarrow \left(\frac{\sigma}{r}\right)^6 = \frac{1}{2} \]

Step 4: Solve for $r$.
\[ \frac{\sigma}{r} = \left(\frac{1}{2}\right)^{1/6} \Rightarrow r = 2^{1/6}\sigma \]

Step 5: Conclusion.
\[ \boxed{2^{1/6}\sigma} \]