Question

The potential energy of a diatomic molecule in terms of the interatomic separation \(R\) is given by \[ U(R)=-\frac{A}{R^2}+\frac{B}{R^{10}} \] where \(A\) and \(B\) are constants. For stable equilibrium at \(R=R_e\), the value of \(R_e\) is:

• A. \(\left(\dfrac{9B}{A}\right)^{1/8}\)
• B. \(\left(\dfrac{5B}{A}\right)^{1/8}\)
• C. \(\left(\dfrac{5B}{A}\right)^{1/6}\)
• D. \(\left(\dfrac{8B}{A}\right)^{1/6}\)

Hint:

At stable equilibrium, differentiate potential energy and put \(\frac{dU}{dR}=0\).

Answer 1

The Correct Option is B

Concept:
For stable equilibrium, potential energy is minimum. Therefore: \[ \frac{dU}{dR}=0 \]

Step 1: Write the given potential energy. \[ U(R)=-\frac{A}{R^2}+\frac{B}{R^{10}} \] \[ U(R)=-AR^{-2}+BR^{-10} \]

Step 2: Differentiate with respect to \(R\). \[ \frac{dU}{dR}=2AR^{-3}-10BR^{-11} \]

Step 3: Apply equilibrium condition. \[ 2AR^{-3}-10BR^{-11}=0 \] \[ 2AR^{-3}=10BR^{-11} \] \[ 2AR^8=10B \] \[ R^8=\frac{5B}{A} \] \[ R=\left(\frac{5B}{A}\right)^{1/8} \] \[ \therefore \text{Correct Answer is (B)} \]