Question

Potential energy as a function of \(r\) is given by \( U = \frac{A}{r^{10}} - \frac{B}{r^5} \), where \(r\) is the interatomic distance, \(A\) and \(B\) are positive constants. The equilibrium distance between the two atoms will be:

• A. \( \left( \frac{A}{B} \right)^{\frac{1}{5}} \)
• B. \( \left( \frac{B}{A} \right)^{\frac{1}{5}} \)
• C. \( \left( \frac{2A}{B} \right)^{\frac{1}{5}} \)
• D. \( \left( \frac{B}{2A} \right)^{\frac{1}{5}} \)

Hint:

To find the equilibrium distance, differentiate the potential energy function and set the derivative equal to zero to find the value of \( r \).

Answer 1

The Correct Option is C

Step 1: The equilibrium position is obtained by differentiating the potential energy function with respect to \( r \) and setting the derivative equal to zero. \[ \frac{dU}{dr} = -\frac{10A}{r^{11}} + \frac{5B}{r^6} \] Setting \( \frac{dU}{dr} = 0 \) for equilibrium: \[ -\frac{10A}{r^{11}} + \frac{5B}{r^6} = 0 \] \[ \frac{10A}{r^{11}} = \frac{5B}{r^6} \] \[ \frac{2A}{r^5} = B \] \[ r^5 = \frac{2A}{B} \] \[ r = \left( \frac{2A}{B} \right)^{\frac{1}{5}} \] 
Step 2: The equilibrium distance between the atoms is \( r = \left( \frac{2A}{B} \right)^{\frac{1}{5}} \).