The potential energy between two atoms in a molecule is given by `U(x)= (1)/(x^(12))-(b)^(x^(6))`, where a and b are positive constants and x is the distance between the atoms. The atom is in stable equilibirum when

Options

A. `x = 6sqrt((11a)/(5b))`
B. `x = 6sqrt((a)/(2b))`
C. `x =0`
D. `x = 6sqrt((2a)/(b))`

Explanation

To determine the condition for stable equilibrium between two atoms in a molecule, we start with the potential energy function given by:

\[ U(x) = \frac{A}{x^{12}} - \frac{B}{x^{6}} \]

where \( A \) and \( B \) are positive constants, and \( x \) is the distance between the atoms.

### Step 1: Find the Force
The force \( F \) between the atoms can be derived from the potential energy \( U(x) \) using the relation:

\[ F = -\frac{dU}{dx} \]

### Step 2: Differentiate \( U(x) \)
Now, we differentiate \( U(x) \):

\[ \frac{dU}{dx} = \frac{d}{dx} \left( \frac{A}{x^{12}} - \frac{B}{x^{6}} \right) \]

Using the power rule for differentiation:

1. The derivative of \( \frac{A}{x^{12}} \) is:
\[ -12A x^{-13} \]
2. The derivative of \( -\frac{B}{x^{6}} \) is:
\[ +6B x^{-7} \]

Putting it together, we have:

\[ \frac{dU}{dx} = -12A x^{-13} + 6B x^{-7} \]

### Step 3: Set the Force to Zero for Equilibrium
For stable equilibrium, the force must be zero:

\[ -\frac{dU}{dx} = 0 \]

Thus, we set the derivative equal to zero:

\[ -(-12A x^{-13} + 6B x^{-7}) = 0 \]

This simplifies to:

\[ 12A x^{-13} = 6B x^{-7} \]

### Step 4: Rearranging the Equation
Rearranging the equation gives:

\[ 12A x^{-13} = 6B x^{-7} \]

Dividing both sides by \( x^{-7} \):

\[ 12A x^{-6} = 6B \]

### Step 5: Solve for \( x \)
Now, we can solve for \( x \):

\[ x^{-6} = \frac{6B}{12A} \]

This simplifies to:

\[ x^{-6} = \frac{B}{2A} \]

Taking the reciprocal gives:

\[ x^{6} = \frac{2A}{B} \]

Finally, taking the sixth root:

\[ x = \left(\frac{2A}{B}\right)^{\frac{1}{6}} \]

### Conclusion
Thus, the distance \( x \) at which the atoms are in stable equilibrium is given by:

\[ x = \left(\frac{2A}{B}\right)^{\frac{1}{6}} \]