The potential energy between two atoms in a molecule is given by `U(x)= (a)/(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 equilibrium when

Options

A. x=0
B. `x=((a)/(2b))^(1//6)`
C. `x=((2a)/(b))^(1//6)`
D. `x=((11a)/(5b))^(1//6)`

Explanation

To determine the distance \( x \) at which two atoms in a molecule are in stable equilibrium, we start with the given potential energy function:

\[ U(x) = \frac{a}{x^{12}} - \frac{b}{x^{6}} \]

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 by taking the negative derivative of \( U(x) \) with respect to \( x \):

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

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

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

Using the power rule for differentiation, we have:

\[ \frac{d}{dx} \left( \frac{a}{x^{12}} \right) = -12a x^{-13} \]
\[ \frac{d}{dx} \left( \frac{b}{x^{6}} \right) = -6b x^{-7} \]

Thus, combining these results, we get:

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

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

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

This leads to the equation:

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

### Step 4: Solve for \( x \)
Setting the derivative equal to zero gives:

\[ 12a x^{-13} = 6b x^{-7} \]

Rearranging this equation, we can multiply both sides by \( x^{13} \):

\[ 12a = 6b x^{6} \]

Now, divide both sides by \( 6b \):

\[ x^{6} = \frac{12a}{6b} = \frac{2a}{b} \]

Taking the sixth root of both sides gives:

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

### Final Answer
Thus, the distance \( x \) at which the atoms are in stable equilibrium is:

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

---