The potential energy between two atoms in a molecule is given by
`U=ax^(2)-bx^(2)`
where `a` and `b` are positive constants and `x` is the distance between the atoms. The atom is in stable equilibrium when `x` is equal to `:-`

Options

A. `0`
B. `(2b)/(3a)`
C. `(3a)/(2b)`
D. `(3b)/(2a)`

Explanation

To find the distance \( x \) at which the atom is in stable equilibrium, we start with the given potential energy equation:

\[ U = ax^2 - bx^3 \]

where \( a \) and \( b \) are positive constants.

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

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

### Step 2: Differentiate the Potential Energy
Now, we differentiate \( U \) with respect to \( x \):

\[ \frac{dU}{dx} = \frac{d}{dx}(ax^2 - bx^3) = 2ax - 3bx^2 \]

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

\[ F = -\frac{dU}{dx} = 0 \implies 2ax - 3bx^2 = 0 \]

### Step 4: Factor the Equation
We can factor out \( x \):

\[ x(2a - 3bx) = 0 \]

This gives us two solutions:

1. \( x = 0 \)
2. \( 2a - 3bx = 0 \)

### Step 5: Solve for \( x \)
From the second equation, we can solve for \( x \):

\[ 2a = 3bx \implies x = \frac{2a}{3b} \]

### Step 6: Check Stability
To ensure that this point is a point of stable equilibrium, we need to check the second derivative of \( U \):

\[ \frac{d^2U}{dx^2} = \frac{d}{dx}(2ax - 3bx^2) = 2a - 6bx \]

Now, substituting \( x = \frac{2a}{3b} \):

\[ \frac{d^2U}{dx^2} = 2a - 6b\left(\frac{2a}{3b}\right) = 2a - 4a = -2a \]

Since \( a > 0 \), \( \frac{d^2U}{dx^2} < 0 \), indicating that the equilibrium is stable.

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

\[ x = \frac{2a}{3b} \]