Step-by-step Solution:

1. Relationship between Force and Potential Energy: The force F acting on a particle moving in a one-dimensional potential V(x) is given by the negative gradient (or derivative) of the potential energy with respect to position: $F = - \frac{dV(x)}{dx}$

2. Analyze the Potential Energy for $|x| > X_0$: The problem states that for the region $|x| > X_0$, the potential energy V(x) is a constant value, $V_0$. $V(x) = V_0 \quad \text{for} \quad |x| > X_0$

3. Calculate the Force in this Region: To find the force on the particle in this region, we take the derivative of the potential energy function with respect to x. $F = - \frac{d}{dx}(V_0)$ Since $V_0$ is a constant, its derivative with respect to x is zero. $F = -0 = 0$ Therefore, the net force acting on the particle for $|x| > X_0$ is zero.

4. Relate Force to Acceleration using Newton's Second Law: According to Newton's Second Law of Motion, the net force on a particle is equal to the product of its mass m and its acceleration a. $F = ma$

5. Determine the Acceleration: We substitute the force we found in step 3 into Newton's Second Law: $0 = ma$ Since the particle has mass m (which is non-zero), for the product ma to be zero, the acceleration a must be zero. $a = 0$

6. Conclusion: The acceleration of the particle for $|x| > X_0$ is zero. Comparing this result with the given options:

   • A: proportional to V0. (Incorrect)
   • B: proportional to V0/mX0. (Incorrect)
   • C: proportional to $\sqrt {{V_0}/m{X_0}}$. (Incorrect)
   • D: zero. (Correct)

Thus, the correct option is D.