Animated Solution for Physics - Work, Energy, and Power: The potential energy ($U$) of a diatomic molecule is a function dependent on $r$ (interatomic distance) as $U=\frac{\alpha }{r^{10}}-\frac{\beta }{r^5}$, where $\alpha$ and $\beta$ are positive constants. The equilibrium distance between two atoms will be ${\left(\frac{2\alpha }{\beta }\right)}^{\frac{a}{b}}$, where $a=\dots \dots$ .

Imagine two atoms in a diatomic molecule. They are engaged in a constant tug-of-war. If they get too close, their electron clouds overlap, creating a massive repulsive force. If they move too far apart, long-range attractive forces (like Van der Waals forces) try to pull them back together.

This delicate balance is perfectly captured by the potential energy function given in the problem:

The positive term $\frac{\alpha }{r^{10}}$ dominates at very small distances, representing the sharp repulsion. The negative term $-\frac{\beta }{r^5}$ dominates at larger distances, representing the attraction. The equilibrium distance is the exact sweet spot where these two forces cancel each other out, resulting in a net force of zero.

In physics, conservative forces are directly related to potential energy through the negative gradient:

For the atoms to be in equilibrium, the net force must be zero ($F=0$). This implies that we need to find the minimum of the potential energy curve, which occurs where the derivative of $U(r)$ with respect to $r$ is zero:

Let's differentiate our potential energy function. To avoid silly mistakes with fractions, it is always best to rewrite the terms using negative exponents:

Now, applying the standard power rule $\frac{d}{dr}(r^n)=n{r}^{n-1}$:

Simplifying the expression by multiplying out the constants and signs:

To isolate $r$, let's move the negative term to the right side of the equation:

Now, let's rewrite the negative exponents back as fractions to make the cross-multiplication visually intuitive:

By cross-multiplying, we can group all the $r$ terms on one side and the constants on the other:

Using the laws of exponents ($r^{11}/r^6=r^{11-6}=r^5$), we simplify the left side. On the right side, $10/5$ simplifies to $2$:

Taking the fifth root of both sides gives us the equilibrium distance:

The problem states that the equilibrium distance is of the form ${\left(\frac{2\alpha }{\beta }\right)}^{\frac{a}{b}}$. By directly comparing our derived expression with the given format, we can see that the exponent is $\frac{1}{5}$.

Therefore, $\frac{a}{b}=\frac{1}{5}$, which means the value of the numerator $a$ is exactly $1$.

As a thought experiment, consider what happens if you slightly perturb the atoms from this equilibrium distance $r_{eq}$. The restoring force will cause them to oscillate, executing Simple Harmonic Motion! The stiffness of this "molecular spring" can be found by taking the second derivative of the potential energy, $\frac{d^{2}U}{d{r}^2}$, evaluated at $r_{eq}$.