The potential energy function for the force between two atoms in a diatomic molecule is approximatel

Q: Why do we differentiate the potential energy to find equilibrium?

In physics, force is the negative gradient of potential energy ($F = -dU/dx$). At equilibrium, the net force acting on the system is zero. Therefore, setting the derivative of potential energy to zero allows us to find the exact distance where the attractive and repulsive forces perfectly balance each other.

Q: Why did we solve for $x^6$ instead of just $x$?

If you look closely at the potential energy function, it only contains terms with $x^{12}$ and $x^6$. Since $x^{12}$ is just $(x^6)^2$, substituting the value of $x^6$ directly into the equation is much faster and avoids messy fractional exponents. It's a smart algebraic shortcut!

Q: What does the negative sign in the minimum potential energy mean?

A negative potential energy indicates a bound state. It means the atoms are trapped in a 'potential well' and are more stable together than they are apart. You would have to add positive energy (work) to the system to break them apart and bring their energy up to zero.

Q: Why is the energy at infinity considered to be zero?

This is a standard convention in physics. When two atoms are infinitely far apart, they no longer interact with each other. The forces between them are zero, so we define this state of no interaction as our reference point, setting its potential energy to zero.

Q: Is the dissociation energy always positive?

Yes! Dissociation energy is defined as the energy *required* to break a bond. Since you have to supply energy to overcome the attractive forces holding the molecule together, this value must always be positive.

Question

The potential energy function for the force between two atoms in a diatomic molecule is approximately given by $U(x) = \frac{a}{x^{12}} - \frac{b}{x^6}$, where $a$ and $b$ are constants and $x$ is the distance between the atoms. If the dissociation energy of the molecule is $D = [U(x = \infty) - U_{	ext{at equilibrium}}]$, $D$ is

Options:

(A) $\frac{b^2}{2a}$

(B) $\frac{b^2}{12a}$

(C) $\frac{b^2}{4a}$

(D) $\frac{b^2}{6a}$

Sigma Percentile · Accepted Answer

### The Tug-of-War Between Atoms

Imagine you are holding two powerful magnets. If you bring them close, they might snap together, but if you try to push them *too* close, their electron clouds will fiercely repel each other. This delicate dance between attraction and repulsion is the heart of chemical bonding, and it is beautifully captured by the **Lennard-Jones potential**.

In our problem, the potential energy function is given by:

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

Here, the positive term $\frac{a}{x^{12}}$ represents the short-range repulsive force (due to the Pauli exclusion principle preventing electron clouds from overlapping), while the negative term $-\frac{b}{x^6}$ represents the long-range attractive van der Waals force. The distance $x$ is the separation between the two atoms.

### Defining Dissociation Energy

The question asks us to find the **dissociation energy**, denoted by $D$. Think of dissociation energy as the ransom you must pay to completely free the atoms from each other's grasp.

Mathematically, it is the difference between the energy of the atoms when they are infinitely far apart (completely free) and their energy when they are resting comfortably at their most stable equilibrium distance:

$$D = U(x = \infty) - U_{	ext{at equilibrium}}$$

First, let's evaluate the energy at infinity. As the distance $x$ approaches $\infty$, both $\frac{a}{x^{12}}$ and $\frac{b}{x^6}$ approach zero. Therefore, the potential energy at infinite separation is simply zero:

$$U(x = \infty) = 0$$

### Finding the Sweet Spot: Equilibrium

To find the dissociation energy, we need to know how deep the potential well is. This requires us to find the equilibrium position.

In physics, equilibrium occurs where the net force acting on the system is zero. Since conservative force is the negative gradient of potential energy ($F = -\frac{dU}{dx}$), we must find the distance $x$ where the derivative of $U(x)$ is zero.

Let's differentiate our potential energy function:

$$\frac{dU}{dx} = \frac{d}{dx}\left(ax^{-12} - bx^{-6}ight)$$

Using the power rule, we get:

$$\frac{dU}{dx} = -12ax^{-13} - (-6)bx^{-7} = -\frac{12a}{x^{13}} + \frac{6b}{x^7}$$

Setting this derivative to zero to find the equilibrium point:

$$-\frac{12a}{x^{13}} + \frac{6b}{x^7} = 0$$

$$\frac{12a}{x^{13}} = \frac{6b}{x^7}$$

By cross-multiplying and simplifying, we can isolate $x^6$:

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

### Calculating the Depth of the Well

Notice a brilliant algebraic shortcut here: we don't actually need to solve for $x$ itself! Our original potential energy function only contains $x^{12}$ and $x^6$. Since $x^{12}$ is simply $(x^6)^2$, we can directly substitute $x^6 = \frac{2a}{b}$ back into the energy equation.

Let's calculate the minimum potential energy, $U_{	ext{eq}}$:

$$U_{	ext{eq}} = \frac{a}{(x^6)^2} - \frac{b}{x^6}$$

Substituting our value for $x^6$:

$$U_{	ext{eq}} = \frac{a}{\left(\frac{2a}{b}ight)^2} - \frac{b}{\left(\frac{2a}{b}ight)}$$

Now, we carefully simplify the fractions:

$$U_{	ext{eq}} = \frac{a}{\left(\frac{4a^2}{b^2}ight)} - \frac{b^2}{2a}$$

$$U_{	ext{eq}} = \frac{ab^2}{4a^2} - \frac{b^2}{2a}$$

$$U_{	ext{eq}} = \frac{b^2}{4a} - \frac{b^2}{2a}$$

To subtract these, we find a common denominator of $4a$:

$$U_{	ext{eq}} = \frac{b^2}{4a} - \frac{2b^2}{4a} = -\frac{b^2}{4a}$$

This negative value confirms that the atoms are in a stable, bound state. They are trapped in a potential well of depth $\frac{b^2}{4a}$.

### The Final Leap to Infinity

Finally, we return to our definition of dissociation energy. We need to calculate the energy required to lift the system from the bottom of this well ($-\frac{b^2}{4a}$) all the way up to zero energy at infinity.

$$D = U(x = \infty) - U_{	ext{eq}}$$

$$D = 0 - \left(-\frac{b^2}{4a}ight)$$

$$D = \frac{b^2}{4a}$$

And there we have it! The dissociation energy of the molecule is exactly **$\frac{b^2}{4a}$**.

Select Answer:

Visualized Solution

$U (x)$ and Dissociation Energy

Energy at Infinity

Condition for Equilibrium

Differentiating $U (x)$

Solving for Equilibrium Distance

Calculating $U_{eq}$

Simplifying $U_{eq}$

Final Dissociation Energy

The Sigma Insight: Kinetic Energy, Potential Energy and Power

The Tug-of-War Between Atoms

Defining Dissociation Energy

Finding the Sweet Spot

Calculating the Depth of the Well

The Final Leap to Infinity

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