Question

The $\mathit{r}$ dependence of pair potential functions. You have two different pair potential functions: $$u_1(r)=-\frac{1}{r}\text{ and }{u}_2(r)=-\frac{1}{r^6}.$$ (a) Plot both functions. (b) At $r=1$, which pair potential has the stronger attraction? (c) At $r=2$, which pair potential has the stronger attraction?

Short answer

(a) Plot the functions. (b) Both have equal attraction at $r=1$. (c) $u_1(r)$ has stronger attraction at $r=2$.

Step-by-step solution

Understand the pair potential functions

Two pair potential functions are given: $$u_1(r)=-\frac{1}{r}\text{and}{u}_2(r)=-\frac{1}{r^6}.$$These functions describe the potential energy as a function of the distance $r$ between two particles.

Plot the functions

For plotting both functions, use a range of $r$ values, such as from 0.1 to 5. Use a graphing tool or software to plot the functions. The plot for $u_1(r)$ will generally show a hyperbolic decay, while $u_2(r)$ will display a much steeper decay.

Evaluate the pair potentials at $r=1$

Calculate each function at $r=1$: $$u_1(1)=-\frac{1}{1}=-1\text{and}{u}_2(1)=-\frac{1}{1^6}=-1.$$Both functions have the same value at $r=1$.

Compare the pair potentials at $r=1$

Since both $u_1(1)$ and $u_2(1)$ are equal to $-1$, both pair potentials have the same strength of attraction at $r=1$.

Evaluate the pair potentials at $r=2$

Calculate each function at $r=2$: $$u_1(2)=-\frac{1}{2}=-0.5\text{and}{u}_2(2)=-\frac{1}{2^6}=-\frac{1}{64}=-0.015625.$$

Compare the pair potentials at $r=2$

Compare the calculated values $u_1(2)=-0.5$ and $u_2(2)=-0.015625$. Since $-0.5$ is less than $-0.015625$, the pair potential $u_1(2)$ has a stronger attraction at $r=2$.

Key concepts

Pair Potential Energy

Pair potential energy describes the potential energy between two particles as a function of the distance between them. It's important because it helps us understand how particles interact in various molecules and systems. For example, consider the two pair potential functions:

• $u_1(r)=-\frac{1}{r}$
• $u_2(r)=-\frac{1}{r^6}$

These functions show how the potential energy between two particles changes with the distance $r$. While both functions describe attraction, they do so differently. Understanding these differences aids in predicting molecular behavior.

Distance Dependence

The distance $r$ between two particles significantly affects their potential energy. In the given functions:

• $u_1(r)=-\frac{1}{r}$ shows a hyperbolic decay. It indicates that as the distance increases, the potential energy decreases relatively slowly.
• $u_2(r)=-\frac{1}{r^6}$ shows a much steeper decay. Here, the potential energy decreases very quickly with an increase in distance.

The difference in how fast the potential energy decays with distance explains how strong and over what range these forces act.

Potential Energy Comparison

To compare the pair potentials at specific distances, we can evaluate them at given points:

• At $r=1$:$u_1(1)=-1$ and $u_2(1)=-1$. Both functions provide the same potential energy. Thus, the attraction strength is equal.
• At $r=2$:$u_1(2)=-0.5$ and $u_2(2)=-0.015625$. Here, $u_1$ has a stronger attraction because $-0.5$ is a larger negative value than $-0.015625$.

These comparisons show that $u_1$ maintains a stronger attraction at larger distances compared to $u_2$.

Molecular Interactions

Molecular interactions are deeply connected to the concept of pair potential energy. Different molecules exhibit different types of forces:

$u_1(r)=-\frac{1}{r}$ can represent long-range interactions such as electrostatic forces.

$u_2(r)=-\frac{1}{r^6}$ may represent short-range forces like van der Waals attractions.

Understanding these interactions is crucial in fields such as chemistry, physics, and material science.By analyzing pair potential functions, scientists can predict how molecular systems will behave under various conditions, helping in designing new materials and drugs.