Chapter 8: Q. 8.9 (page 338)

Show that the Lennard-Jones potential reaches its minimum value at $r = r_{0}$ , and that its value at this minimum is $- u_{0}$ . At what value of $r$ does the potential equal zero?

Short Answer

Expert verified

At, $r = \frac{r_{0}}{2^{\frac{1}{6}}}$ the value of Lennard-Jones potential become 0.

Step by step solution

Step 1. Given information

Lennard- Jones potential=

$u (r) = u_{0} [{(\frac{r_{0}}{r})}^{12} - 2 {(\frac{r_{0}}{r})}^{6}]$

Step 2. Differentiating both sides with respect to r

We get,

$\frac{d u}{d r} = u_{0} [r_{0}^{12} \times - 12 \frac{1}{r^{13}} - 2 r_{0}^{6} \times - 6 \frac{1}{r^{7}}]$

= u_{0} [- 12 \frac{r_{0}^{12}}{r^{13}} + 12 \frac{r_{0}^{6}}{r^{7}}]

For extreme points,

$\frac{d u}{d r} = 0$

$∴ u_{0} [- 12 \frac{r_{0}^{12}}{r^{13}} + 12 \frac{r_{0}^{6}}{r^{7}}] = 0$

$\frac{12 r_{0}^{6} u_{0}}{r^{13}} [r^{6} - r_{0}^{6}] = 0$

$r^{6} - r_{0}^{6} = 0$

$r = r_{0}$

Thus the Lennard-Jones potential has extremum point at $r = r_{0}$

Step 3. To check that the potential at r=r0 is maximum or minimum take double derivative

We get,

\frac{d^{2} u}{d r^{2}} = u_{0} [- 12 \times \frac{- 13 r_{0}^{12}}{r^{14}} + 12 \times \frac{- 7 r_{0}^{6}}{r^{8}}]

$= 12 u_{0} [\frac{13 r_{0}^{12}}{r^{14}} - \frac{7 r_{0}^{6}}{r^{8}}]$

${\frac{d^{2} u}{d r^{2}}|}_{r = s_{0}} = 12 u_{0} [\frac{13 r_{0}^{12}}{r_{0}^{14}} - \frac{7 r_{0}^{6}}{r_{0}^{8}}]$

$= 12 u_{0} [\frac{13}{r_{0}^{2}} - \frac{7}{r_{0}^{2}}]$

$= \frac{12 u_{0}}{r_{0}^{2}} [6]$

$= \frac{72 u_{0}}{r_{0}^{2}} > 0$

${∴ \frac{d^{2} u}{d r^{2}}|}_{r = r_{0}} > 0$

So the Lennard- Jones potential has minimum value at $r^{'} = r_{0}$

${u (r)|}_{r = r_{0}} = u_{0} {[{(\frac{r_{0}}{r})}^{12} - 2 {(\frac{r_{0}}{r})}^{6}]}_{r = r_{0}}$

$= u_{0} [{(\frac{r_{0}}{r})}^{12} - 2 {(\frac{r_{0}}{r})}^{6}]$

$= u_{0} [1 - 2]$

${u (r)|}_{r = r_{0}} = - u_{0}$

Step 4. If the potential becomes 0.

$u (r) = 0$

$u_{0} [{(\frac{r_{0}^{*}}{r})}^{12} - 2 {(\frac{r_{0}}{r})}^{6}] = 0$

$\frac{u_{0}}{r^{12}} [r_{0}^{12} - 2 r_{0}^{6} r^{6}] = 0$

$r_{0}^{12} - 2 r_{0}^{6} r^{6} = 0$

$r^{6} = \frac{r_{0}^{12}}{2 r_{0}^{6}} = \frac{r_{0}^{6}}{2}$

$r^{6} = \frac{r_{0}^{12}}{2 r_{0}^{6}}$

$r = \frac{1}{2^{\frac{1}{6}}} r_{0}$

Thus the Lennard-Jones potential becomes zero at $r = \frac{1}{2^{\frac{1}{6}}} r_{0}$

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Show that the Lennard-Jones potential reaches its minimum value at $r = r_{0}$ , and that its value at this minimum is $- u_{0}$ . At what value of $r$ does the potential equal zero?

Short Answer

Step by step solution

Step 1. Given information

Step 2. Differentiating both sides with respect to r

Step 3. To check that the potential at r=r0 is maximum or minimum take double derivative

Step 4. If the potential becomes 0.

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Show that the Lennard-Jones potential reaches its minimum value at r=r0, and that its value at this minimum is -u0. At what value of rdoes the potential equal zero?

Short Answer

Step by step solution

Step 1. Given information

Step 2. Differentiating both sides with respect to r

Step 3. To check that the potential at r=r0 is maximum or minimum take double derivative

Step 4. If the potential becomes 0.

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Physics Textbooks

Waves Physics

Engineering Physics

Turning Points in Physics

Fields in Physics

Physics of Motion

Torque and Rotational Motion

Study anywhere. Anytime. Across all devices.

Show that the Lennard-Jones potential reaches its minimum value at $r = r_{0}$ , and that its value at this minimum is $- u_{0}$ . At what value of $r$ does the potential equal zero?