Question

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

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=

$u\left(r\right)=u_0\left[{\left(\frac{r_0}{r}\right)}^{12}-2{\left(\frac{r_0}{r}\right)}^6\right]$

Step 2. Differentiating both sides with respect to r

We get,

$\frac{du}{dr}=u_0\left[r_0^{12}\times -12\frac{1}{r^{13}}-2r_0^6\times -6\frac{1}{r^7}\right]$

$=u_0\left[-12\frac{r_0^{12}}{r^{13}}+12\frac{r_0^6}{r^7}\right]$

For extreme points,

$\frac{du}{dr}=0$

$\therefore u_0\left[-12\frac{r_0^{12}}{r^{13}}+12\frac{r_0^6}{r^7}\right]=0$

$\frac{12r_0^6{u}_0}{r^{13}}\left[r^6-r_0^6\right]=0$

$r^6-r_0^6=0$

$r=r_0$

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,

$\frac{d^{2}u}{d{r}^2}=u_0\left[-12\times \frac{-13r_0^{12}}{r^{14}}+12\times \frac{-7r_0^6}{r^8}\right]$

$=12u_0\left[\frac{13r_0^{12}}{r^{14}}-\frac{7r_0^6}{r^8}\right]$

${\left.\frac{d^{2}u}{d{r}^2}\right|}_{r=s_0}=12u_0\left[\frac{13r_0^{12}}{r_0^{14}}-\frac{7r_0^6}{r_0^8}\right]$

$=12u_0\left[\frac{13}{r_0^2}-\frac{7}{r_0^2}\right]$

$=\frac{12u_0}{r_0^2}\left[6\right]$

$=\frac{72u_0}{r_0^2}>0$

${\left.\therefore \frac{d^{2}u}{d{r}^2}\right|}_{r=r_0}>0$

So the Lennard- Jones potential has minimum value at $r^{\text{'}}=r_0$

${\left.u\left(r\right)\right|}_{r=r_0}=u_0{\left[{\left(\frac{r_0}{r}\right)}^{12}-2{\left(\frac{r_0}{r}\right)}^6\right]}_{r=r_0}$

$=u_0\left[{\left(\frac{r_0}{r}\right)}^{12}-2{\left(\frac{r_0}{r}\right)}^6\right]$

$=u_0[1-2]$

${\left.u\left(r\right)\right|}_{r=r_0}=-u_0$

Step 4. If the potential becomes 0.

$u\left(r\right)=0$

$u_0\left[{\left(\frac{r_0^{*}}{r}\right)}^{12}-2{\left(\frac{r_0}{r}\right)}^6\right]=0$

$\frac{u_0}{r^{12}}\left[r_0^{12}-2r_0^6{r}^6\right]=0$

$r_0^{12}-2r_0^6{r}^6=0$

$r^6=\frac{r_0^{12}}{2r_0^6}=\frac{r_0^6}{2}$

$r^6=\frac{r_0^{12}}{2r_0^6}$

$r=\frac{1}{2^{\frac{1}{6}}}{r}_0$

Thus the Lennard-Jones potential becomes zero at$r=\frac{1}{2^{\frac{1}{6}}}{r}_0$