91Ó°ÊÓ

The given equation of the potential is,

$U\left(r\right)=U_0\left[{\left(\frac{R_0}{r}\right)}^{12}−2{\left(\frac{R_0}{r}\right)}^6\right]$ …… (1)

The force F ( r ) is given as,

$\mathrm{F}\left(\mathrm{r}\right)=−\frac{\mathrm{dU}\left(\mathrm{r}\right)}{\mathrm{dr}}$ …… (2)

From equation one, solve for the first derivative of the potential energy,

localid="1668311432381" $$\begin{gathered} \frac{\mathrm{dU}\left(\mathrm{r}\right)}{\mathrm{dr}}={\mathrm{U}}_0\frac{\mathrm{d}}{\mathrm{dr}}\left[{\left(\frac{{\mathrm{R}}_0}{\mathrm{r}}\right)}^{12}−2{\left(\frac{{\mathrm{R}}_0}{\mathrm{r}}\right)}^6\right] \\ ={\mathrm{U}}_0\left[−12\frac{{\mathrm{R}}_0^{12}}{{\mathrm{r}}^{13}}−(−12)\frac{{\mathrm{R}}_0^6}{{\mathrm{r}}^7}\right] \\ =12\frac{{\mathrm{U}}_0}{{\mathrm{R}}_0}\left[−\frac{{\mathrm{R}}_0^{13}}{{\mathrm{r}}^{13}}+\frac{{\mathrm{R}}_0^7}{{\mathrm{r}}^7}\right] \end{gathered}$$

Substitute this value in equation (2),

$F=12\left(\frac{U_0}{R_0}\right)\left[\frac{R_0^{13}}{r^{13}}−\frac{R_0^7}{r^7}\right]$

Therefore, the required graphs are plotted as:-

The graphs in part (a) exactly shows the location of points $r_1$and$r_2$ . For equilibrium of the system, F = 0 . Thus, the point of equilibrium is r₂ because that is where the potential energy U is minimum value.

(c)Put equation (i) equal to zero, i.e. U = 0 at point$r=r_1$.

localid="1668311453049" $\begin{array}{r}{U}_0\left[{\left(\frac{R_0}{r_1}\right)}^{12}−2{\left(\frac{R_0}{r_1}\right)}^6\right]=0\\ {\left(\frac{R_0}{r_1}\right)}^{12}=2{\left(\frac{R_0}{r_1}\right)}^6\\ r_1=\frac{R_0}{(2{)}^{\frac{1}{6}}}\end{array}$

At $r−r_2,F=0$

$\begin{array}{r}12\left(\frac{U_0}{R_0}\right)\left[\frac{R_0^{13}}{{r_2}^{13}}−\frac{R_0^7}{{r_2}^7}\right]=0\\ {\left(\frac{r_2}{R_0}\right)}^6=1\\ r_2=R_0\end{array}$

So, $r_1$and $r_2$in terms of $R_0$is

$r_2=R_0$and

$r_1=\frac{R_0}{(2{)}^{\frac{1}{6}}}$

The ratio of $r_1$and $r_2$is,

$\begin{array}{r}\frac{r_1}{r_2}=\frac{(2{)}^{\frac{1}{6}}}{R_0}\\ =(2{)}^{−\frac{1}{6}}\end{array}$

The work done is equal to the change in potential energy and here the molecule’s potential energy has to be calculated at infinity and then at a distance $r=R_0$ .So, at.

localid="1668227337785" $\begin{array}{r}\mathrm{r}→\mathrm{∞},\mathrm{U}=0\\ \mathrm{W}=\mathrm{Δ\pm «}\\ =\mathrm{U}\left(\mathrm{∞}\right)−\mathrm{U}\left({\mathrm{R}}_0\right)\\ =−{\mathrm{U}}_0\left[{\left(\frac{{\mathrm{R}}_0}{{\mathrm{R}}_0}\right)}^{12}−2{\left(\frac{{\mathrm{R}}_0}{{\mathrm{R}}_0}\right)}^6\right]\\ =+{\mathrm{U}}_0\end{array}$

Thus, the work done in this case is equal to the depth of the potential well $+U_0$.