Figure 8-73a shows a molecule consisting of two atoms of masses mand m(with$\mathit{m}\mathbf{\ll }\mathit{M}$) and separation r. Figure 8-73b shows the potential energy $\mathit{U}\left(r\right)$of the molecule as a function of r. Describe the motion of the atoms (a) if the total mechanical energy Eof the two-atom system is greater than zero (as is${\mathit{E}}_{\mathbf{1}}$), and (b) if Eis less than zero (as is${\mathit{E}}_{\mathbf{2}}$). For ${\mathit{E}}_{\mathbf{1}}\mathbf{=}\mathbf{1}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{19}}\mathit{J}$and r=0.3nm, find (c) the potential energy of the system, (d) the total kinetic energy of the atoms, and (e) the force (magnitude and direction) acting on each atom. For what values of ris the force (f) repulsive, (g) attractive, and (h) zero?

Mechanical energy of system is$1\times {10}^{-19}Jatr=0.3nm$

Here we can use the formula for total energy to find kinetic energy and potential energy with the given conditions and graph in the problem.

Formula:

The total energy of the system, E=K+U (i)

From the given figure, we observe that the horizontal line representing intersects the potential energy curve at an approximate value ofr=0.07nm and it seems that the horizontal line will not intersect the curve at larger. Therefore, if m were thrown towards M from large r with energy$E_1$_, it would turn around at 0.07nm and head back in the direction from which it came.

Also from the figure, we observe that the line representing has two intersections points $r_1=0.16nm$and $r_2=0.28nm$with$U\left(t\right)$ plot. Therefore, if m starts in region$r_1<r<r_2$ with energy $E_2$, it will bounce back and forth between these two points.

From the Figure 8.73 (b), we observe that at r=0.3 nm the potential energy is roughly equal to

$U=-1.1\times {10}^{-19}J$.

Using equation (i) and the above values, we can get the kinetic energy of the system as:

$$\begin{gathered} 1\times {10}^{-19}=K-1\times {10}^{-19} \\ K=2\times {10}^{-19}J \end{gathered}$$

Hence, the value of the energy is $2\times {10}^{-19}J$.

The slope of the distance-potential curve represents force. From the curve, the estimated force at r=0.3 nm is$1\times {10}^{19}N$

From the equation,$F\left(x\right)=-\frac{dU}{dx}$ we can conclude that force is negatively valued and atoms are attracted towards each other.

We observe from the figure that the slope of the curve is negative when $r<0.2nm$, so force is positively valued and repulsive in this range.

We observe from the figure that the slope of the curve is positive when $0.2nm<r<0.4nm$, so force is negatively valued and attractive in this range.

At r=0.2nm the slope is zero and force is zero.