91Ó°ÊÓ

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

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

$U=-1.1×{10}^{-19}J$.

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

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

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

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

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.