Answer to Question #141173 in Mechanics | Relativity for Darren

Question

The force between two molecules may be regarded as an attractive force which increases as their separation decreases and a repulsive force which is only important at small separations and which there varies very rapidly. Draw sketch graphs
(a) for force-separation.
(b) for potential-energy separation.
On each graph, mark the equilibrium distance and on (c) indicate the energy which would be needed to separate two molecules initially at the equilibrium distance. With the help of your graphs, discuss briefly the resulting motion if the molecules are displaced from the equilibrium position.

Answer 1

The potential energy (PE) vs separation graph is shown in figure.

 From the graph-

 (i) the nuclear force is repulsive when the separation is less 1 fm (1fermi=10−15m).

 (ii) It is attractive when the separation is greater than 1 fm. 

The characteristic properties of nuclear force:

i) The nuclear force is short range force.

iii) The nuclear force is independent of electric charge. 

Here As you show in the graphh, When the potential energy is minimum, This is the point of equillibrium. Hence Equillibrium will occur nearly at $1fm=10^{-15}m$ .

The force of before 1fm Is repulsive and after It is attractive, Similarly the potential energy first decreases to nearly 0, And after equillibrium point It again goes on increasing with increasing distance.