# What happens when the PE equals to zero in the potential energy vs intermolecular distance graph? [closed]

In the potential energy versus inter molecular distance graph, we know that atoms/molecules/particles want to be at optimum distance from each other ie $$r_0$$ and to the left of this position in the graph, we get the repulsive force and to the right would be the attractive force acting between them.

Note: Inter-molecular distance is denoted by ‘r’.

However, I have the following questions regarding this graph:

1. Why does graph represent negative Potential energy after a certain inter-molecular distance ?

2. What does negative potential energy mean in this context since the repulsive energy at r=0 was positive?

3. What happens at the point when P.E. becomes zero for a certain inter-molecular distance?

4. If the P.E. at that point has already reached zero, why is there a further need to go into negative potential energy?

• Hello! Please only ask one question per post – otherwise it might get closed due to lack of focus. You can always edit your question or ask a new one. Thanks! Commented Sep 11, 2021 at 9:37

First keep in mind that it is only differences in potential energy that influence physical behaviour.

The potential energy curve can be shown with any constant added to it and it would still imply the same interatomic force.

The way the curve is normally drawn is such that $$V(r)$$ tends to zero as $$r \rightarrow \infty$$. Another way of saying this is that we are plotting $$V(r) - V(\infty)$$.

The point at $$V(r) = V(\infty)$$ is the point $$V(r)=0$$ on the graph. This is a rough indication of the collision radius, since if two atoms released from rest at infinity are attracted to one another by this force, then in a classical picture they will reach a distance of closest approach such that $$V(r) = V(\infty)$$. But of course this is just one collision among many possibilities for the initial conditions. Usually atoms don't start from at rest when far away, instead (e.g. in a gas) they have higher relative speeds than this so they approach closer in a collision.

The main message then is that positive / negative here is simply a statement of how $$V(r)$$ compares with $$V(\infty)$$. There is nothing particularly special happening at $$V(r) = 0$$.

• Your answer makes a lot of sense but could you elaborate a little more in your 4th paragraph? I didn't quite understand what you meant with several possibilities of collision. Commented Sep 11, 2021 at 9:00
• @SnehalSaurabh done Commented Sep 11, 2021 at 9:45

The slope of the potential simply indicates the magnitude and direction of the force acting between the two atoms. You might be able to understand if you consider a simple analogy of two rubber balls, one of which has a magnet inside it and the other a piece of steel inside it. When they are very far apart, the effect of the magnet is negligible, so they don't attract each other- the slope of the potential is zero. When they are put side by side, the magnet will attract the steel, so the two balls will stick together. However, if the balls are pressed tightly together, their rubbery nature will force them apart, as the force of their elasticity will be greater than the magnetic attraction between them. So you will see there is a position of equilibrium at which the magnetic force pulling them together is just balances by the repulsive effect of their elasticity, where again the slope of the potential is zero.

The absolute value of the potential is in some ways irrelevant, as it is the change in potential over space that determines the forces involved. An analogy to illustrate that is a ball on a ramp- it is the slope of the ramp that determines how the ball will roll down it- whether the ramp is on the ground floor of a building or the first floor (ie its height) doesn't really make a difference.

• I found your answer useful. Thanks.... Commented Mar 29, 2023 at 10:13