TBR CBT discrete atomic PE

[member232]

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Quick question about the picture attached to this post:

The answer is B, that the attraction exceeds the repulsion.

For some reason, my intuition led me to answer C. Looking at the graph, I noticed the negative potential energy at the Bohr's radius, and thought that would mean the two atoms are just close enough to feel the repulsion of each other's electrons. I assumed a positive potential energy would mean an attractive force...

Can someone please analyze/break down this graph for me and how the Potential Energy of the atoms plays in with the Bohr's radius?

Thanks in advance!

 

[kasho11]

If the repulsion energy were greater than the attraction energy the atoms would want to move away from each other and would have a high potential energy.

The question is slightly misleading as it describes the situation as two atoms approaching each other but the x axis is labeled as increasing distance.

 

[member232]

Yeah, that threw me off a bit--

If I'm getting this right, the graph starts at a high potential energy, and very small radius--meaning that the positive charges of the nucleus are repelling each other ?-- then the PE becomes negative until the Bohr's radius (does this low negative PE mean that it is attractive at this point?) -- and then finally as you further increase the distance from the radius, the electrons will repel each other?

It's kind of a strange way to represent the concept on the graph.... I just want to know if I understand this properly...

 

[circulus vitios]

At a small nuclear distance, the nuclear force will strongly repel the two nuclei. If you look at the graph, a small distance equals +PE. So +PE must mean repulsion. At a great nuclear distance, the attractive force is going to fall off to 0 like 1/r^2. -PE must then correspond to attraction.

With this, it's easy to rule out C & D. I'm not sure how you can rule out A?

 

[LoLCareerGoals]

It says nothing about approaching or anything dynamic in the question itself. It just asks, what happens when d = Rb? (not as it approaches Rb from left or right side).

In general I've come to associate high energy as instability/undesirability. To create high energy you need to do work, hence I would've picked B since energy is negative even though I don't really remember how to compute potential energy for an electrostatic force.

 

[circulus vitios]

It says nothing about approaching or anything dynamic in the question itself. It just asks, what happens when d = Rb? (not as it approaches Rb from left or right side).

In general I've come to associate high energy as instability/undesirability. To create high energy you need to do work, hence I would've picked B since energy is negative even though I don't really remember how to compute potential energy for an electrostatic force.

I get that it's a static problem. But what makes you rule out A?

edit: Maybe we're supposed to guess that they are referring to the fact that electrons repel each other at any distance (1/r^2) and that they aren't talking about a net attraction. I hate these questions.

 

[LoLCareerGoals]

At a small nuclear distance, the nuclear force will strongly repel the two nuclei. If you look at the graph, a small distance equals +PE. So +PE must mean repulsion. At a great nuclear distance, the attractive force is going to fall off to 0 like 1/r^2. -PE must then correspond to attraction.

With this, it's easy to rule out C & D. I'm not sure how you can rule out A?

Good point. Only thing I can think of is that A is a little too all encompassing. As in how do we know how heavy these 2 atoms are? Maybe they are so massive that G attraction overcomes electrostatic repulsion. (i know very unlikely, but no info is given and this is MCAT). I'd a lot more trouble eliminating A if it said "electrostatically repel" instead of just "repel".
Edit: Pretty sure it is net, since they say "atoms" in A.

 

[circulus vitios]

Good point. Only thing I can think of is that A is a little too all encompassing. As in how do we know how heavy these 2 atoms are? Maybe they are so massive that G attraction overcomes electrostatic repulsion. (i know very unlikely, but no info is given and this is MCAT).

See my edit. It's probably a case of picking the "most correct" answer. Oh, MCAT. 😡

 

[LoLCareerGoals]

See my edit. It's probably a case of picking the "most correct" answer. Oh, MCAT. 😡

I disagree. See mine. I can't see how they can mean anything other than net by using the word atom (both nucleous and el-ons). There are forces beyond el-static that could be in play. A encompasses all of those forces that we don't have info on, hence we eliminate A.
What's the book's explanation?

 

[member232]

At a small nuclear distance, the nuclear force will strongly repel the two nuclei. If you look at the graph, a small distance equals +PE. So +PE must mean repulsion. At a great nuclear distance, the attractive force is going to fall off to 0 like 1/r^2. -PE must then correspond to attraction.

With this, it's easy to rule out C & D. I'm not sure how you can rule out A?

I ruled out A because its too absolute to say there is NO repulsive force... Even though the net force is attractive, the like charges should still repel each other...

When you say "the nuclear force" strongly repels the two nuclei... are you talking about the positive charges in each nucleus? And if so, where does the repelling force of the electrons come in (edit: on the graph)?

 

[circulus vitios]

I ruled out A because its too absolute to say there is NO repulsive force... Even though the net force is attractive, the like charges should still repel each other...

When you say "the nuclear force" strongly repels the two nuclei... are you talking about the positive charges in each nucleus? And if so, where does the repelling force of the electrons come in (edit: on the graph)?

http://hyperphysics.phy-astr.gsu.edu/hbase/forces/funfor.html#c1

http://en.wikipedia.org/wiki/Nuclear_force

The force is powerfully attractive between nucleons at distances of about 1 femtometer (fm) between their centers, but rapidly decreases to insignificance at distances beyond about 2.5 fm. At very short distances less than 0.7 fm, it becomes repulsive, and is responsible for the physical size of nuclei, since the nucleons can come no closer than the force allows.

If you look at the HyperPhysics page, the electromagnetic force (the force that is responsible for nucleus-electron attraction; electron-electron repulsion; and nucleus-nucleus repulsion at distances >0.7 fm) is much weaker and operates over a much larger range. I'm not sure where the exact cutoffs are in your chart, but you can clearly see the nuclear force by the potential energy approaching infinity at small distances.

 

[member232]

ahh, makes sense... i was always taught the nuclear strong force is attractive, never understood the other side of it

thanks!!