TBR test 1 physical sciences number 37

[2010premed]

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The following graph depicts the combined potential energy of two atoms as they approach one another. What occurs when the distance separating the atoms equals the Bohr radius?

A. The atoms do not repel each other.
B. The attraction of the electrons to each nucleus exceeds the internuclear repulsion and interelectronic repulsion.
C. The repulsion energy is greater than the attraction energy.
D. The repulsion energy is equivalent to the attraction energy.
A: B

I'm very confused about this one... I thought that since the PE is decreaseing as they approach each other, then they were unlike charges?​

 

[donaldduck]

Is the answer "A"?

 

[fizzgig]

no it's B and i've seen this posted before and i didnt get it then either. if the atoms stop moving towards each other then forces have to be balanced in some way. unbalanced forces would yield continued movement of some sort... yeah, i don't get it.

 

[2010premed]

yeah I chose A, but it's good to know I'm not the only one who's having a problem with this

 

[dextor2003]

heres my insight...

i immediately ruled out A cuz i thought that atoms are always constantly repelling and attracting each other, albeit at different strengths at different distances. its just a matter of competition between the two forces when the energy is the lowest. so i picked D...but apparently my reasoning is flawed since you guys didnt choose D..

but yea, still stumped overall by this questions

 

[fizzgig]

i was on board with D. that is what i would pick on an exam. i mean unless it was this exact question with these exact answers, in which case we've established i'm wrong...

 

[brood910]

Bump. Can anyone clarify this?

 

[t5Nitro]

This one is weird. I originally thought I had an explanation and then lost my thought process. But can use some logic for this one. If A is correct, then D is also correct therefore neither can be correct. In A, the atoms do not repel which is true of choice D, considering the atoms are neither repelled nor are attracted. Between B and C, B is the better answer because in C the atoms would be too close together, a point left of the Bohr radius (the equilibrium position).

I don't know; I originally wanted to pick D and then thought A and D said the same thing in a way, but I cannot fully understand the reasoning behind B. What I tried to pick apart in B was the attraction of the electrons to each nucleus. There are 2 nuclei pulling in on an electron cloud. It's like the electrons experience 2 attractive forces (one from each nucleus). The total attractive force exerted by each nucleus on the electron cloud exceeds a single repulsive force between two nuclei. Edit: also, assuming the atoms are identical, the electrons of each atom should be equally attracted to each nucleus, putting them at an equilibrium distance between the two nuclei (this is where I mean they experience double the attractive force, one from each nucleus, but there exists only one repulsive force between the nuclei).

edit 2: but then there is this set of lecture slides. Scroll down to page 2 and read the last bullet point under "Bonding Models" http://web.eng.fiu.edu/wangc/EGN3365-2b.pdf

This basically says D should be the answer.

I'm just throwing ideas out there; maybe one of you can expand on it or alter something so that it makes sense. Otherwise process of elimination answers the question in about 5 seconds.

 

[inasensegone]

This question really bugged me, and just like t5Nitro, I thought I figured it out at first only to be stumped again. Here is my initial reasoning:

Notice on the y axis, potential energy is actually negative at the bohr radius. This would imply that the atoms are attracted to each other (which is actually the electrons attracted to the opposite nuclei) more strongly than they are repelled. How do we know this? Well, as the atoms get closer and closer together, we definitely know that they will eventually repel each other and the resultant bond length is a measure of the balance of attractive and repulsive forces. This is why at the far left side of the graph, potential energy is at a maximum. So a net repulsive force must be synonymous with a positive potential energy and conversely, a negative potential energy must be synonymous with a net attractive force.

Here is the problem though:
The Bohr radius is defined as the radius of the lowest energy electron orbit. If the interNUCLEAR distance were equivalent to this distance, then it would imply that the atoms are actually too close together (the bond distance should be 2 Bohr radii) and would experience a net repulsive force.

I suppose you could also reason that the attraction of the electrons to each nucleus, exceeds the internuclear repulsion, and also exceeds the interelectronic repulsion, but combined (internuclear repulsion and interelectronic repulsion) exceed the attraction of the electrons to each nucleus.

Meh, I'm stumped.

Here is a picture from a scientific article I found in the International Journal of Sciences (this problem is really bugging me):

This article basically says that the aB,H is the Bohr radius, and therefore the bond length is 2 times this distance (1.060 Angstroms). Perhaps this question is flawed?

Also, the Bohr radius is considered an obsolete model.

EDIT: Can someone please post how TBR explained it?

EDIT 2: I just realized that I made a mistake concerning the bond length. It is just slightly under 2 bohr radii, but still greater than 1 bohr radii. The picture above indicates an H2 cation. A neutral H2 molecule would have the following relationship:

Basically, the Bohr radius is 0.53 Angstroms and the bond length is 0.748 Angstroms. This is still consistent with our dispute with this problem, although it may validate my 2nd reasoning above:

"...you could also reason that the attraction of the electrons to each nucleus, exceeds the internuclear repulsion, and also exceeds the interelectronic repulsion, but combined (internuclear repulsion and interelectronic repulsion) exceed the attraction of the electrons to each nucleus."

(The Article is titled Bond Lengths, Bond Angles, and Bohr Radii from Ionization Potentials Related via the Golden Ratio for H2+ , O2, O3, H2O, SO2, NO2, and CO2, and the author is Raji Heyrovska)

 

[patrickmbyrne]

A. The atoms do not repel each other. (No because they still repel or the two nuclei would fuse.)
B. The attraction of the electrons to each nucleus exceeds the internuclear repulsion and interelectronic repulsion.
C. The repulsion energy is greater than the attraction energy. (Penergy is negative at the Bohr radius so attractive forces are greater)
D. The repulsion energy is equivalent to the attraction energy. (Penergy is negative at the Bohr radius so attractive forces are greater)
A: B

I also disagree with the graph the lowest PE should be at 2xBohr radius no?

 

[brood910]

Gawd, some of these TBR questions dont make any sense. They should have better answer explanations.

 

[Odi]

The description says the atoms approach one another.

The graph shows they are losing potential energy. Don't let the negative potential energy throw you off. When it dips down into the negative numbers you just treat it normally. A potential energy of 0 is an arbitrarily chosen reference frame.

At the maximum negative potential energy (Bohr radius distance away from eachother)--the two atoms have maximum kinetic energy and momentum. At the Bohr radius distance away from each other their instantaneous velocity is at its maximum pointing towards each other. They are at maximum attraction amidst a lesser repulsion force. This makes B true and all other options false.