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Calculating the bonding energy between two ions

  • Chemistry
  • Thread starter jisbon
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  • #1
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Homework Statement:

The repulsive potential energy of 2 ions ##A^{2+}## and ##B^{2-}## is given as ##3.6*10^{-2} r^{-5}## eV.
i) Derive the attractive force between the two ions, and subsequently the attractive potential energy.

ii) Calculate the bonding energy of the 2 ions

Relevant Equations:

-
I seem to be having problems with part two in general, but here's my solution.

Attractive force:
##F_{A}=\dfrac {k\left( 2\right) \left( -2\right) }{r^{2}}=\dfrac {-4k}{r^{2}} ##
##E_{A}=-\int ^{r}_{\infty }F_{A}\cdot dr ##
##\begin{aligned}E_{A}=\int ^{r}_{\infty }\dfrac {4k}{r^{2}}dr\\ =\left[ \dfrac {4k}{r}\right] ^{r}_{\infty }=\dfrac {4k}{r}\end{aligned} ##

(ii) This is where I started having trouble. I understand that to find the bonding energy, I need to find the condition where the attractive force + repulsive force = 0 and find out the radius from there. However, I can't seem to find the radius.
This is what I attempted,

Bonding energy when F =0,

##\begin{aligned}E_{r}=\dfrac {3.6\times 10^{-2}}{r^{5}}\\ F_{r}=\int E_{r}dr\end{aligned} ##
##\begin{aligned}F_{r}=-\dfrac {0\cdot 18}{r^{6}}\\ F_{0}=0=-\dfrac {0.18}{r^{6}}+\dfrac {-4k}{r^{2}}\end{aligned} ##


Can anyone show me where I did wrong? Thanks
 

Answers and Replies

  • #2
BvU
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##
F_{r}=\int E_{r}dr
## Isn't right (And it isn't what you do to get from ##E## to ##F## either.... ) what should it have been ?
 
  • #3
mjc123
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You're not being careful with signs. The integral of 1/r2 is -1/r. F = -dE/dr, so Fr is +0.18/r6.
 
  • #4
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Don't give it all away :mad: ! That robs @jisbon from the exercise !
 
  • #5
TSny
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The repulsive potential energy of 2 ions ##A^{2+}## and ##B^{2-}## is given as ##3.6*10^{-2} r^{-5}## eV.

Attractive force:
##F_{A}=\dfrac {k\left( 2\right) \left( -2\right) }{r^{2}}=\dfrac {-4k}{r^{2}} ##
What units are you using for the distance ##r##? What is the value of ##k## in these units?
 
  • #6
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##
F_{r}=\int E_{r}dr
## Isn't right (And it isn't what you do to get from ##E## to ##F## either.... ) what should it have been ?
Oops, differntating the potential energy should give force instead, and there should be a negative sign infront. I assume when integrating from force to potential energy, there should also be a negative sign in front?

What units are you using for the distance ##r##? What is the value of ##k## in these units?
r is in nano meters. k is simply the constant which is
1579600681756.png


In this case, I should convert my constant to nm?

Thanks
 
  • #7
mjc123
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What about the eV?
 
  • #8
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What about the eV?
eV is in joules. So.. what would that actually imply? Sorry, abit confused here
 
  • #9
mjc123
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eV is in eV. That is a unit, equal to 1.602 x 10-19 J. You have to convert 3.6 x 10-2 eV into Joules. (By the way, you're given that figure to 2 sig figs, so there's no justification for taking k to 17 sig figs!)
 
  • #10
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eV is in eV. That is a unit, equal to 1.602 x 10-19 J. You have to convert 3.6 x 10-2 eV into Joules. (By the way, you're given that figure to 2 sig figs, so there's no justification for taking k to 17 sig figs!)
So it will be...

##E_{r}=\dfrac {\left( 3.6\times 10^{-2}\right) \left( 1.602\times 10^{-19}\right) }{r^{5}}##
##\begin{aligned}F_{r}=-\dfrac {d}{dr}E_{r}\\ =\dfrac {2.8836\times 10^{-20}}{r^{6}}\end{aligned} ##
?
 
  • #11
mjc123
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That should be 10-21
 
  • #12
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That should be 10-21
Because of the unit of distance?
 
  • #13
mjc123
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Sorry, my mistake. I was multiplying 10-2 by 10-19, and forgot about the factor of 5 when you differentiate.
 
  • #14
BvU
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Because the exercise uses eV for energy, I would stick to eV for the answer as well ... :oldeyes:
 
  • #15
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Because the exercise uses eV for energy, I would stick to eV for the answer as well ... :oldeyes:
Which part are you referring to? Sorry :/

Sorry, my mistake. I was multiplying 10-2 by 10-19, and forgot about the factor of 5 when you differentiate.
Oh okay.
Since
##\begin{aligned}F_{r}=-\dfrac {d}{dr}E_{r}\\ =\dfrac {2.8836\times 10^{-20}}{r^{6}}\end{aligned} ##
##\dfrac {2.8836\times 10^{-20}}{r^{6}}= \dfrac{4k}{r}##
##2.88\times 10^{-20}=4kr^{5} ##
where k =9.0*10^9
r=0.000000956m?

Isn't it a bit too short? Or is this actually correct?
 
  • #16
mjc123
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You're using the wrong expression. FA = -4k/r2.
 
  • #17
BvU
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Which part are you referring to?
I am referring to
is given as ##3.6*10^{-2} r^{-5}## eV.
Isn't it a bit too short?
Actually, 1 ##\mu##m is huge on an atomic scale
 
  • #18
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I am referring to


Actually, 1 ##\mu##m is huge on an atomic scale
Oh.. I realised, it may a bit too large? I recalled seeing my notes that states the avg is around 0.3nm. In this case, is 1nm still correct?


Also regarding the eV units, if I don't convert it to joules, won't I be able to proceed?
 
  • #19
mjc123
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Also regarding the eV units, if I don't convert it to joules, won't I be able to proceed?
Yes, but you'd have to convert everything else (e.g. k) into consistent units. I find it easier to convert everything to SI, then convert back at the end if necessary.

BTW, when you said above "r is in nanometres", did you mean that the question says that Er = 3.6 x10-2/r5 eV when r is in nm? If so, what units are you using for r in the expression FA = -4k/r2, if you are using a k value in SI units? And what about the charge units? You've got to make sure everything's consistent, whichever units system you're working in.
 
  • #20
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r is in nano meters
How do you know this ? Can you render the full and complete problem statement, litterally ?
 
  • #21
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Yes, but you'd have to convert everything else (e.g. k) into consistent units. I find it easier to convert everything to SI, then convert back at the end if necessary.

BTW, when you said above "r is in nanometres", did you mean that the question says that Er = 3.6 x10-2/r5 eV when r is in nm? If so, what units are you using for r in the expression FA = -4k/r2, if you are using a k value in SI units? And what about the charge units? You've got to make sure everything's consistent, whichever units system you're working in.
How do you know this ? Can you render the full and complete problem statement, litterally ?
Regarding:
BTW, when you said above "r is in nanometres", did you mean that the question says that Er = 3.6 x10-2/r5 eV when r is in nm?
Yes, it is.

Getting back into the question, and decided to do it in terms of metre and joules first (SI units)

Since ##E_{r}=\dfrac {3.6\times 10^{-2}}{r^{5}\left[ nm\right] }eV ##
##E_{r}=3.6\times 10^{-2}\times \left( 1.6\times 10^{-19}\right) \times \left( 10^{9}\right) ^{5} ##
to convert eV to joules and r to m. Since it is in the denominator it will be positive power.

##E_{r}=\dfrac {5.76\times 10^{24}}{r^{5}\left[ m\right] }J ##

Hence differentiating it with a negative sign will give me the repulsive force:

##\begin{aligned}F_{R}=\dfrac {2.88\times 10^{25}}{r^{6}}N\\ \dfrac {-4\left( 1.6\times 10^{-19}\right) }{4\pi \varepsilon _{0}r^{2}}+\dfrac {2.88\times 10^{25}}{r^{6}}=0\end{aligned} ##

where r is in metres. Is this now correct? Or did I make something wrong again?
 
  • #22
mjc123
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The product of the charges is -4*(1.6x10-19)2. Otherwise it looks correct.
 
  • #23
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The product of the charges is -4*(1.6x10-19)2. Otherwise it looks correct.
Oh yes, forgot to put that in. Thanks for checking
 
  • #24
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Just wanted to bump this and apparently the interatomic distance seems to be VERY big? Why is this so?
 
  • #25
mjc123
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In your expression for Er, you should have multiplied by (10-9)5. I missed that first time.
 

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