 #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
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