Answer
$E_{A}=\frac{-2.34}{r}$
$E_{B}=\frac{3.34\times 10^{-4}}{r^8}$
Work Step by Step
Given that $ r_{0}=0.38nm$,$E_{0}=-5.37eV$, $n=8$
We find that:
$r_{0}=(\frac{A}{nB})^{(\frac{1}{1-n})}$
$E_{0}=\frac{-A}{(\frac{A}{nB})^{(\frac{1}{1-n})}}+\frac{B}{(\frac{A}{nB})^{(\frac{n}{1-n})}}$
$0.38=(\frac{A}{8B})^{(\frac{1}{1-8})}$
${A}=(0.38)^{-7}({8B})$
$-5.37=\frac{-A}{(\frac{A}{8B})^{(\frac{1}{1-8})}}+\frac{B}{(\frac{A}{8B})^{(\frac{8}{1-8})}}$
$-5.37=\frac{-(0.38)^{-7}({8B})}{(\frac{(0.38)^{-7}({8B})}{8B})^{(\frac{1}{1-8})}}+\frac{B}{(\frac{(0.38)^{-7}({8B})}{8B})^{(\frac{8}{1-8})}}$
$B=3.34\times10^{-4}$
$A=(0.38)^{-7}({8B})=(0.38)^{-7}({8\times3.34\times10^{-4}})=2.34$
Thus:
$E_{A}=\frac{-2.34}{r}$
$E_{B}=\frac{3.34\times 10^{-4}}{r^8}$
