Respuesta :
Answer:
The following are the solution to this question:
Explanation:
Total energy capacity EN from adjacent ions is:
[tex]\to EN= - \frac{C}{r}+De^{(\frac{-r}{p})}.............(1)[/tex]
where the value of "C, D, and "r is constants.
In point a:
Differentiating EN:
[tex]\to \frac{d(EN)}{dr} = \frac{C}{r^2}+D e^-\frac{r}{p}(^\frac{-1}{p})[/tex]
[tex]=\frac{C}{r^2}-\frac{D}{p}e^\frac{-r}{p}[/tex]
In point b:
The above equation is equal to zero to get the value of C:
[tex]\to \frac{C}{r^2}-\frac{D}{p}e^\frac{-r}{p}=0 \\\\\to C= \frac{D r^2}{p}e^\frac{-r}{p} \ \ \ \ or \ \ \ \ D= \frac{C p}{r^2}e^\frac{-r}{p}[/tex]
The [tex]E_0[/tex] value is replaced by the C value in (1):
[tex]\to E_0= \frac{1}{r_0} \frac{Dr_0^2}{p}.e^-\frac{r_0}{p} +D e^-\frac{r_0}{p}\\\\[/tex]
[tex]= D e^-\frac{r_0}{p}( {1-\frac{r_0}{p}})[/tex]
The [tex]E_0[/tex] value is replaced by the D value in (1):
[tex]\to E_0= -\frac{C}{r_0} \frac{C p }{r_0^2}.e^\frac{r_0}{p}[/tex]
[tex]= \frac{C}{r_0}( {\frac{p} {r_0} -1})[/tex]
