Chemistry, 21.09.2020 09:01, mariaveliz2201

The net potential energy EN between two adjacent ions, is sometimes represented by the expression EN = -(C/r) + D exp(-r/rho)

in which r is the interionic separation and C, D, and rho are constants whose values depend on the specific material.

Required:
a. Derive an expression for the bonding energy E0 in terms of the equilibrium interionic separation r0 and the constants D and rho using the following procedure:

1. Differentiate EN with the respect to r and set the resulting expression equal to zero.
2. Solve for C in terms of D, rho, and ro.
3. Determine the expression for Eo by substitution for C in the equation EN = -(C/r) + D exp(-r/rho)

b. Derive another expression for Eo in terms of ro, C, and rho using a procedure analogous to the one outlined in part (a).

Answers: 3

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Answers

answered: niki209

The following are the solution to this question:

Explanation:

Total energy capacity EN from adjacent ions is:

$\to EN= - \frac{C}{r}+De^{(\frac{-r}{p})}.............(1)$

where the value of "C, D, and "r is constants.

In point a:

Differentiating EN:

$\to \frac{d(EN)}{dr} = \frac{C}{r^2}+D e^-\frac{r}{p}(^\frac{-1}{p})$

$=\frac{C}{r^2}-\frac{D}{p}e^\frac{-r}{p}$

In point b:

The above equation is equal to zero to get the value of C:

$\to \frac{C}{r^2}-\frac{D}{p}e^\frac{-r}{p}=0 \\\\\to C= \frac{D r^2}{p}e^\frac{-r}{p} \ \ \ \ or \ \ \ \ D= \frac{C p}{r^2}e^\frac{-r}{p}$

The E_0 value is replaced by the C value in (1):

$\to E_0= \frac{1}{r_0} \frac{Dr_0^2}{p}.e^-\frac{r_0}{p} +D e^-\frac{r_0}{p}\\\\$

$= D e^-\frac{r_0}{p}( {1-\frac{r_0}{p}})$

The E_0 value is replaced by the D value in (1):

$\to E_0= -\frac{C}{r_0} \frac{C p }{r_0^2}.e^\frac{r_0}{p}$

$= \frac{C}{r_0}( {\frac{p} {r_0} -1})$

answered: Guest

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explanation:

answered: Guest

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