Question

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).

Answer 1

The following are the solution to this question:

Step-by-step explanation:

Total energy capacity EN from adjacent ions is:

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

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

In point a:

Differentiating EN:

`\to (d(EN))/(dr) = (C)/(r^2)+D e^-(r)/(p)(^(-1)/(p))`

`=(C)/(r^2)-(D)/(p)e^(-r)/(p)`

In point b:

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

`\to (C)/(r^2)-(D)/(p)e^(-r)/(p)=0 \\\\\to C= (D r^2)/(p)e^(-r)/(p) \ \ \ \ or \ \ \ \ D= (C p)/(r^2)e^(-r)/(p)`

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

`\to E_0= (1)/(r_0) (Dr_0^2)/(p).e^-(r_0)/(p) +D e^-(r_0)/(p)\\\\`

`= D e^-(r_0)/(p)( {1-(r_0)/(p)})`

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

`\to E_0= -(C)/(r_0) (C p )/(r_0^2).e^(r_0)/(p)`

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