# Bulk modulus as a function of $U(r)$ at equilibrium

PART I

The Bulk modulus equation $$B=-V\left(\frac{\partial P}{\partial V}\right)\tag{eq 1}$$ can be transformed into a similar equation as a function of $r$ (interionic equilibrium distance in a ionic crystal).

Considering (at constant N number of ions) $$P=-\frac{dE}{dV}\tag{eq 2}$$

Bulk modulus then becomes $$B=V\frac{\partial}{\partial V}\left(\frac{\partial E}{\partial V}\right)\tag{eq 3}$$

where $$V=8r^3\tag{in a fcc ionic crystal cell}$$ $$\frac{\partial}{\partial V}=\frac{1}{24r^2}\frac{\partial}{\partial r}$$ $$E=4u\tag{in a fcc cell with 4 pairs of ions}$$ $$u\tag{enegy per ion pair}$$

Pluggin in values and simplifying the equation then gives: $$B=\frac{1}{18}r\frac{\partial}{\partial r}\frac{1}{r^2}\frac{\partial}{\partial r}u\tag{eq 4}$$

The equilibrium separation r0 is that which minimizes the energy between two ions (u).Therfore du/dr vanishes in equilibrium and eq (4) reduces to: $$B_0=\frac{1}{18r}\frac{\partial^2u}{\partial r^2}\Bigg|_{r=r_0}\tag{eq 5}$$

PART II

The energy per pair ion as a function of r is: $$u(r)=-\frac{\alpha e^2}{r}+\frac{C}{r^m}\tag{eq 6}$$ where all are experimentally derived constants (e=q) except for r.

OBJECTIVE:

THE OBJECTIVE is to determine the exponential $m$, but m can not be determined by solving $m$ in eq (6), because any little variation in $u(r)$ affects tremendously our variable $m$. So m can be derived using the Bulk modulus. If $B_0$ and $r_0$ are the equilibrium bulk modulus and the interionic distance at equilibrium, then we can show that: $$m=1+\frac{18B_0 r_0 ^3}{u^{coul}}\tag{target equation}$$ $$u^{coul} =\frac{\alpha e^2}{r_0}$$

I know if we minimize eq (6) u(r)=0 so we can have an expression r0 at "equilibrium" interionic distance: $$u_0=u\left(r_0\right)=-\frac{\alpha e^2}{r_0}\frac{m-1}{m}\tag{eq 7}$$ $$r_0=\left(\frac{m C}{\alpha e^2}\right)^\frac{1}{m-1}\tag{eq 8}$$

So we already have an expression for B0 (eq 5) and an equation for r0 (eq 8)

How do you get to target eq?

I tried to plug in eq (6) into eq (5) but did not work for me.

I tried to plug in eq (7) into eq (5) but did not work for me.

If you notice, the target equation has $r_0$ in it. So in equation 8, instead of solving for $r_0$, solve for C. Taking the first derivative of u and assuming that it goes to zero at $r_0$ gives

$$\frac{\alpha e^2r_0^{m-1}}{m} = C .$$

Next, take the second derivative of u with respect to r to get

$$\frac{d^2u}{dr^2}= \frac{-2\alpha e^2}{r^3} + \frac{m(m+1)C}{r^{m+2}}.$$

Equation 5 says to evaluated this derivative at $r_0$. You also want to plug in the value you found for C. $$B_0 = \frac{1}{18r_0}\left(\frac{-2\alpha e^2}{r_0^3} + \frac{\alpha e^2(m+1)}{r_0^3}\right)$$

A bit of simplification will get you to the target equation.

• I'm sorry I am not getting it ! What part should I plug into eq (5)? C ? But there's no C in eq (5)... Feb 20, 2015 at 0:10
• Why the first derivative ? isn't the second derivative as eq (5) says ??? Feb 20, 2015 at 0:11
• Sorry about that, added in some more steps so it should be clear. Feb 20, 2015 at 0:29
• OMG it works !!! thanks a lot...you've just saved 3 students from not turning in their homework. Feb 20, 2015 at 0:50
• Then at least what he deserves is a +1! Jul 25, 2017 at 5:47