The total energy E of an ionic solid is given by the expression:

$E = -\frac{a e^2}{4\pi\epsilon_o r} + \frac{B}{r^9}$

Here, $a$ represents the Madelung constant, $e$ is the elementary charge, $\epsilon_o$ is the permittivity of free space, $r$ is the inter-ionic distance, and $B$ is a constant. The question uses 'a' for Madelung constant, but options use '$\alpha$'. Assuming they are equivalent, we use $\alpha$.

$E = -\frac{\alpha e^2}{4\pi\epsilon_o r} + \frac{B}{r^9}$

Deriving Constant B at Equilibrium

Equilibrium separation ($r_o$) occurs when the total energy E is at a minimum. This condition is met when the derivative of E with respect to r is zero ($dE/dr = 0$).

Step 1: Differentiate the Energy Expression

First, rewrite the energy expression using negative exponents for easier differentiation:

$E = -\frac{\alpha e^2}{4\pi\epsilon_o} r^{-1} + B r^{-9}$

Now, differentiate E with respect to r:

$\frac{dE}{dr} = \frac{d}{dr} \left(-\frac{\alpha e^2}{4\pi\epsilon_o} r^{-1}\right) + \frac{d}{dr} \left(B r^{-9}\right)$

$\frac{dE}{dr} = -\frac{\alpha e^2}{4\pi\epsilon_o} (-1) r^{-2} + B (-9) r^{-10}$

$\frac{dE}{dr} = \frac{\alpha e^2}{4\pi\epsilon_o r^2} - \frac{9B}{r^{10}}$

Step 2: Apply Equilibrium Condition

At the equilibrium separation $r = r_o$, the derivative $dE/dr$ must be zero:

$\frac{\alpha e^2}{4\pi\epsilon_o r_o^2} - \frac{9B}{r_o^{10}} = 0$

Step 3: Solve for B

Rearrange the equation to solve for $B$:

$\frac{\alpha e^2}{4\pi\epsilon_o r_o^2} = \frac{9B}{r_o^{10}}$

Multiply both sides by $r_o^{10} / 9$:

$B = \frac{\alpha e^2}{4\pi\epsilon_o r_o^2} \times \frac{r_o^{10}}{9}$

Simplify the expression:

$B = \frac{\alpha e^2 r_o^{(10-2)}}{4 \times 9 \pi\epsilon_o}$

$B = \frac{\alpha e^2 r_o^8}{36\pi\epsilon_o}$

This value matches the first option.