Alumina is polymorphic, meaning more than one crystal structure is commonly found. The

temperature, pressure, processing parameters, and other conditions can all affect which crystal

structure is observed. Most often, alumina adopts the crystal structure named corundum or

"alpha" alumina, $-A{l}_2{O}_3.$ Sapphire is a large single crystal of corundum. A common polymorph

of alumina is called "gamma" phase, $-A{l}_2{O}_3.$ A really interesting nanomaterials experiment has

shown that large particles are usually $-A{l}_2{O}_3,$ but very small particles are usually $-A{l}_2{O}_3.$

The average surface energy of $-A{l}_2{O}_3$ is $2\mathrm{.}04\frac{J}{m^2},$ and for $-A{l}_2{O}_3$ it is $1\mathrm{.}53\frac{J}{m^2}.$ The lattice

$($i$.$e$.,$ Born$-$Land$)$ energy of $-A{l}_2{O}_3$ is $-16356k\frac{J}{m}ol,$ and for $-A{l}_2{O}_3$ it is $-17700k\frac{J}{m}ol.$ The

density of $-A{l}_2{O}_3$ is $3\mathrm{.}95\frac{g}{c}{m}^3,$ but for $-A{l}_2{O}_3$ it is $3\mathrm{.}65\frac{g}{c}{m}^3.$ Given these numbers, below

what particle size do we expect the $-A{l}_2{O}_3$ phase to be more stable? Please explain how you

determined$/$calculated your answer. You can assume the particle is a perfect sphere.