Ammoina dissociates into `N_(2)` and `H_(2)` such that degree of dissociation `alpha` is very less than 1 and equilibrium pressure is `P_(0)` then the value of `alpha` is [if `K_(p)` for `2NH_(3)(g)hArrN_(2)(g) + 3H_(2)(g) + 3H_(2)(g)` is `27xx10^(-8)P_(0)^(2)]` :