l=√(q²εn^(1/3)kBT)
l=√(εkBTnq²)
l=√(q²εn^(2/3)kBT)
l=√(nq²εkBT)
[ε₀kBT/q²] = [q²F/r²F/q] = [1/r] = [L⁻¹][q²/ε₀kBT] = [L]Option A: [√(nq²ε₀kBT)] = [√L⁻¹L] = [L⁻¹]Option B: [√(ε₀kBTnq²)] = √[L⁻¹L] = [L]Option C: [√(q²ε₀kBTn^(2/3))] = [√L⁻¹L] = [L^(3/2)]Option D: [√(q²ε₀kBTn^(2/3))] = [√L⁻¹L] = [L]Answer is BD.