Question

The empirical mass formula (neglecting a term representing the odd – even effect) is M(A, Z) = Z(mp + me) + (A − Z)mn − αA + βA2/3 + γ(A − 2Z)2/A + εZ2A−1/3 where α, β, γ and ε are constants. By finding the minimum in M(A, Z) for constant A obtain the expression Zmin = 0.5A(1 + 0.25A2/3ε/γ)−1 for the value of Z which corresponds to the most stable nucleus for a set of isobars of mass number A.

Answer 1

The empirical mass formula is

M(A, Z) = Z(mp+me)+(A−Z)mn−αA+βA2/3 +γ (A−2Z)2 /A+εZ2A−1/3

where α, β, γ and ε are constants. Holding A as constant, differentiate M(A, Z) with respect to Z and set ∂M/∂ Z = 0.

∂ M/∂Z = mp + me − mn − 4γ(A − 2Z)/A + 2εZA−1/3 = 0

The terms mp+me−mn ≅ mH, the mass of hydrogen atom which is neglected. Rearranging the remaining terms, we obtain

Zmin = A /(2 + (ε/2γ )A2/3)

which is identical with the given expression