Beginning with the Weizsäcker semi-empirical mass formula, showthat the minimum in a mass parabola occurs at a value of atomicnumber, Zmin, given by
Note: The Weizsäcker semi-empirical mass formula for the bindingenergy, B, of a nucleus is where A is the atomic mass number of thenucleus and Z is the atomic number of the nucleus. Hint: Althoughatomic number, Z, only takes integer values, assume it is acontinuous variable for the purpose of this exercise.
The expression for atomic mass, m, at the bottom of Page 4/10 ofthe lecture notes can then be regarded as a quadratic function ofZ.
Take ¶m/¶Z, the partial derivative of the expression as afunction of Z. ¶m/¶Z is zero at the minimum value of m. Therefore,setting ¶m/¶Z to zero and solving for Z gives you Zmin. Zmin = mn−m p ( −me )c2 +aC A−1/3 +4asym 2aC A−1/3 +8asym A−1 B=+ aV A − aSA2/3 − aC Z(Z −1) A1/3 − asym (A−2Z ) 2 A −aP 1 A3/4
