Question

Beginning with the Weizsäcker semi-empirical mass formula, show that the minimum in a mass parabola occurs at a value of atomic number, Zmin, given by

Note: The Weizsäcker semi-empirical mass formula for the binding energy, B, of a nucleus is where A is the atomic mass number of the nucleus and Z is the atomic number of the nucleus. Hint: Although atomic number, Z, only takes integer values, assume it is a continuous variable for the purpose of this exercise.

The expression for atomic mass, m, at the bottom of Page 4/10 of the lecture notes can then be regarded as a quadratic function of Z.

Take ¶m/¶Z, the partial derivative of the expression as a function 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 − aS A2/3 − aC Z(Z −1) A1/3 − asym (A−2Z ) 2 A −aP 1 A3/4

Answer 1

WU (9 we know that wizsachors nomi empirical mass formula are, M4, ) = 2Mp + N Man - B.E. -0 M1A, 2) atomic mass of isolofe Also, ZNA z> murmber of trotomia, N> meignten of neutams. B.E. (Binding energj >-1-20A +08A% A4_, ZZ) , > mas poton moos di matton A So, we can to write aan aro, m/n, 2) - Zm+(A-2) Mm - QUAT A 2 1 @ 2120 + om 1A-22) moso neglecting 's 2 toring 2(2-1) = 2², we got, M(A2) 2mp+ (1-2) Min-03 +09+34.922%va + am 1,72237 Now, to for min'z, (for constant A egn @ toon con lo rodunod an . Red, FAZ ACMm-ag+am) a At $ 2-4am - (Mon-imp) 92 1 / CQ A3 - 4m - MCA,Z) = PA +62 +922 - So, (dm = $ + 2q2 at Z = Knion 1 Zimin = -39,22 m A lan) Bu substituting values of Mary Mp sam, ac we gel Zmin = A/(1.98 +0.015 A2) Are