Use the SEMF to derive the following analytical expression for the most stable ratio of N/ Z: N/Z 0.98 +0.015 A2/3 [Hint: After finding the lowest mass for a given value of A, substitute Z by A/(1+N/Z) and solve for the quantity (N/Z)] Is the above expression in good agreement with experimental observation? Justify your answer. [Hint: Compare some of the N/Z ratios predicted by the above formula with the observed ratios in stable nuclides as shown in a Segre chart below] Proton number 2 100 50 proton rich neutros rich 50 Neutron number N 100 130 140 160 Determine which nuclide with A = 111 is predicted to be the most stable. Does this prediction agree with observation (see graph below)? Mass excess (atomic mass units) -0.086 -0.088 A 111 -0.090- -0.092 -0.094 -0.096 45 46 47 48 49 Rh Pd Ag Cd In Ная 50 51 Sn Sb Atomic number (Z) For each combination of Z and N (with A = Z + N), the atomic mass M(Z,A) is approximately given by the semi-empirical mass formula (SEMF): M(Z,A) Zm +(A - Z)m B(Z,A) c² where m is the atomic mass of the hydrogen atom, m, is the neutron mass, and = B(Z,A) a A-a, A²/3. Z² (A-2Z)² ap ac A1/3 aasym + A A3/4 is the binding energy of the nucleus. In this equation, a₁ = 15.6 MeV, a = 17.2 MeV, a = 0.70 MeV, a asym = 23.3 MeV, and a = 34.0 MeV (for A even, with the positive sign for even-even nuclides and the negative sign for odd-odd nuclides) and a = 0 (for A odd). Also, my, c² = 938.8 MeV and m, c² = 939.6 MeV.

Use the SEMF to derive the following analytical expression for the most stable ratio of N/ Z: N/Z 0.98 +0.015 A2/3 [Hint: After finding the lowest mass for a given value of A, substitute Z by A/(1+N/Z) and solve for the quantity (N/Z)] Is the above expression in good agreement with experimental observation? Justify your answer. [Hint: Compare some of the N/Z ratios predicted by the above formula with the observed ratios in stable nuclides as shown in a Segre chart below] Proton number 2 100 50 proton rich neutros rich 50 Neutron number N 100 130 140 160 Determine which nuclide with A = 111 is predicted to be the most stable. Does this prediction agree with observation (see graph below)? Mass excess (atomic mass units) -0.086 -0.088 A 111 -0.090- -0.092 -0.094 -0.096 45 46 47 48 49 Rh Pd Ag Cd In Ная 50 51 Sn Sb Atomic number (Z) For each combination of Z and N (with A = Z + N), the atomic mass M(Z,A) is approximately given by the semi-empirical mass formula (SEMF): M(Z,A) Zm +(A - Z)m B(Z,A) c² where m is the atomic mass of the hydrogen atom, m, is the neutron mass, and = B(Z,A) a A-a, A²/3. Z² (A-2Z)² ap ac A1/3 aasym + A A3/4 is the binding energy of the nucleus. In this equation, a₁ = 15.6 MeV, a = 17.2 MeV, a = 0.70 MeV, a asym = 23.3 MeV, and a = 34.0 MeV (for A even, with the positive sign for even-even nuclides and the negative sign for odd-odd nuclides) and a = 0 (for A odd). Also, my, c² = 938.8 MeV and m, c² = 939.6 MeV.