Exercise 2.10: The semi-empirical mass formula In nuclear physics, the semi-empirical mass formula is a formula for calculating the approximate nuclear binding energy B of an atomic nucleus with atomic number Z and mass number A: A-a2A2/3-a370-a,-. A (A-2Z)2 as A1/2 17.23 where, in units of millions of electron volts, the constants are a = 15.67, a2 A3 = 0.75, a4 = 93.2, and if A is odd, if A and Z are both even, if A is even and Z is odd a5 12.0 12.0 a) Write a program that takes as its input the values of A and Z, and prints out the binding energy for the corresponding atom. Use your program to find the binding energy of an atom with A 58 and Z = 28, (Hint: The correct answer is around 490 MeV) b) Modify your program to print out not the total binding energy B, but the binding energy per nucleon, which is B/A c) Now modify your program so that it takes as input just a single value of the atomic number Z and then goes through all values of A from A = Z to A = 32, to find the one that has the largest binding energy per nucleon. This is the most stable nucleus with the given atomic number. Have your program print out the value of A for this most stable nucleus and the value of the binding energy per nucleon d) Modify your program again so that, instead of taking Z as input, t runs through all values of Z from 1 to 100 and prints out the most stable value of A for each one. At what value of Z does the maximum binding energy per nucleon occur? (The true answer, in real life, is Z = 28, which is nickel. You should find that the semi-empirical mass formula gets the answer roughly right, but not exactly.)