In nuclear physics the semiempirical mass formula is a formu

In nuclear physics, the semi-empirical mass formula is a formula for calculating the approximate nucelar binding energy D of an atomic nucleus with atomic number Z and mass number A. The formula looks like: where, in units of millions of electron volts (MeV), the constants are ai = 15.67, a_2 = 17.23, a_3 = 0.75, a_4 = 93.2 and Write a program nuclear.m that takes as its input the values of A and Z, and prints out the binding energy D for the corresponding atom. You may find the if, elseif, else conditional block useful for this problem. When testing for equality of two variables in the conditional statement, make sure to use the double-equals sign ==, and not the single-equal sign =. The function mod may also be of use. Use your program to find the binding energy of an atom with A = 58 and Z = 28.

Solution

%nuclear.m
clc;
clear all;
a1=15.67; a2=17.23;a3=0.75; a4=93.2
A=input(\'Enter the mass number A = \')% asking input of A from user
Z=input( \'Enter the atomic number Z = \')%asking input of Z from user

%%Checking whether A is a odd /even number
%Calculating the remainder when divided by 2; If it is a
% even number remainder will be zero otherwise non zero
z1=rem(Z,2);
a1=rem(A,2);
if a1~=0 % when remainder is a non zero, i.e when A is an odd number
a5=0
disp(\'A is an odd number\')
elseif a1==0 & z1==0 % when when A is an even and Z is a even number
a5= 12
disp(\'A and Z both are even numbers\')
elseif a1==0 & z1~=0 %when when A is an even and Z is a odd number
a5= -12
disp(\'A is an even and Z is a odd number\')
end

%Calculating binding energy
B=a1*A- a2*A^(2/3)-a3*Z^2*A^(-1/3)-a4*(A-2*Z)^2*A^(-1)+a5/A^(1/2)
disp(\'meV\')

%%%%%%%%%%%%%%%% End%%%%%%%%%%%

Out put of the program:

Enter the mass number A = 58

A =

58

Enter the atomic number Z = 28

Z =

28

a5 =

12

A and Z both are even numbers

B =

-414.9244

meV