Calculate the mass defect of aluminum 30 13.with a.u.m. of aluminum 29.99817

The nucleus consists of nucleons: positively charged protons and neutral particles – neutrons. The number of protons Z is determined by the ordinal number of the atom in the periodic table, the number of neutral particles – neutrons N is determined by the atomic mass of the nucleus A, we obtain A = Z + N. From the condition of the problem it is known that A = 30, Z = 13, M (nuclei) = 29 , 99817 amu Then N = A – Z; N = 30-13; N = 17. We take the masses of the proton p and the neutron n in the reference book: m (p) = 1.00727 amu; m (n) = 1.00866 amu To find the mass defect Δm, equal to the difference between the total mass of free nucleons entering the atomic nucleus and the mass of the atomic nucleus itself, we use the formula: Δm = (Z ∙ m (p) + N ∙ m (n)) – M (nucleus). Substitute the values ​​of physical quantities in the calculation formula and make calculations: Δm = (13 ∙ 1.00727 + 17 ∙ 1.00866) – 29.99817; Δm = 0.2435 amu
Answer: the defect in the mass of the isotope of aluminum – 30 is 0.2435 amu.