(a) Define (i) proton number; (ii) nucleon number; (iii) isotopes. (b) A nuclide \(^A_ZX\) emits \(\beta\)-particle to form a daughter nuclide Y. Write a nuclear equation to illustrate the charge conservation. (c) The radioactive nuclei \(^{210}_{84}P_o\) emits an \(\alpha\) - particle to produce \(^{206}_{82}P_b\). Calculate the energy, in MeV, released in each disintegration.
Take the masses of \(^{210}_{84}P_o\) = 209.936730 u; \(^{206}_{82}P_b\) = 205.929421 u; \(^{4}_{2}He\) = 4.001504 u; and that 1u = 931 MeV
Explanation
(a)(i) Proton number — the number of protons in the nucleus of an atom (ii) Nucleon number — the total number of protons and neutrons in the nucleus of are atom. (iii) Isotopes — are atoms or nuclides of the samc element having the same number of protons or atomic number but different neutrons or mass numbers (b) \(^A_Z \to ^A_{Z+}Y + ^0_{-1}\) \(^{210}_{84}Po \to ^{206}_{82}Pb + ^4_2He + Energy\) Mass detect = 209.936730u - (205.929421 + 4.001504)u 209.9367330u - 209.930925u = 5.805 x 10\(^{-3}\)U Energy released = 5.805 x 10\(^{-3}\) x 931 = 5.40MeV
