(a) Deduce the expression, N = N0 e−λt, for the law of radioactive decay. (b) (i) Write symbolically the process expressing the β+ decay of Na1122. Also write the basic nuclear process underlying this decay. (ii) Is the nucleus formed in the decay of the nucleus Na1122, an isotope or isobar?



(a) According to the law of radioactive decay, we have: where, N = Number of nuclei in the sample ΔN = Amount undergoing decay Δt = Time where, λ = Decay constant or disintegration constant Δt = 0 On integrating both sides, we get: ln N − ln N0 = −λ (t − t0) At t0 = 0: (b) (i) The β+ decay for Na1122 is given below: Na1122→Ne1022+β++ν If the unstable nucleus has excess protons than required for stability, a proton converts itself into a neutron. In the process, a positron e+ (or a β+) and a neutrino ν are created and emitted from the nucleus. p→n+β++ν This process is called beta plus decay.
(ii) The nucleus so formed is an isobar of Na1122 because the mass number is same, but the atomic numbers are different.