Question

(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?

Answer 1

(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.