Question

The radionuclide${}^{\mathbf{196}}\mathbf{Ir}$decays by emitting an electron. (a) Into which square in Fig. 42-6 is it transformed? (b) Do further decays then occur?

Short answer

a) It is transformed into beige square in Fig.42-6.

b) Further decays do occur.

Step-by-step solution

Write given data

In Fig.42-6, the radioactive decay of nuclide ${}^{196}\mathrm{Ir}$by emitting an electron is given.

Determine the concept of decay  

In beta decay, a nuclide decays while emitting an electron with a neutrino particle. This implies that the proton number of the nuclide is added by 1 more proton.

a) Calculate the square of the transformed radionuclide

In the figure 42-6, observe that the radionuclide${}^{196}\mathrm{Ir}$ is present in the beige square that resembles it radioactive nature with the given half-life of 52s.

b) Determine whether it continues decaying

By emitting an electron, the radionuclide${}^{196}\mathrm{Ir}$ undergoes a beta decay process. That can be given as follows:

${}^{196}\mathrm{Ir}\to {}^{197}\mathrm{Pt}+{\mathrm{e}}^{-1}+\mathrm{v}$

The product of the decay process is a radionuclide ${}^{197}\mathrm{Pt}$, which also lies in the beige region with a half-life of 18.3h.

Hence, even after the decay process, the decay still occurs.