Figure 42-20 shows part of the decay scheme ofon a plot of mass number Aversus proton number Z; five lines that represent either alpha decay or beta-minus decay connect dots that represent isotopes. What is the isotope at the end of the five decays (as marked with a question mark in Fig. 42-20)?

Figure 42-20 shows the part of the decay scheme of the isotope through five lines of decay.

There are two decays that the radionuclides mostly undergo. That is alpha decay and beta decay.

In alpha decay, a highly radioactive substance due to its unstable and radioactive nature undergoes a change in its atomic as well as a mass number such that the atomic number of the new daughter isotope decreases by a value of 2 from the parent isotope and the mass number of the daughter nucleus is decreased by a factor of 4 compared to the parent nucleus.

In a beta decay process, the parent isotope due to its radioactive property emits a daughter isotope or a nucleus with a decreased atomic number than the parent nucleus. Here, the mass value of the isotope that is newly formed remains constant while the atomic number differs by one value from the parent nucleus.

The lines that lead toward the lower left indicate that the isotope is undergoing alpha decays, thus having a change in their atomic number as:$\Delta Z_{\alpha }=-2$and a change in their mass number as:$\Delta A_{\alpha }=-4$.

Again, the shorter horizontal lines toward the right being parallel to the x-axis that is the atomic number axis indicate the beta decays of the isotopes (involving electrons, not positrons). This implies that the mass number of the isotopeAstays the same but the atomic number of the isotope changes by $\Delta Z_{\beta }=+1$.

Thus, the given figure 42-20 shows three alpha decays and two beta decays.

This now implies that the atomic number of the isotope got changed three times due to alpha decays and two times due to beta decay and the mass number of the isotope is changed in total of three times due to the alpha decay.

This can be given as:

$Z_f=Z_i+3\Delta Z_{\alpha }+2\Delta Z_{\beta }\text{and}{A}_f=A_i+3\Delta A_{\alpha }$

Referring to Appendix F or G, we get $Z_i=93$for neptunium, so, the atomic number of the new isotope is given as:

$\begin{array}{rcl}{Z}_f& =& 93+3(-2)+2(+1)\\ & =& 89\end{array}$

Consider the above value indicates the element actinium.

Given,$A_i=237$ , so the mass number of the isotope will become

$\begin{array}{rcl}{A}_f& =& 237+3(-4)\\ & =& 225\end{array}$.

Therefore, the final isotope is ${}^{225}\text{Ac}$.