Answer
$^{137}\text{Cs}$
Work Step by Step
First we need to understand each process we have:
For $^{212}\text{Po} \rightarrow ^{208}\text{Pb} + \alpha$
In this decay, an alpha particle is emitted. Alpha particles are relatively heavy and carry a +2 charge, but they do not penetrate matter well. In biological applications, alpha particles would be absorbed by body tissues and would not be able to escape the body to be detected externally.
For $^{137}\text{Cs} \rightarrow ^{137}\text{Ba} + e^- + \gamma$
In this decay, a beta particle (electron) is emitted, along with a gamma ray. Beta particles (electrons) penetrate slightly better than alpha particles but are still easily absorbed by tissue and would not travel far. However, the important part of this decay is the emission of gamma rays. Gamma rays are high-energy electromagnetic waves that can penetrate tissues much more effectively than alpha or beta particles. This ability to pass through tissue allows gamma rays to escape the body and be detected externally, making this isotope suitable for imaging or tracing in biological systems.
For $^{90}\text{Sr} \rightarrow ^{90}\text{Y} + e^-$
This decay only emits a beta particle (electron), which, like in the case of $^{137}\text{Cs}$, would not penetrate tissue well enough to escape the body and be detected.
From all the above, it is obvious that $\underline{\color{red}{\text{$^{137}\text{Cs}$ is the most useful isotope for biological tracing}}}$ because, after undergoing beta decay, it emits a gamma ray where gamma rays can penetrate body tissues and escape, making them detectable from outside the body, which is essential for tracing applications.
