Chapter 18: Problem 54

Write the nuclear equation for the release of an alpha particle by \(_{70}^{157} \mathrm{Yb} .\)

Short Answer

Expert verified

The nuclear equation for the release of an alpha particle by \(_{70}^{157} \mathrm{Yb}\) is \(_{70}^{157} \mathrm{Yb} \rightarrow _{68}^{153} \mathrm{Tm} + _{2}^{4}\mathrm{He}\)

Step by step solution

Understand the Components of the Nuclear Equation

In a nuclear equation, we start with a parent nucleus that undergoes some type of radioactive decay, resulting in a different daughter nucleus and some type of emitted particle or radiation. In this case, \(_{70}^{157} \mathrm{Yb}\) is the parent nucleus, and the emitted particle is an alpha particle, \(_{2}^{4}\mathrm{He}\). As the parent nucleus loses an alpha particle, it will transform into a new daughter nucleus which we need to identify. Conservation laws are important here, and they stipulate that both atomic numbers (denoted by subscript) and nucleon numbers (denoted by superscript) must be conserved in the transformation.

Determine the Daughter Nucleus

To find the daughter nucleus, subtract the atomic and nucleon numbers of the alpha particle (\(_{2}^{4}\mathrm{He}\)) from the parent nucleus (\(_{70}^{157} \mathrm{Yb}\)). The atomic number becomes \(70 - 2 = 68\), and the nucleon number becomes \(157 - 4 = 153\). Therefore, our daughter nucleus will be \(_{68}^{153} \mathrm{X}\), where X is the element with atomic number 68, in this case Thulium (Tm).