Chapter 30, Problem 44P

To determine

In a series of decays, the nuclide ${}_{92}{}^{235}U$ becomes ${}_{82}{}^{207}Pb$. How many $\alpha$ and ${\beta }^{-1}$ particles are emitted in this series?

Answer to Problem 44P

Solution:

In this series7 $\alpha$ and 4 ${\beta }^{-1}$ are emitted.

Explanation of Solution

Given info:

The nuclide ${}_{92}{}^{235}U$ becomes ${}_{82}{}^{207}Pb$.

For a $\alpha$ decay the atomic number decreases by two (2)and atomic number mass decreases by four (4), therefore we have:

${}_Z{}^{A}X\to {}_{Z-4}{}^{A-2}Y+{}_2{}^{4}He$ (1)

Similarly, for a ${\beta }^{-1}$ decay the atomic number increases by one (1)and the atomic number mass does not change. Therefore we have:

${}_Z{}^{A}X\to {}_{Z+1}{}^{A}Y+{\beta }^{-1}+\nu$ (2)