Chapter 19, Problem 19.1P

(a)

Interpretation Introduction

To determine:

Write a balanced nuclear equation.
Beta emission from ruthenium $-106$.

  ${}_{44}^{106}Ru{\to }_{-1}^{0}e+\mathrm{..........}$

Explanation of Solution

In beta $\left({\beta }^{-}\right)$ decay, the atom emits ${}_{-1}^{0}e$ particle.
Let the complete space be filled with ${}_Z^{A}X$.

  ${}_{44}^{106}Ru{\to }_Z^{A}X{+}_{-1}^{0}e$

The incomplete nuclear is mentioned above. Now, balance the equation.

  $\begin{array}{l}Z-1=44\\ A=106\\ Z=45,A=106\end{array}$

Therefore, mass number $=106$ and atomic number $=45$

  ${}_{44}^{106}Ru{\to }_{45}^{106}X{+}_{-1}^{0}e$

From periodic table, element X is rhodium $\left(Rh\right)$.

  ${}_{44}^{106}Ru{\to }_{45}^{106}Rh{+}_{-1}^{0}e$

(b)

Interpretation Introduction

To determine:

Write a balanced nuclear equation.

  ${}_{83}^{189}Bi{\to }_2^{4}He+\mathrm{.........}?$

Explanation of Solution

In alpha $\left(\alpha \right)$ decay, the atom emits a helium $\left({}_2^{4}He\right)$ nucleus. So, emitting an alpha particle reduces the nucleus by two proton and two neutrons.
Let the blank element be ${}_Z^{A}X$.

  ${}_{83}^{189}Bi{\to }_Z^{A}X{+}_2^{4}He$

Balance the equation.

  $\begin{array}{c}Z+2=83\\ Z=83-2=81\\ A+4=189\\ A=189-4\\ =185\end{array}$

  ${}_{83}^{189}Bi{\to }_{81}^{185}Ti{+}_2^{4}He$

(c)

Interpretation Introduction

To determine:

  ${}_{84}^{204}Po{+}_{-1}^{0}e\to$

Explanation of Solution

In case of electron capture, the mass number remains same and atomic number decreases by $1$.

  ${}_{84}^{204}Po{+}_{-1}^{0}e{\to }_Z^{A}X$

Balance the chemical reaction.

  $\begin{array}{c}84-1=Z\\ Z=83\\ A=204\end{array}$

Therefore, the element ${}_Z^{A}X{\text{will be }}_{83}^{204}X$.
From periodic table, element will be bismuth.
Therefore,

  ${}_{84}^{204}Po{+}_{-1}^{0}e{\to }_{83}^{204}Bi$

Conclusion

Atomic and mass number should be balance.