Chapter 30, Problem 67GP

(a)

To determine

To find: The daughter nucleus.

Answer to Problem 67GP

The daughter nucleus is $\text{I}\text{r}$

Explanation of Solution

Given:

  ${\beta }^{-}$ energy is $0.14\text{ MeV}$

  $\gamma$ energy is $0.042\text{ MeV}$ and $0.129\text{ MeV}$

Calculation:

The decay reaction can be written as:

  $\text{O}\text{s}\to \text{X}+{\beta }^{-}$

So, the atomic number of the daughter nucleus can be given as:

  $=76-\left(-1\right)=77$

And the mass number of daughter nucleus is given by:

  $=191-0=191$

So, the daughter nucleus is $\text{I}\text{r}$

Conclusion:

The daughter nucleus is $\text{I}\text{r}$

(b)

To determine

To draw:

An energy level diagram showing the ground states of parent and daughter and excited states of the daughter.

Answer to Problem 67GP

  ${\beta }^{-}$ decay of $\text{O}\text{s}$ occur into the daughter state of $\text{I}\text{r}$

Explanation of Solution

Given:

  ${\beta }^{-}$ energy is $0.14\text{ MeV}$

  $\gamma$ energy is $0.042\text{ MeV}$ and $0.129\text{ MeV}$

  

Firstly, the ${\beta }^{-}$ decay of $\text{O}\text{s}$ into the excited state $\text{I}{\text{r}}^{\ast }$ then, the first gamma decay occurs with energy $0.042\text{ MeV}$ and then the second gamma decay occurs with energy of $0.129\text{ MeV}$

Hence, ${\beta }^{-}$ decay of $\text{O}\text{s}$ occur into the daughter state of $\text{I}\text{r}$

Conclusion:

Hence, ${\beta }^{-}$ decay of $\text{O}\text{s}$ occur into the daughter state of $\text{I}\text{r}$