Chapter 18, Problem 18E

Interpretation Introduction

Interpretation:

The nuclides generated in each step of the $\text{11-}$ step decay of radioactive neptunium $\text{-237}$ is to be stated.

Concept introduction:

Heavy atoms have unstable nuclei. Due to which they are radioactive and disintegrates into smaller nuclei. The decay of radioactive nuclei can be done by alpha particle emission, beta-particle emission, positron decay, gamma decay, and electron capture.

Answer to Problem 18E

The decaying nuclides in the series are (a) ${}_{91}^{\text{233}}\text{Pa}$, (b) ${}_{92}^{\text{233}}\text{U}$, (c) ${}_{90}^{\text{229}}\text{Th}$, (d) ${}_{\text{88}}^{\text{225}}\text{Ra}$, (e) ${}_{\text{89}}^{\text{225}}\text{Ac}$, (f) ${}_{\text{87}}^{\text{221}}\text{Fr}$, (g) ${}_{\text{85}}^{\text{217}}\text{At}$, (h) ${}_{\text{83}}^{\text{213}}\text{Bi}$, (i) ${}_{\text{84}}^{\text{213}}\text{Po}$, and (j) ${}_{\text{82}}^{\text{209}}\text{Pb}$.

Explanation of Solution

Radioactive decay series is the stepwise disintegration of unstable radioactive nuclei into stable nuclei.

In the alpha decay, the mass number of parent nuclei decreases by four units and the atomic number decreases by two units.

In the beta decay, the mass number remains the same and the atomic number of the nucleus increases by one unit.

Therefore, following these changes the following are the equation mentioning the particle decaying.

The disintegration of series reactions to give nuclides (a), (b), (c), (d) are shown below.

$\begin{array}{l}{}_{\text{93}}^{\text{237}}\text{Np}{\to }_{91}^{\text{233}}\text{Pa}{+}_{\text{2}}^{\text{4}}\text{He}\\ {}_{91}^{\text{233}}\text{Pa}{\to }_{92}^{\text{233}}\text{U}{+}_{-1}^0\text{e}\\ {}_{92}^{\text{233}}\text{U}{\to }_{90}^{\text{229}}\text{Th}{+}_{\text{2}}^{\text{4}}\text{He}\\ {}_{90}^{\text{229}}\text{Th}{\to }_{\text{88}}^{\text{225}}\text{Ra}{+}_{\text{2}}^{\text{4}}\text{He}\end{array}$

The disintegration series reactions to give nuclides (e), (f), (g), (h) are shown below.

$\begin{array}{l}{}_{\text{88}}^{\text{225}}\text{Ra}{\to }_{\text{89}}^{\text{225}}\text{Ac}{+}_{-1}^0\text{e}\\ {}_{\text{89}}^{\text{225}}\text{Ac}{\to }_{\text{87}}^{\text{221}}\text{Fr}{+}_{\text{2}}^{\text{4}}\text{He}\\ {}_{\text{87}}^{\text{221}}\text{F}{\to }_{\text{85}}^{\text{217}}\text{At}{+}_{\text{2}}^{\text{4}}\text{He}\\ {}_{\text{85}}^{\text{217}}\text{At}{\to }_{\text{83}}^{\text{213}}\text{Bi}{+}_{\text{2}}^{\text{4}}\text{He}\end{array}$

The disintegration series reactions to give nuclides (i), (j) are shown below.

$\begin{array}{l}{}_{\text{83}}^{\text{213}}\text{Bi}{\to }_{\text{84}}^{\text{213}}\text{Po}{+}_{-1}^0\text{e}\\ {}_{\text{84}}^{\text{213}}\text{Po}{\to }_{\text{82}}^{\text{209}}\text{Pb}{+}_2^4\text{He}\\ {}_{\text{82}}^{\text{209}}\text{Pb}{\to }_{\text{83}}^{\text{209}}\text{Bi}{+}_{-1}^0\text{e}\end{array}$

Therefore, decaying nuclides in the series are (a) ${}_{91}^{\text{233}}\text{Pa}$, (b) ${}_{92}^{\text{233}}\text{U}$, (c) ${}_{90}^{\text{229}}\text{Th}$, (d) ${}_{\text{88}}^{\text{225}}\text{Ra}$, (e) ${}_{\text{89}}^{\text{225}}\text{Ac}$, (f) ${}_{\text{87}}^{\text{221}}\text{Fr}$, (g) ${}_{\text{85}}^{\text{217}}\text{At}$, (h) ${}_{\text{83}}^{\text{213}}\text{Bi}$, (i) ${}_{\text{84}}^{\text{213}}\text{Po}$, and (j) ${}_{\text{82}}^{\text{209}}\text{Pb}$.

Conclusion

The nuclides generated in the sequence are rightfully stated above.