24.136 CP - Chemistry: Molecular Nature Of Matter And Change

Question

In the naturally occurring thorium-232 decay series, the steps emit this sequence of particles: $\alpha , {\beta }^{-}, {\beta }^{-}, \alpha , \alpha , \alpha , \alpha , {\beta }^{-}, \alpha .$ Write a balanced equation for each step.

Step-by-Step Solution

Answer

$$\begin{gathered} {}_{90}^{232}Th{\to }_2^4\alpha {+}_{88}^{228}Ra{ \\ }_{88}^{228}Ra{\to }_{-1}^0\beta {+}_{89}^{228}Ac{ \\ }_{89}^{228}Ac{\to }_{-1}^0\beta {+}_{90}^{228}Th{ \\ }_{90}^{228}Th{\to }_2^4\alpha {+}_{88}^{224}Ra{ \\ }_{88}^{224}Ra{\to }_2^4\alpha {+}_{86}^{220}Rn{ \\ }_{86}^{220}Rn{\to }_2^4\alpha {+}_{84}^{216}Po{ \\ }_{84}^{216}Po{\to }_2^4\alpha {+}_{82}^{212} Pb{ \\ }_{82}^{212} Pb{\to }_{-1}^0\beta {+}_{83}^{212}Bi{ \\ }_{83}^{212}Bi{\to }_{-1}^0\beta {+}_{84}^{212}Po{ \\ }_{84}^{212}Po{\to }_2^4\alpha {+}_{82}^{208} Pb \end{gathered}$$

Step 1: Definition of Nuclear Chemistry

Nuclear chemistry is a branch of chemistry concerned with the “study of changes” in the nucleus of atoms. These nuclear alterations produce nuclear power and radioactivity, and the energy generated by nuclear reactions has a wide range of uses.

Step 2: Explanation of the balanced equation.

$\alpha$ decay is a radioactive process in which an alpha particle is emitted from a nucleus. $\beta$ decay A radioactive process in which a beta particle is emitted from a nucleus.

In the naturally occurring thorium-232  decay series, the steps emit this sequence of particles:

First decay process reaction 

$\left({}_{90}{}^{232}Th\right. emits \alpha emission):{ }_{90}^{232}Th{\to }_2^4\alpha {+}_{88}^{228}Ra$

Second decay process reaction $\left({}_{88}{}^{228}Ra\right. emits {\beta }^{-}emission):$

${}_{88}{}^{228}Ra{\to }_{-1}^0\beta {+}_{89}^{228}Ac$

Third decay process reaction $\left({}_{89}{}^{228}Ac\right. emits {\beta }^{-}emission):$

${}_{89}{}^{228}Ac{\to }_{-1}^0\beta {+}_{90}^{228}Th$

Forth decay process reaction $\left({}_{90}{}^{228}Th\right. emits \alpha emission):$

${}_{90}{}^{228}Th{\to }_2^4\alpha {+}_{88}^{224}Ra$

Fifth decay process reaction ${(}_{88}^{224}Ra emits \alpha emission):$

${}_{88}{}^{224}Ra{\to }_2^4\alpha {+}_{86}^{220}Rn$

Step 3: Explanation of the balanced equation.

Sixth decay process reaction $\left({}_{86}{}^{220}Rn\right. emits \alpha emission):$

${}_{86}{}^{220}Rn{\to }_2^4\alpha {+}_{84}^{216}P{o}_o$

Seventh decay process reaction $\left({}_{84}{}^{216}Po\right. emits \alpha emission):$

${}_{84}{}^{216}Po{\to }_2^4\alpha {+}_{82}^{212} Pb$

Eighth decay process reaction $\left({}_{82}{}^{212} Pb\right. emits {\beta }^{-} emission):$

${}_{82}{}^{212} Pb{\to }_{-1}^0\beta {+}_{83}^{212}Bi$

Ninth decay process reaction $\left({}_{83}{}^{212}Bi\right. emits {\beta }^{-} emission):$

${}_{83}{}^{212}Bi{\to }_{-1}^0\beta {+}_{84}^{212}{P}_o$

Tenth decay process reaction $\left({}_{84}{}^{212}Po\right. emits \alpha emission):$

${}_{84}{}^{212}Po{\to }_2^4\alpha {+}_{82}^{208} Pb$