Question

1. thorium - 232 undergoes a radioactive decay until a stable isotope is reached. its radioactive decay is the largest single contributor to the earth’s internal heat. there are ten steps, beginning with alpha decay. write the reactions for its decay, and circle the final stable product. 1. alpha 2. beta 3. beta 4. alpha 5. alpha 6. alpha 7. alpha 8. beta 9. beta 10. alpha

Explanation:

Step1: Recall alpha - decay formula

An alpha - particle is $_{2}^{4}\text{He}$. In alpha - decay, the mass number ($A$) of the parent nucleus decreases by 4 and the atomic number ($Z$) decreases by 2. For beta - decay ($\beta^-$), a neutron in the nucleus is converted into a proton, an electron, and an antineutrino. The mass number remains the same and the atomic number increases by 1. Thorium - 232 has $A = 232$ and $Z=90$.

Step2: First alpha - decay

$_{90}^{232}\text{Th}
ightarrow_{88}^{228}\text{Ra}+_{2}^{4}\text{He}$

Step3: First beta - decay

$_{88}^{228}\text{Ra}
ightarrow_{89}^{228}\text{Ac}+_{- 1}^{0}\text{e}+\bar{
u}$

Step4: Second beta - decay

$_{89}^{228}\text{Ac}
ightarrow_{90}^{228}\text{Th}+_{-1}^{0}\text{e}+\bar{
u}$

Step5: Second alpha - decay

$_{90}^{228}\text{Th}
ightarrow_{88}^{224}\text{Ra}+_{2}^{4}\text{He}$

Step6: Third alpha - decay

$_{88}^{224}\text{Ra}
ightarrow_{86}^{220}\text{Rn}+_{2}^{4}\text{He}$

Step7: Fourth alpha - decay

$_{86}^{220}\text{Rn}
ightarrow_{84}^{216}\text{Po}+_{2}^{4}\text{He}$

Step8: Third beta - decay

$_{84}^{216}\text{Po}
ightarrow_{85}^{216}\text{At}+_{-1}^{0}\text{e}+\bar{
u}$

Step9: Fourth beta - decay

$_{85}^{216}\text{At}
ightarrow_{86}^{216}\text{Rn}+_{-1}^{0}\text{e}+\bar{
u}$

Step10: Fifth alpha - decay

$_{86}^{216}\text{Rn}
ightarrow_{84}^{212}\text{Po}+_{2}^{4}\text{He}$

Answer:

The final stable product is $_{84}^{212}\text{Po}$. The decay reactions are:
$_{90}^{232}\text{Th}
ightarrow_{88}^{228}\text{Ra}+_{2}^{4}\text{He}$
$_{88}^{228}\text{Ra}
ightarrow_{89}^{228}\text{Ac}+_{- 1}^{0}\text{e}+\bar{
u}$
$_{89}^{228}\text{Ac}
ightarrow_{90}^{228}\text{Th}+_{-1}^{0}\text{e}+\bar{
u}$
$_{90}^{228}\text{Th}
ightarrow_{88}^{224}\text{Ra}+_{2}^{4}\text{He}$
$_{88}^{224}\text{Ra}
ightarrow_{86}^{220}\text{Rn}+_{2}^{4}\text{He}$
$_{86}^{220}\text{Rn}
ightarrow_{84}^{216}\text{Po}+_{2}^{4}\text{He}$
$_{84}^{216}\text{Po}
ightarrow_{85}^{216}\text{At}+_{-1}^{0}\text{e}+\bar{
u}$
$_{85}^{216}\text{At}
ightarrow_{86}^{216}\text{Rn}+_{-1}^{0}\text{e}+\bar{
u}$
$_{86}^{216}\text{Rn}
ightarrow_{84}^{212}\text{Po}+_{2}^{4}\text{He}$ (Circle $_{84}^{212}\text{Po}$)