Question

22.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: Start with Thorium - 232

Thorium - 232 has atomic number 90, written as $_{90}^{232}Th$. Alpha decay emits a $_{2}^{4}He$ particle.
$_{90}^{232}Th
ightarrow_{88}^{228}Ra + _{2}^{4}He$

Step2: First beta decay

Beta decay converts a neutron to a proton and emits an electron. For $_{88}^{228}Ra$, $_{88}^{228}Ra
ightarrow_{89}^{228}Ac+_{ - 1}^{0}e$

Step3: Second beta decay

For $_{89}^{228}Ac$, $_{89}^{228}Ac
ightarrow_{90}^{228}Th + _{ - 1}^{0}e$

Step4: Second alpha decay

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

Step5: Third alpha decay

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

Step6: Fourth alpha decay

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

Step7: Fifth alpha decay

For $_{84}^{216}Po$, $_{84}^{216}Po
ightarrow_{82}^{212}Pb+_{2}^{4}He$

Step8: Third beta decay

For $_{82}^{212}Pb$, $_{82}^{212}Pb
ightarrow_{83}^{212}Bi+_{ - 1}^{0}e$

Step9: Fourth beta decay

For $_{83}^{212}Bi$, $_{83}^{212}Bi
ightarrow_{84}^{212}Po+_{ - 1}^{0}e$

Step10: Sixth alpha decay

For $_{84}^{212}Po$, $_{84}^{212}Po
ightarrow_{82}^{208}Pb+_{2}^{4}He$

Answer:

The final stable product is $_{82}^{208}Pb$. The full - set of decay reactions are:
$_{90}^{232}Th
ightarrow_{88}^{228}Ra + _{2}^{4}He$
$_{88}^{228}Ra
ightarrow_{89}^{228}Ac+_{ - 1}^{0}e$
$_{89}^{228}Ac
ightarrow_{90}^{228}Th + _{ - 1}^{0}e$
$_{90}^{228}Th
ightarrow_{88}^{224}Ra+_{2}^{4}He$
$_{88}^{224}Ra
ightarrow_{86}^{220}Rn + _{2}^{4}He$
$_{86}^{220}Rn
ightarrow_{84}^{216}Po+_{2}^{4}He$
$_{84}^{216}Po
ightarrow_{82}^{212}Pb+_{2}^{4}He$
$_{82}^{212}Pb
ightarrow_{83}^{212}Bi+_{ - 1}^{0}e$
$_{83}^{212}Bi
ightarrow_{84}^{212}Po+_{ - 1}^{0}e$
$_{84}^{212}Po
ightarrow_{82}^{208}Pb+_{2}^{4}He$ (circle $_{82}^{208}Pb$)