QUESTION IMAGE
Question
super practice. (extra credit)
uranium - 238 undergoes a series of 14 nuclear decay reactions to ultimately become lead - 206. for each reaction in the sequence the main product (not the alpha or beta decay particle) of the reaction becomes the reactant of the next reaction. the table below lays out the 14 steps. complete each reaction.
| decay type | reactant | products |
|---|---|---|
| beta | ||
| beta | ||
| alpha | ||
| alpha | ||
| alpha | ||
| alpha | ||
| alpha | ||
| beta | ||
| beta | ||
| alpha | ||
| beta | ||
| beta | ||
| alpha |
Step1: Recall alpha - decay formula
In alpha - decay, the general formula is $^{A}_{Z}X
ightarrow^{4}_{2}He + ^{A - 4}_{Z - 2}Y$. For the first reaction $^{238}_{92}U
ightarrow^{4}_{2}He+^{234}_{90}Th$.
Step2: Recall beta - decay formula
In beta - decay ($\beta^-$), the general formula is $^{A}_{Z}X
ightarrow^{0}_{- 1}e+^{A}_{Z + 1}Y$.
- For the second reaction, the reactant is $^{234}_{90}Th$. So, $^{234}_{90}Th
ightarrow^{0}_{-1}e + ^{234}_{91}Pa$.
- For the third reaction, the reactant is $^{234}_{91}Pa$, and $^{234}_{91}Pa
ightarrow^{0}_{-1}e+^{234}_{92}U$.
- For the fourth reaction, with reactant $^{234}_{92}U$, using alpha - decay formula, $^{234}_{92}U
ightarrow^{4}_{2}He+^{230}_{90}Th$.
- For the fifth reaction, with reactant $^{230}_{90}Th$, $^{230}_{90}Th
ightarrow^{4}_{2}He+^{226}_{88}Ra$.
- For the sixth reaction, with reactant $^{226}_{88}Ra$, $^{226}_{88}Ra
ightarrow^{4}_{2}He+^{222}_{86}Rn$.
- For the seventh reaction, with reactant $^{222}_{86}Rn$, $^{222}_{86}Rn
ightarrow^{4}_{2}He+^{218}_{84}Po$.
- For the eighth reaction, with reactant $^{218}_{84}Po$, $^{218}_{84}Po
ightarrow^{4}_{2}He+^{214}_{82}Pb$.
- For the ninth reaction, with reactant $^{214}_{82}Pb$, $^{214}_{82}Pb
ightarrow^{0}_{-1}e+^{214}_{83}Bi$.
- For the tenth reaction, with reactant $^{214}_{83}Bi$, $^{214}_{83}Bi
ightarrow^{0}_{-1}e+^{214}_{84}Po$.
- For the eleventh reaction, with reactant $^{214}_{84}Po$, $^{214}_{84}Po
ightarrow^{4}_{2}He+^{210}_{82}Pb$.
- For the twelfth reaction, with reactant $^{210}_{82}Pb$, $^{210}_{82}Pb
ightarrow^{0}_{-1}e+^{210}_{83}Bi$.
- For the thirteenth reaction, with reactant $^{210}_{83}Bi$, $^{210}_{83}Bi
ightarrow^{0}_{-1}e+^{210}_{84}Po$.
- For the fourteenth reaction, with reactant $^{210}_{84}Po$, $^{210}_{84}Po
ightarrow^{4}_{2}He+^{206}_{82}Pb$.
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| Decay Type | Reactant | Products |
|---|---|---|
| Beta | $^{234}_{90}Th$ | $^{0}_{-1}e + ^{234}_{91}Pa$ |
| Beta | $^{234}_{91}Pa$ | $^{0}_{-1}e+^{234}_{92}U$ |
| Alpha | $^{234}_{92}U$ | $^{4}_{2}He+^{230}_{90}Th$ |
| Alpha | $^{230}_{90}Th$ | $^{4}_{2}He+^{226}_{88}Ra$ |
| Alpha | $^{226}_{88}Ra$ | $^{4}_{2}He+^{222}_{86}Rn$ |
| Alpha | $^{222}_{86}Rn$ | $^{4}_{2}He+^{218}_{84}Po$ |
| Alpha | $^{218}_{84}Po$ | $^{4}_{2}He+^{214}_{82}Pb$ |
| Beta | $^{214}_{82}Pb$ | $^{0}_{-1}e+^{214}_{83}Bi$ |
| Beta | $^{214}_{83}Bi$ | $^{0}_{-1}e+^{214}_{84}Po$ |
| Alpha | $^{214}_{84}Po$ | $^{4}_{2}He+^{210}_{82}Pb$ |
| Beta | $^{210}_{82}Pb$ | $^{0}_{-1}e+^{210}_{83}Bi$ |
| Beta | $^{210}_{83}Bi$ | $^{0}_{-1}e+^{210}_{84}Po$ |
| Alpha | $^{210}_{84}Po$ | $^{4}_{2}He+^{206}_{82}Pb$ |