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39. The transmutation of a radioactive uranium isotope, [tex]${ }_{92}^{234} U$[/tex], into a radon isotope, [tex]${ }_{86}^{222} Rn$[/tex], involves a series of three nuclear reactions. At the end of the first reaction, a thorium isotope, [tex][tex]${ }_{90}^{230} Th$[/tex][/tex], is formed, and at the end of the second reaction, a radium isotope, [tex]${ }_{88}^{226} Ra$[/tex], is formed. In both reactions, an alpha particle is emitted.

Write the balanced equations for the three successive nuclear reactions.

Please show all work.

(Reference: p. 808-813)



Answer :

GinnyAnswer
### Step-by-Step Solution

To write the balanced equations for the nuclear transmutation of uranium-234 ([tex]\( {}^{234}_{92}U \)[/tex]) into radon-222 ([tex]\( {}^{222}_{86}Rn \)[/tex]), we need to identify the products of each step based on the provided information that the intermediate products and an alpha particle (which is a helium nucleus, [tex]\( {}^{4}_{2}He \)[/tex]) are involved in each reaction.

Step 1: Uranium to Thorium
- Starting nucleus: [tex]\( {}^{234}_{92}U \)[/tex]
- First intermediate product: [tex]\( {}^{230}_{90}Th \)[/tex]
- An alpha particle is emitted: [tex]\( {}^{4}_{2}He \)[/tex]

First Reaction:
[tex]\[ {}^{234}_{92}U \rightarrow {}^{230}_{90}Th + {}^{4}_{2}He \][/tex]

Step 2: Thorium to Radium
- Starting nucleus: [tex]\( {}^{230}_{90}Th \)[/tex]
- Second intermediate product: [tex]\( {}^{226}_{88}Ra \)[/tex]
- An alpha particle is emitted: [tex]\( {}^{4}_{2}He \)[/tex]

Second Reaction:
[tex]\[ {}^{230}_{90}Th \rightarrow {}^{226}_{88}Ra + {}^{4}_{2}He \][/tex]

Step 3: Radium to Radon
- Starting nucleus: [tex]\( {}^{226}_{88}Ra \)[/tex]
- Final product: [tex]\( {}^{222}_{86}Rn \)[/tex]
- An alpha particle is emitted: [tex]\( {}^{4}_{2}He \)[/tex]

Third Reaction:
[tex]\[ {}^{226}_{88}Ra \rightarrow {}^{222}_{86}Rn + {}^{4}_{2}He \][/tex]

### Summary of Balanced Reactions

Putting all the steps together, here are the three balanced nuclear reactions for the transmutation:

1. First Reaction:
[tex]\[ {}^{234}_{92}U \rightarrow {}^{230}_{90}Th + {}^{4}_{2}He \][/tex]

2. Second Reaction:
[tex]\[ {}^{230}_{90}Th \rightarrow {}^{226}_{88}Ra + {}^{4}_{2}He \][/tex]

3. Third Reaction:
[tex]\[ {}^{226}_{88}Ra \rightarrow {}^{222}_{86}Rn + {}^{4}_{2}He \][/tex]

These balanced equations show the process of alpha decay where the uranium nucleus loses two protons and two neutrons in each step to ultimately form the radon nucleus.

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