Avogadro's Number: `6.022xx10^(23)`
An atom of `..^(238)U` disontegates by a series of `alpha`-decays and `beta`-decays until it becomes `.^(206)Pb`, which is stable.
(i) How many `alpha`-decays and many `beta`-decays does an atom starting as `..^(238)U` undergo before it becomes stable?
(ii) One of the following ten nuclides is formed from a series of disintegrations starting at `..^(238)U`. Which one?
`.^(235)U, .^(234)U, .^(228)Ac, .^(224)Ra, .^(224)Rn, .^(220)Rn, .^(215)Po, .^(212)Po, .^(212)Pb, .^(211)Pb`.

Answer

To solve the problem step by step, we will analyze the decay process of Uranium-238 (U-238) to lead-206 (Pb-206) and determine the number of alpha and beta decays involved.

### Step 1: Understand the Decay Process
Uranium-238 (U-238) undergoes a series of radioactive decays to transform into a stable isotope, lead-206 (Pb-206). The decay can involve two types of emissions:
- **Alpha decay**: This process reduces the atomic mass by 4 units and the atomic number by 2 units.
- **Beta decay**: This process does not change the atomic mass but increases the atomic number by 1 unit.

### Step 2: Determine the Changes in Atomic Number and Mass
- The initial isotope is U-238, which has:
- Atomic number (Z) = 92
- Mass number (A) = 238

- The final stable isotope is Pb-206, which has:
- Atomic number (Z) = 82
- Mass number (A) = 206

### Step 3: Calculate the Total Changes
1. **Change in Mass Number (A)**:
- From 238 (U-238) to 206 (Pb-206):
- Change in mass number = 238 - 206 = 32

2. **Change in Atomic Number (Z)**:
- From 92 (U-238) to 82 (Pb-206):
- Change in atomic number = 92 - 82 = 10

### Step 4: Set Up Equations for Decays
Let:
- \( x \) = number of alpha decays
- \( y \) = number of beta decays

From the decay processes, we can set up the following equations:
1. For mass number:
\[ 238 - 4x = 206 \]
Rearranging gives:
\[ 4x = 32 \implies x = 8 \]

2. For atomic number:
\[ 92 - 2x + y = 82 \]
Substituting \( x = 8 \):
\[ 92 - 16 + y = 82 \implies y = 6 \]

### Step 5: Conclusion
(i) The number of alpha decays \( x \) is 8, and the number of beta decays \( y \) is 6.

(ii) To determine which nuclide is formed from the decay of U-238, we can look at the sequence of decays. After 8 alpha decays and 6 beta decays, the final product is Pb-206. However, we need to check the intermediate products.

### Step 6: Identify the Nuclide Formed
After 8 alpha decays (which reduce the atomic number by 16) and 6 beta decays (which increase the atomic number by 6), we can conclude that the intermediate nuclides formed include:
- U-238 → Th-234 (after 1 alpha decay)
- Th-234 → Pa-234 (after 1 beta decay)
- Pa-234 → U-234 (after 1 beta decay)
- U-234 → Th-230 (after 1 alpha decay)
- Th-230 → Ra-226 (after 1 alpha decay)
- Ra-226 → Rn-222 (after 1 alpha decay)
- Rn-222 → Po-218 (after 1 alpha decay)
- Po-218 → Pb-214 (after 1 beta decay)
- Pb-214 → Bi-214 (after 1 beta decay)
- Bi-214 → Po-214 (after 1 beta decay)
- Po-214 → Pb-210 (after 1 beta decay)
- Pb-210 → Bi-210 (after 1 beta decay)
- Bi-210 → Po-210 (after 1 beta decay)
- Po-210 → Pb-206 (after 1 alpha decay)

From the list of nuclides given, the one that can be formed from U-238 through its decay series is **Pb-206**.

### Final Answers:
(i) 8 alpha decays and 6 beta decays.
(ii) The nuclide formed is **Pb-206**.