If `._92 U^238` undergoes successively `8 alpha-`decays and `6 beta-`decays, then resulting nucleus is.

Options

A. `Z=84,A=206`
B. `Z=84,A=224`
C. `Z=82,A=206`
D. `Z=82,A=200`

Answer

To solve the problem of determining the resulting nucleus after 8 alpha decays and 6 beta decays of Uranium-238 (U-238), we will follow these steps:

### Step 1: Understand the initial conditions
- The initial nucleus is Uranium-238 (U-238).
- The atomic number (Z) of Uranium is 92.
- The mass number (A) of U-238 is 238.

### Step 2: Calculate the effect of alpha decays
- An alpha decay reduces the mass number by 4 and the atomic number by 2.
- For 8 alpha decays:
- Total reduction in mass number = 8 * 4 = 32
- Total reduction in atomic number = 8 * 2 = 16

### Step 3: Update the mass number and atomic number after alpha decays
- New mass number (A') = 238 - 32 = 206
- New atomic number (Z') = 92 - 16 = 76

### Step 4: Calculate the effect of beta decays
- A beta decay increases the atomic number by 1 but does not change the mass number.
- For 6 beta decays:
- Total increase in atomic number = 6 * 1 = 6

### Step 5: Update the atomic number after beta decays
- New atomic number (Z'') = 76 + 6 = 82
- The mass number remains the same (A'') = 206.

### Step 6: Identify the resulting nucleus
- The resulting nucleus has an atomic number of 82 and a mass number of 206.
- The element with atomic number 82 is Lead (Pb).

### Final Result
The resulting nucleus after 8 alpha decays and 6 beta decays from U-238 is Lead-206 (Pb-206).

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