A radioactive element X emits six `alpha`-particles and four `beta`-particles leading to end product `""_(82)^(208)Pb`. X is

Options

A. `""_(92)^(238)U`
B. `""_(90)^(230)Th`
C. `""_(90)^(232)Th`
D. `""_(92)^(239)U`

Answer

To find the radioactive element X that emits six alpha particles and four beta particles leading to the end product of lead (Pb) with atomic number 82 and mass number 208, we can follow these steps:

### Step 1: Understand the effects of alpha and beta decay
- **Alpha decay**: Each alpha particle emission decreases the mass number (A) by 4 and the atomic number (Z) by 2.
- **Beta decay**: Each beta particle emission does not change the mass number (A) but increases the atomic number (Z) by 1.

### Step 2: Calculate the changes due to alpha emissions
Since X emits 6 alpha particles:
- Decrease in mass number due to alpha emissions:
\[ 6 \times 4 = 24 \]
- Decrease in atomic number due to alpha emissions:
\[ 6 \times 2 = 12 \]

### Step 3: Define the intermediate element Y
Let the mass number and atomic number of element X be A and Z, respectively. After emitting 6 alpha particles, the intermediate element Y will have:
- Mass number of Y:
\[ A - 24 \]
- Atomic number of Y:
\[ Z - 12 \]

### Step 4: Calculate the changes due to beta emissions
Since Y emits 4 beta particles:
- Mass number remains unchanged:
\[ A - 24 \]
- Increase in atomic number due to beta emissions:
\[ 4 \times 1 = 4 \]

### Step 5: Define the final element Pb
After emitting 4 beta particles, the final element (Pb) will have:
- Mass number of Pb:
\[ A - 24 \]
- Atomic number of Pb:
\[ Z - 12 + 4 = Z - 8 \]

### Step 6: Set up equations based on the final product
Given that the final product is Pb with atomic number 82 and mass number 208:
1. From mass number:
\[ A - 24 = 208 \implies A = 208 + 24 = 232 \]
2. From atomic number:
\[ Z - 8 = 82 \implies Z = 82 + 8 = 90 \]

### Step 7: Identify the element X
Now we have:
- Mass number (A) = 232
- Atomic number (Z) = 90

The element with atomic number 90 is Thorium (Th). Therefore, the radioactive element X is:
\[ \text{Th}_{90}^{232} \]

### Final Answer:
The radioactive element X is \( \text{Th}_{90}^{232} \).

---